<table><caption>Table 55: Estimated positive and negative flows to biodiversity conservation (in $ bn)</caption><thead><tr><th rowspan="2">Financial flows</th><th colspan="2">2019</th><th colspan="2">2030</th></tr><tr><th>Lower limit</th><th>Upper limit</th><th>Lower limit</th><th>Upper limit</th></tr></thead><tbody><tr><td>Harmful subsidy reform</td><td>-542.0</td><td>-273.9</td><td>-268.1</td><td>0.0</td></tr><tr><td>Domestic budgets and tax policy</td><td>74.6</td><td>77.7</td><td>102.9</td><td>155.4</td></tr><tr><td>Natural infrastructure</td><td>26.9</td><td>26.9</td><td>104.7</td><td>138.6</td></tr><tr><td>Biodiversity offsets</td><td>6.3</td><td>9.2</td><td>162.0</td><td>168.0</td></tr><tr><td>Sustainable supply chains</td><td>5.5</td><td>8.2</td><td>12.3</td><td>18.7</td></tr><tr><td>Official development assistance (ODA)</td><td>4.0</td><td>9.7</td><td>8.0</td><td>19.4</td></tr><tr><td>Green financial products</td><td>3.8</td><td>6.3</td><td>30.9</td><td>92.5</td></tr><tr><td>Philanthropy & conservation NGOs</td><td>1.7</td><td>3.5</td><td></td><td></td></tr><tr><td>Nature-based solutions & carbon markets</td><td>0.8</td><td>1.4</td><td>24.9</td><td>39.9</td></tr><tr><td>Total</td><td>123.6</td><td>142.9</td><td>445.7</td><td>632.5</td></tr></tbody></table>

Source: Deutz et al (2020, Figure 5.1, page 64).

## 5.2 IPBES

Established in 2012, the Intergovernmental Science-Policy Platform on Biodiversity and Ecosystem Services (IPBES) is an international organization that assesses the state of the world's biodiversity and ecosystem services. It plays a similar role for biodiversity as the Intergovernmental Panel on Climate Change (IPCC) does for global warming. The current staff of the IPBES Secretariat is employed by UNEP and located in Bonn, Germany. In addition, about 2900 experts participate or have participated in the work of IPBES. Its main functions are to provide comprehensive scientific assessments of the state of biodiversity and ecosystem services, to support policy development and implementation, and to identify key scientific information needed for policymakers. The organization has gained international recognition primarily through its influential assessment reports. As of March 2025, it has produced 13 assessment reports:

* Global reports

- Global assessment report on biodiversity and ecosystem services (2019)

* Thematic reports

- Pollinators, pollination and food production (2016)

- Land degradation and restoration (2018)

- Sustainable use of wild species (2022)

- Invasive alien species and their control (2023)

- Interlinkages among biodiversity, water, food and health (2024)

* Methodological reports

- Scenarios and models of biodiversity and ecosystem services (2016)

- The diverse values and valuation of nature (2022)

- Underlying causes of biodiversity loss and the determinants of transformative change and options for achieving the 2050 Vision for Biodiversity (2024)
* Regional reports

- Biodiversity and ecosystem services for Africa (2018)

- Biodiversity and ecosystem services for the Americas (2018)

- Biodiversity and ecosystem services for Asia and the Pacific (2018)

- Biodiversity and ecosystem services for Europe and Central Asia (2018)

In addition, four new assessment reports will be published in the coming years: Impact and dependence of business on biodiversity and nature's contributions to people (2025); Monitoring biodiversity and nature's contributions to people (2026); Integrated biodiversity-inclusive spatial planning and ecological connectivity (2027); Second global assessment of biodiversity and ecosystem services (2028).

<figure><img src="image_116.png" /><figcaption>Figure 80: Some IPBES assessment reports</figcaption></figure>

Source: IPBES & www.ipbes.net/assessing-knowledge.

## 5.3 TNFD

The Taskforce on Nature-related Financial Disclosures is a global initiative to develop a framework for companies and financial institutions to report and act on nature-related risks and opportunities. Officially launched in 2021 by four founding organizations (Global Canopy, UNDP, UNEP FI and WWF), the TNFD differs from the TCFD in that it is not supported by the Financial Stability Board (FSB) or any other international regulatory body. Instead, it is a market-driven initiative supported by national governments, corporations and financial institutions. The 40 members of the Taskforce are senior executives from financial institutions (17), corporations (17) and market service providers (6). In April 2024, the ISSB and TNFD announced joint research projects on risks and opportunities related to nature and human capital. However, there are currently no plans to integrate the TNFD framework into the ISSB's sustainability standards (IFRS S1 and IFRS S2).

TNFD uses the same structure as TCFD, organized into four pillars: Governance, Strategy, Risk & impact management, and Metrics & targets. The TNFD framework is based on 14 recommended disclosures, 11 of which are carried over from the TCFD framework (Table 56). The three additional recommended disclosures are (3) engagement with indigenous peoples, local communities (IPLC) and affected stakeholders, (7) disclosure of the location of assets and activities in direct operations and, where possible, in the upstream/downstream value chain, and (9) identification and assessment of
nature-related dependencies in the upstream/downstream value chain. According to TNFD (2023), the framework is consistent with ISSB and GRI standards, and is aligned with Target 15 of the Kunming-Montreal GBF. In addition, the TNFD has developed guidance for some specific sectors (e.g., aquaculture, beverages, metals and mining, chemicals) and the following biomes: tropical and sub-tropical forests, savannas and grasslands, intensive land-use systems, urban and industrial ecosystems, rivers and streams, and marine shelf. Two types of indicators are included in TNFD:

* A small set of core indicators (global core indicators applicable to all sectors and sector core indicators for each sector)

* A large set of additional disclosure and assessment indicators (optional)

In TNFD (2023), there are 14 global core indicators, including 9 indicators for nature-related dependencies and impacts (Table 57) and 5 indicators for nature-related risks and opportunities (Table 58). In addition, TNFD (2023) introduces several additional global and sector-specific indicators<fnref n="165" />.

<table><caption>Table 56: The 14 recommended disclosures (TNFD, 2023)</caption><thead><tr><th>Pillar</th><th>#</th><th>Recommended Disclosure</th></tr></thead><tbody><tr><td rowspan="3">Governance</td><td>1</td><td>Board oversight</td></tr><tr><td>2</td><td>Management's role</td></tr><tr><td>3</td><td>Human rights policies (IPLC)</td></tr><tr><td rowspan="5">Strategy</td><td>4</td><td>Risks and opportunities</td></tr><tr><td>5</td><td>Impact on organization</td></tr><tr><td>6</td><td>Resilience of strategy</td></tr><tr><td>7</td><td>Locations of assets/activites/value chain</td></tr><tr><td>8</td><td>Risk identification and assessment processes</td></tr><tr><td rowspan="4">Risk management</td><td>9</td><td>Dependencies in the value chain</td></tr><tr><td>10</td><td>Risk management processes</td></tr><tr><td>11</td><td>Integration into overall risk management</td></tr><tr><td>12</td><td>Nature-related metrics</td></tr><tr><td rowspan="2">Metrics and targets</td><td>13</td><td>Metrics used to manage impacts and risks</td></tr><tr><td>14</td><td>Nature-related targets</td></tr></tbody></table>

Source: TNFD (2023) & https://tnfd.global.

As of October 2024, there are 502 TNFD adopters, including 318 corporations and 129 financial institutions. The majority are located in Asia (236) and Europe (183). These adopters represent $17.7 trillion in assets under management and $6.5 trillion in market capitalization.

## 5.4 European biodiversity framework

The EU Biodiversity Strategy for 2030 was published in May 2020 and is part of the European Green Deal to tackle the ongoing biodiversity loss crisis. It replaces the Biodiversity strategy for 2020. The overall goal is to restore biodiversity and healthy ecosystems across the EU. Examples of quantitative targets include planting 3 billion trees by 2030, restoring 25 000 km of free-flowing rivers by removing barriers, increasing organic farming to 25% of agricultural land, and reducing pesticide use by 50%. In June 2022, the European Commission also adopted a proposal for a Nature Restoration Law. EU member states will have to develop their national restoration plans by 2026

<footnote n="165">¹⁶⁵These can be found in TNFD (2023, Annex 2, pages 89–99) and Section 3 of the various sector guidance documents.</footnote>
<table><caption>Table 57: TNFD core global disclosure indicators for nature-related dependencies and impacts</caption><thead><tr><th>#</th><th>Indicator</th><th>Unit</th><th>GBF Targets</th></tr></thead><tbody><tr><td>C1.0</td><td>Total spatial footprint</td><td>km<sup>2</sup></td><td>1, 2, 5, 11</td></tr><tr><td>C1.1</td><td>Extent of land/freshwater/ocean-use change</td><td>km<sup>2</sup></td><td>1, 2, 5, 11</td></tr><tr><td>C2.0</td><td>Pollutants released to soil split by type</td><td>tonne</td><td>7, 11</td></tr><tr><td>C2.1</td><td>Wastewater discharged</td><td>m<sup>3</sup></td><td>7, 11</td></tr><tr><td>C2.2</td><td>Waste generation and disposal</td><td>m<sup>3</sup></td><td>7, 11</td></tr><tr><td>C2.3</td><td>Plastic pollution</td><td>tonne</td><td>7, 11</td></tr><tr><td>C2.4</td><td>Non-GHG air pollutants</td><td>PM<sub>2.5</sub>, etc.</td><td>7, 11</td></tr><tr><td>C3.0</td><td>Water withdrawal and consumption from areas of water scarcity</td><td>m<sup>3</sup></td><td>11</td></tr><tr><td>C3.1</td><td>Quantity of high-risk natural commodities sourced from land/ocean/freshwater</td><td>tonne</td><td>5, 9, 11</td></tr><tr><td>C4.0</td><td>Measures against unintentional introduction of invasive alien species</td><td></td><td>6, 11</td></tr><tr><td>C5.0</td><td>Ecosystem condition</td><td></td><td>1, 2, 3, 4, 11</td></tr><tr><td>C5.0</td><td>Species extinction risk</td><td></td><td>1, 2, 3, 4, 11</td></tr></tbody></table>

<table><caption>Table 58: TNFD core global disclosure indicators for nature-related risks and opportunities</caption><thead><tr><th>#</th><th>Indicator</th></tr></thead><tbody><tr><td>C7.0</td><td>Value of assets, liabilities, revenue and expenses that are assessed as vulnerable to nature-related transition risks (total and proportion of total)</td></tr><tr><td>C7.1</td><td>Value of assets, liabilities, revenue and expenses that are assessed as vulnerable to nature-related physical risks (total and proportion of total)</td></tr><tr><td>C7.2</td><td>Description and value of significant fines/penalties received/litigation action in the year due to negative nature-related impacts</td></tr><tr><td>C7.3</td><td>Amount of capital expenditure, financing or investment deployed towards nature-related opportunities, by type of opportunity, with reference to a government or regulator green investment taxonomy or third-party industry or NGO taxonomy, where relevant</td></tr><tr><td>C7.4</td><td>Increase and proportion of revenue from products and services producing demonstrable positive impacts on nature with a description of impacts</td></tr></tbody></table>

<table><caption>Table 59: Examples of TNFD additional global disclosure indicators</caption><thead><tr><th>#</th><th>Indicator</th></tr></thead><tbody><tr><td>A2.3</td><td>Light and noise pollution</td></tr><tr><td>A3.4</td><td>Area used for the production of natural commodities</td></tr><tr><td>A3.5</td><td>Use of wild species</td></tr><tr><td>A4.0</td><td>Number/extent of unintentionally introduced species, varieties or strains</td></tr><tr><td>A7.0</td><td>Value of write-offs and early retirements of assets due to nature-related risks</td></tr><tr><td>A8.4</td><td>Capital expenditure on adaption due to nature-related physical risks</td></tr><tr><td>A20.0</td><td>Proportion of sites that have active engagement with local stakeholders on nature-related issues</td></tr></tbody></table>

Source: TNFD (2023, Tables 6–10, pages 83–99).
with the following objectives: restore at least 30% of habitats in poor condition by 2030, 60% by 2040, and 90% by 2050. Since then, two guidelines on forest and soil monitoring have been published in 2023. In the coming years, we can expect new biodiversity laws and regulations in the European Union, following the establishment of the EU Biodiversity Platform (EUBP) in 2022. The platform currently consists of 10 working groups addressing various biodiversity issues<fnref n="166" />.

# 6 Investment approaches

While ESG and climate investing are now two mainstream investment approaches in asset management, the concept of biodiversity investing is relatively new. In fact, the term ‘biodiversity finance’ is generally more appropriate, as it encompasses a broader range of financial instruments beyond investment practices. Nevertheless, the field of biodiversity investing is developing rapidly. For example, with the exception of BIOFIN, which was launched by UNDP in 2012 following the CBD COP10 in Nagoya in 2010, most of the major biodiversity-focused financial initiatives and alliances have been created more recently, particularly following the One Planet Summit or the CBD COP15 in 2022 (Table 60). This also explains why biodiversity investing is a relatively new topic for academics<fnref n="167" />, as most of the existing literature is in the gray area (Hutchinson and Lucey, 2024).

<table><caption>Table 60: Biodiversity finance initiatives</caption><thead><tr><td>Acronym</td><td>Name</td><td>Website</td><td>Year</td></tr></thead><tbody><tr><td>BCA</td><td>Biodiversity Credit Alliance</td><td>www.biodiversitycreditalliance.org</td><td>2022</td></tr><tr><td>BIOFIN</td><td>Biodiversity Finance Initiative</td><td>www.biofin.org</td><td>2012</td></tr><tr><td>BfN</td><td>Business for Nature</td><td>www.businessfornature.org</td><td>2019</td></tr><tr><td>FfB</td><td>Finance for Biodiversity Foundation</td><td>www.financeforbiodiversity.org</td><td>2021</td></tr><tr><td>NCIA</td><td>Natural Capital Investment Alliance</td><td>www.sustainable-markets.org</td><td>2021</td></tr><tr><td>NA 100</td><td>Nature Action 100</td><td>www.natureaction100.org</td><td>2023</td></tr></tbody></table>

**Remark 17** *Biodiversity finance can include economic and financial instruments that are not related to the concepts of biodiversity investment. For example, most OECD research and publications on biodiversity finance focus on public finance, policy instruments such as taxes, subsidies or payments for ecosystem services.*

## 6.1 Financial instruments

The Global Biodiversity Framework considers that the private sector is essential to achieve the 2050 Vision for Biodiversity, not just the public sector. For instance, the private sector is mentioned in several targets. Thus, Target 15 requires large companies and financial institutions to monitor, assess, and disclose biodiversity risks. Target 19 states that private resources must be mobilized by “leveraging private finance, promoting blended finance, implementing strategies for raising new and additional resources, and encouraging the private sector to invest in biodiversity, including through impact funds and other instruments.” To better understand biodiversity investing from the

<footnote n="166">¹⁶⁶These are (1) Working group on forests and nature; (2) Invasive alien species expert group; (3) Sub-group on monitoring and assessment; (4) Working group on green infrastructure; (5) Working group on invasive alien species; (6) Commission expert group on the birds and habitats directives; (7) Marine issues; (8) Reporting under the nature directives; (9) Expert group on the nature restoration regulation; (10) Working group on pollinators.</footnote>

<footnote n="167">¹⁶⁷Recent academic research on biodiversity investing includes Giglio et al (2023b), Flammer et al (2025), and Cherief et al (2025).</footnote>
perspective of private investors, one approach is to list the different instruments that fit into this category. Here is a classification of the main instruments:

## Fixed-income instruments

- Blue bonds

*Definition* Debt securities designed to raise capital for marine and ocean conservation projects (e.g., protecting marine biodiversity, restoring coastal ecosystems, financing sustainable fisheries)

*Example* Seychelles Blue Bond (2018), which raised $15 million to support sustainable marine areas and fisheries

- Debt-for-nature swaps

*Definition* Financial transactions in which a portion of a country's debt is forgiven in exchange for environmental commitments

*Example* Gabon debt-for-nature swap (2023), which restructured $500 million of debt to protect 30% of marine and forest ecosystems

- Green and sustainable bonds

*Definition* Debt instruments that target environmental projects and sustainable land use

*Example* Colombia Biodiversity Bond (BBVA/IFC), which raised $50 million to finance projects focused on reforestation and wildlife habitat restoration

- Natural capital bonds, nature performance bonds and conservation performance bonds

*Definition* Bonds that directly finance the protection and restoration of natural capital, with returns linked to specific ecological performance metrics and biodiversity outcomes, or that monetize the value of ecosystem services and biodiversity

*Example* Voluntary carbon credit-linked bonds, such as the IFC Forest Bond

- Sustainability-linked bonds and pay-for-success financial instruments

*Definition* Bonds whose financial characteristics can change based on the achievement of sustainability targets

*Example* Rhino Bond (2022), issued by the World Bank ($150 million), where returns are linked to the growth of the black rhino population in Africa

## Market-based instruments

- Biodiversity credits/offsets

*Definition* Marketable units representing positive biodiversity outcomes (market mechanisms to promote biodiversity conservation)

*Example* UK Biodiversity Net Gain (BNG) policy (developers must ensure a 10% net gain in biodiversity by funding conservation projects or purchasing biodiversity credits)

- Nature-based insurance products

*Definition* Insurance mechanisms to protect natural capital and ecosystem services

*Example* Parametric insurance for coral reef protection in the Caribbean and Central America (provides insurance coverage for coastal infrastructure and triggers payouts for reef restoration after hurricanes)

- Payments for ecosystem services (PES)
*Definition* Schemes that provide financial incentives to landowners or communities to manage their land in a way that maintains or enhances ecosystem services

*Example* Vittel (Nestlé Waters) offers PES to farmers in the Vosges mountains in France to maintain water quality

## Investment funds

- Biodiversity impact funds

*Definition* Specialized investment vehicles focused on biodiversity conservation

*Example* The African Forestry Impact Platform (AFIP) managed by Norfund, which invests in sustainable forestry and conservation projects

- Blended finance

*Definition* Combines public and private capital to attract more capital to biodiversity projects

*Example* The Land Degradation Neutrality (LDN) Fund initiated by the United Nations Convention to Combat Desertification (UNCCD) and Mirova, which finances the rehabilitation of degraded land

- Conservation trust funds

*Definition* Long-term financing mechanisms for conservation and sustainable development

*Example* Bhutan Trust Fund for Environmental Conservation (BTFEC)

- Private equity and venture capital funds

*Definition* Investment funds focused on companies and technologies that support biodiversity

*Example* Regeneration VC Fund

## 6.2 The avoid-minimize-restore-offset approach

To effectively integrate biodiversity into investment processes, it is essential to understand the mitigation hierarchy, a fundamental principle of conservation biology (Arlidge et al., 2018). The mitigation hierarchy is a structured framework to guide decision-making in managing environmental impacts. It prioritizes actions to avoid, minimize, restore, and offset negative impacts on biodiversity (Figure 81). Avoid prevents negative impacts on biodiversity before they occur. This step emphasizes avoiding environmentally harmful activities and encourages decision-makers to choose alternative actions or project designs that have no or minimal adverse effects on biodiversity. For example, this might involve selecting a project site that is not located in a critical habitat or a biodiversity hotspot. Minimize aims to reduce the impact of unavoidable harm to biodiversity as much as possible by decreasing the magnitude, duration and/or intensity of potential impacts when they cannot be avoided completely. An example is implementing construction practices that reduce habitat destruction, such as minimizing land clearing or avoiding key wildlife breeding seasons. Restore focuses on rehabilitating ecosystems that have been damaged by development activities. This can include replanting trees in areas affected by deforestation, restoring native vegetation, or rebuilding habitat structures to support local ecological systems. Finally, if residual impacts remain after avoid, minimize, and restore measures, offsets are used to compensate for the remaining losses. This involves creating or enhancing biodiversity elsewhere to achieve no net loss or a net gain in ecological value.
<figure><img src="image_117.png" /><figcaption>Figure 81: Mitigation hierarchy for nature conservation</figcaption></figure>

By construction, the mitigation hierarchy is an impact approach where we assess how companies and investments impact biodiversity. From a financial perspective, it is equally important to consider how companies and investments are affected by biodiversity loss. In this context, we can adopt a dependency approach based on four key risk dimensions: transition risk, physical risk, litigation risk and systemic risk. For instance, some sectors or companies are highly dependent on raw materials, soil productivity, or freshwater. As a result, loss of biodiversity can create physical risk directly or indirectly through the supply chain. Transitional risk can arise from new or emerging regulations and changes in consumer preferences. Litigation risk arises when companies face potential legal challenges related to environmental damage, failure to protect biodiversity, or non-compliance with environmental regulations. Such risks can result in significant financial penalties, reputational damage and lawsuits. The final category of risk is systemic risk, which affects companies indirectly. For example, they may be affected by severe pandemics that disrupt global ecosystems and economic systems. This dual perspective of biodiversity risk refers to the double materiality principle:

<table><thead><tr><th colspan="2">Double materiality principle</th></tr></thead><tbody><tr><td>Dependency</td><td>How does biodiversity affect companies and investments?</td></tr><tr><td>Impact</td><td>How do companies and investments affect biodiversity?</td></tr></tbody></table>

Applying the avoid dimension of the mitigation hierarchy to investments requires the adoption of negative screening and the establishment of exclusion criteria for issuers or projects that significantly harm biodiversity. The minimize dimension can be effectively implemented through best-in-class strategies or selective approaches that prioritize companies with stronger biodiversity practices. A key challenge in this dimension is the development and use of biodiversity metrics that appropriately
account for both sector-specific impacts and geographic contexts, as the importance of biodiversity varies widely across ecosystems. For the restore/offset dimension, the focus shifts to creating positive impacts. Traditional avoidance and minimization metrics prove inadequate here, as these final stages require the identification and selection of companies and projects that have a positive biodiversity footprint. This requires a different set of metrics that focus on restoration outcomes, additionality and measurable biodiversity gains. For example, an MSA metric is not relevant to this dimension. As a result, each progression through the hierarchy requires increasingly sophisticated measurement approaches, moving from harm reduction to positive contribution assessment.

<figure><img src="image_118.png" /><figcaption>Figure 82: Bridging ESG investment strategies with biodiversity mitigation dimensions</figcaption></figure>

## 6.3 The impact investing approach

According to Phenix Capital (2024b), the biodiversity impact fund landscape appears robust at first glance, with 1 080 biodiversity-related funds and a market size of €129 billion in open-end investment funds. At first glance, these figures appear substantial, especially when compared to the investment required to meet the targets of the GBF. However, a deeper analysis reveals a more nuanced picture. A striking 69% of these funds primarily address food-related needs rather than direct biodiversity conservation. When focusing specifically on funds targeting SDG 14 (life below water) and SDG 15 (life on land), the numbers drop to 162 and 278 impact funds, respectively. Real assets and private equity dominate these specialized sectors, accounting for 33% and 30% of these biodiversity-focused funds, respectively. However, the geographic and thematic distribution
reveals significant gaps. Only 11 real asset impact funds focus on timber and forests in Asia, 12 in Africa, and 26 in South America — regions that contain much of the world's remaining primary biodiversity. This creates a paradox. While the aggregate numbers suggest impressive financial mobilization for biodiversity, an examination of specific critical issues, such as forest conservation, reveals low investment volumes.
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# A Exercises

## A.1 Calculating the prevalence of undernourishment

Let $X$ and $R$ be the random variables representing energy intake and energy requirement, respectively.

1. We assume that the random vector $(X, R)$ follows a bivariate log-normal distribution: $(X, R) \sim \mathcal{LN}(\mu_x, \sigma_x^2, \mu_r, \sigma_r^2, \rho)$.

(a) Find the probability distribution of $D = X/R$. Then, calculate the prevalence of under-nourishment, denoted by $PoU^* = \text{Pr}\{X < R\}$.

(b) Assume that $\mu_x = 7.50$, $\sigma_x = 0.20$, $\mu_r = 7.20$, and $\sigma_r = 0.05$. Plot the density functions of $X$ and $R$.

(c) Plot the function of $PoU^*$ as $\rho$ varies in the interval $[-1, 1]$. Comment on the results.

(d) Compute the prevalence of undernourishment defined by $PoU = \text{Pr}\{X < r_L\}$. Plot the relationship between $r_L$ and $PoU$ when $r_L \in [1000, 1600]$.

(e) Find the value of $r_L^*$ such as $PoU = PoU^*$. Calibrate the parameter $r_L^*$ for the prevalence of undernourishment calculated in Question 1.c.

2. We want to calibrate the probability distribution function of $X$. We assume that $X \sim \mathcal{LN}(\mu_x, \sigma_x^2)$.

(a) Give the first two moments of $X$. We will denote them by $\mu(X)$ and $\sigma^2(X)$.

(b) Deduce the coefficient of variation $CV(X)$.

(c) Find the moment estimators of $\mu_x$ and $\sigma_x$ from $\mu(X)$ and $CV(X)$.

(d) We consider an hypothetical country with a population of 1 million and two food components (cereals and fruits/vegetables), whose the food balance sheet is as follows:

<table><thead><tr><th></th><th>Production</th><th>Imports</th><th>Exports</th><th>ΔStocks</th><th>Feed</th><th>Waste</th></tr></thead><tbody><tr><td>#1</td><td>349 000</td><td>1 025</td><td>40 000</td><td>-1 000</td><td>45 000</td><td>9 000</td></tr><tr><td>#2</td><td>50 000</td><td>9 010</td><td>1 000</td><td>6 000</td><td>500</td><td>500</td></tr></tbody></table>

All the figures are expressed in tonnes. Calculate the food available for human consumption. Assuming that the average energy density is 3500 and 500 Calories/kg for cereals and fruits/vegetables respectively, find the average dietary energy consumption $\mu(X)$ expressed in kcal/capita/day.

(e) The average dietary energy consumption by household expenditure deciles are 1650 (first decile), 1985, 2150, 2350, 2530, 2550, 2650, 2750, 3100 and 3630 (last decile). Calculate the coefficient of variation $CV(X | Y)$. Check that $\mu(X | Y) = \mu(X)$. Assuming that $CV(X | R) = 0.20$, calculate $CV(X)$.

(f) Calibrate the parameters $\mu_x$ and $\sigma_x$ using the previous figures. What is the prevalence of undernourishment if we assume that $r_L = 1850$? Give an estimate of the number of undernourished people.

3. We seek to estimate the minimum dietary energy requirement.
(a) We recall that the body mass index (BMI) is expressed in kg/m² and is defined as the ratio of the weight (in kg) to the square of the height (in meter):

$$BMI = \frac{\text{weight}}{\text{height}^2}$$

The ideal value of BMI is 22. To define undernourished people, we assume that their weight is below a reference value:

$$\text{weight} \leq \text{weight}^*$$

where:

$$\text{weight}^* = \text{BMI}(p) \cdot \text{height}^2$$

where *p* is a percentile value that depends on age. If the age is less than ten years, *p* is set to 50%, while for individuals 10 years and older, it is set to 5%. Below, we give the values of BMI (*p*) and the average height per age and sex:

<table><thead><tr><th></th><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><th rowspan="2">BMI (p)</th><td>Female</td><td>15.5</td><td>15.5</td><td>17.0</td><td>17.5</td><td>17.5</td><td>17.5</td></tr><tr><td>Male</td><td>15.5</td><td>15.5</td><td>17.0</td><td>18.5</td><td>18.5</td><td>18.5</td></tr><tr><th rowspan="2">Height</th><td>Female</td><td>0.80</td><td>1.20</td><td>1.55</td><td>1.60</td><td>1.60</td><td>1.60</td></tr><tr><td>Male</td><td>0.88</td><td>1.24</td><td>1.58</td><td>1.72</td><td>1.72</td><td>1.72</td></tr></tbody></table>

Calculate the reference value weight* per age and sex to determine the undernourishment.

(b) We assume that the basal metabolic rate (BMR) is given by the Schofield equation:

$$\text{BMR} = \alpha + \beta \cdot \text{weight}^*$$

where $\alpha_j$ and $\beta_j$ are the estimates of the linear regression between weight and BMR:

<table><thead><tr><th></th><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><th rowspan="2">α</th><td>Female</td><td>-31.1</td><td>485.9</td><td>692.6</td><td>486.6</td><td>845.6</td><td>658.5</td></tr><tr><td>Male</td><td>-30.4</td><td>504.3</td><td>658.2</td><td>692.2</td><td>873.1</td><td>587.7</td></tr><tr><th rowspan="2">β</th><td>Female</td><td>58.317</td><td>20.315</td><td>13.384</td><td>14.818</td><td>8.126</td><td>9.082</td></tr><tr><td>Male</td><td>59.512</td><td>22.706</td><td>17.686</td><td>15.057</td><td>11.472</td><td>11.711</td></tr></tbody></table>

Calculate the basal metabolic rate for each group.

(c) The minimum dietary energy requirement is equal to the physical activity level (PAL) times the basal metabolic rate:

$$\text{MDER} = \text{PAL} \cdot \text{BMR}$$

We assume that the population is on average lightly active, implying that PAL = 1.55. Calculate the minimum dietary energy requirement for the different groups.

(d) We assume that the proportion of females and males is the same, while the distribution of the population by age is as follows:

<table><thead><tr><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><th>Frequency</th><td>9.9%</td><td>15%</td><td>16%</td><td>18%</td><td>29%</td><td>12.1%</td></tr></tbody></table>

Calculate the minimum dietary energy requirement of the population.
4. We want to calculate the depth of the food deficit:

$$FD = \int_{x<r_L} (\bar{r} - x) f_x(x) dx$$

where $r_L$ is the minimum dietary energy requirement (MDER), $\bar{r}$ is the average dietary energy requirement (ADER), and $f_x(x)$ is the probability density function of the dietary energy consumption $X$.

(a) What is the interpretation of the indicator FD?

(b) Find the probabilistic expression of the indicator FD.

(c) Find the analytical value of FD when $X \sim \mathcal{LN}(\mu_x, \sigma_x^2)$.

(d) Calculate the depth of the food deficit in the case of Question 2 if we assume that the average dietary energy requirement is equal to 2500 Calories per person per day.

## A.2 Calculating the species-area relationship using the theory of island biogeography

1. We consider the model of island biogeography developed by Beaugrand et al. (2024). We denote species richness by $S(t)$. At the initial date $t_0$, we have $S(t_0) = S_0$ (we can assume that $S_0 = 0$). The authors assume the existence of a saturation date $t_s$, i.e., the species richness cannot exceed a limit value:

$$0 \le S(t) \le S_s = S(t_s)$$

(a) Let $f(x) = ae^{b(x/c)^d}$ where $a > 0$, $c > 0$ and $d > 0$. The parameter $b$ can take two values: -1 or +1. We also assume that $0 \le x \le c$. In which case do we get a decreasing, increasing, concave and convex function?

(b) Let $\lambda(t)$ represent the immigration rate. We assume that:

$$\lambda(t) = \lambda_0 \frac{e^{-\left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0}} - e^{-1}}{1 - e^{-1}} \quad \text{for } \lambda_0 \ge \lambda(t) \ge \lambda_s = \lambda(t_s)$$

where $\lambda_0 \ge 0$ is the initial immigration rate, and $\beta_1 > 0$ is a parameter. Prove that $\lambda(t)$ is a decreasing function of $S(t)$, with $\lambda(t_0) = \lambda_0$ and $\lambda(t_s) = 0$. Determine whether $\lambda(t)$ is a convex or concave function. Plot the function $\lambda(t)$ for the following parameter sets ($\lambda_0, \beta_1, S_s$): (1.0, 0.2, 200), (1.0, 0.5, 200), (1.0, 1.0, 200), and (0.9, 2.0, 150). Analyze and comment on these results.

(c) Let $\mu^{\text{long}}(t)$ be the long-term extinction rate. We assume that:

$$\mu^{\text{long}}(t) = \mu_s \frac{e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} - 1}{e^1 - 1} \quad \text{for } \mu_0^{\text{long}} \le \mu^{\text{long}}(t) \le \mu_s = \mu^{\text{long}}(t_s)$$

where $\mu_s \ge 0$ is the extinction rate at the saturation date, and $\beta_2 > 0$ is a parameter. Prove that $\mu^{\text{long}}(t)$ is an increasing function of $S(t)$, with $\mu^{\text{long}}(t_0) = 0$ and $\mu^{\text{long}}(t_s) = \mu_s$. Determine whether $\mu^{\text{long}}(t)$ is a convex or concave function. Plot the function $\mu^{\text{long}}(t)$ for the following parameter sets ($\mu_s, \beta_2, S_s$): (1.0, 0.2, 200), (1.0, 0.5, 200), (1.0, 1.0, 200), and (0.5, 2.0, 150). Analyze and comment on these results.
(d) Let $\mu^{\text{short}}(t)$ be the short-term extinction rate. We assume that:

$$ \mu^{\text{short}}(t) = \beta_3 \lambda(t) e^{-\beta_4 S(t)} \quad \text{for } \mu_0^{\text{short}} \ge \mu^{\text{short}}(t) \ge \mu_s^{\text{short}} = \mu^{\text{short}}(t_s) $$

where $\beta_3$ and $\beta_4$ are two positive parameters. Prove that $\mu^{\text{short}}(t)$ is a decreasing function of $S(t)$, with $\mu^{\text{short}}(t_0) \le \beta_3\lambda_0$ and $\mu^{\text{short}}(t_s) = 0$. Plot the function$^{168}$ $\mu^{\text{short}}(t)$ for the following parameter sets $(\beta_3, \beta_4, S_s)$: (0.2, 0.02, 200), (0.5, 0.02, 200), (0.8, 0.02, 200), and (1.0, 0.10, 150).

(e) Let $\mu(t) = \mu^{\text{short}}(t) + \mu^{\text{long}}(t)$ be the extinction rate. Show that $\mu(t_0) = \mu^{\text{short}}(t_0)$ and $\mu(t_s) = \mu_s$. Plot the function $\mu(t)$ for the following set of parameters: $\lambda_0 = 1, \beta_1 = 0.5, \mu_s = 1, \beta_2 = 2, \beta_3 = 0.70, \beta_4 = 0.01$, and $S_s = 200$. Comment on these results.

(f) The dynamics of species richness is governed by the following differential equation:

$$ \frac{dS(t)}{dt} = \delta(t) = \lambda(t) - \mu(t) $$

What is the condition for $S(t)$ to reach an equilibrium? Illustrate and determine the equilibrium $S^*$ using the set of parameters provided in Question 1.e.

(g) Under what conditions can two equilibria exist? Let $S_1^*$ and $S_2^*$ be the two equilibria, with $S_1^* \le S_2^*$. Show that $S_1^*$ is an unstable steady state, while $S_2^*$ is a stable steady state. Deduce that there is a third equilibrium $S_0^*$, with $S_0^* \le S_1^* \le S_2^*$. Illustrate the three-equilibrium case with the following set of parameters: $\lambda_0 = 1, \beta_1 = 0.5, \mu_s = 1, \beta_2 = 2, \beta_3 = 1.70, \beta_4 = 0.02$, and $S_s = 200$.

(h) Using the two sets of parameters defined in Questions 1.e and 1.g, simulate the process $S(t)$ when the initial species richness $S_0$ is 0, 25, 30, 50 and 120, respectively. Comment on these results.

2. We aim to analyze the dynamics of the TIB steady state $S^*$ under the assumption that there is no short-term extinction rate, meaning $\beta_3 = 0$ or $\mu^{\text{short}}(t) = 0$. The default parameter values are $\lambda_0 = 1, \beta_1 = 0.5, \mu_s = 1, \beta_2 = 1$, and $S_s = 200$.

(a) What is the impact of $\beta_2$ on the equilibrium $S^*$? Compare the solutions for $\beta_2 = 0.2, \beta_2 = 1$, and $\beta_2 = 5$.

(b) Plot the relationship between $\beta_2$ and $S^*$.

(c) What is the impact of $\mu_s$ on the equilibrium $S^*$?

(d) Plot the relationship between $\mu_s$ and $S^*$.

(e) What conclusion can we draw about the relationship between the area $A$ and the equilibrium $S^*$.

(f) We assume that the area $A$ (expressed in km$^2$) is related to the parameter $\beta_2$ as follows: $A = \beta_2^{0.75}$. A sampling of $\beta_2$ is taken between 0.01 and 5.00 with a step size of 0.01. Use nonlinear least squares to estimate the power model $S = cA^z$, the exponential model $S = c + z \ln(A)$, and the Kobayashi model $S = c \ln(1 + zA)$. Compare the TIB equilibrium $S^*$ with the forecasts of the fitted models. Comment on these results.

<footnote>$^{168}$We assume $\lambda_0 = 1.0$ and $\beta_1 = 0.5$ to define $\lambda(t)$.</footnote>
## A.3 Species abundance models

*Remark: This exercise was inspired by the research articles of Coleman (1981), and He and Legendre (2002).*

1. We consider an area (or an ecosystem) A containing S species. Let $n_i$ denote the abundance of the i-th species and let $n = \sum_{i=1}^{S} n_i$ represent the total abundance in the area. We focus on a subarea $a \subseteq A$ and denote by $\tilde{S}_a$ the random variable representing the number of species in this subarea. We introduce the notation $S_a = \mathbb{E}[\tilde{S}_a]$ to represent the expected number of species in the subarea.

(a) Let $p_i$ be the probability that species i is present in the subarea a, and $\tilde{S}_i \sim \mathcal{B}(p_i)$ the random variable indicating its presence ($\tilde{S}_i = 1$) or its absence ($\tilde{S}_i = 0$). What is the probability distribution of $\tilde{S}_a$? Deduce the value of $S_a$. Application: Given $p = (p_1, \dots, p_5) = (20\%, 30\%, 10\%, 65\%, 10\%)$, calculate $\tilde{S}_a$ and $S_a$.

(b) We assume a random placement of the species within the area A. What is the probability of observing k individuals of species i in the subarea a? Show that:

$$ \tilde{S}_i \sim \mathcal{PB} \left( p_i = 1 - \left(1 - \frac{a}{A}\right)^{n_i} \right) $$

Find the value of $S_a$. Application: Given $(n_1, \dots, n_5) = (10, 4, 25, 6, 8)$, $A = 10 \text{ km}^2$, and $a = 2 \text{ km}^2$, calculate $\tilde{S}_a$ and $S_a$.

(c) Calculate $S_a$ for the following models:

i. Most even model:

$$ n_i = \frac{n}{S} \quad \text{for } i = 1, \dots, S $$

ii. Most uneven model:

$$ n_i = 1 \quad \text{for } i < S \quad \text{and} \quad n_S = n - S + 1 $$

Calculate $\tilde{S}_a$ and $S_a$.

iii. Mixed even-uneven model:

$$ n_i = 1 \quad \text{for } i \le s \quad \text{and} \quad n_i = \frac{n-s}{S-s} = n_{s+1} \quad \text{for } i > s $$

iv. He-Gaston model:

$$ p_i = 1 - \left(1 - \frac{a}{A}\right) \left(1 + \frac{n_i a}{\kappa_i A}\right)^{-\kappa_i} $$

where $\kappa_i \in (-\infty, m_i) \cup [0, \infty)$ is a parameter which describes the spatial pattern of species i and $m_i = n_i a/A$ is the mean density of species i in the subarea a.

v. Broken-stick model<fnref n="169" />:

$$ n_i = \frac{n}{S} \sum_{k=i}^{S} \frac{1}{k} $$

<footnote n="169">¹⁶⁹Hint: Approximate the harmonic sum by $\ln(\frac{S}{i})$, and replace the sum $\sum_{i=1}^{S} (1 - \frac{a}{A})^{n_i}$ with its integral form.</footnote>
(d) The species abundance distribution is defined as the series {$s(1), s(2), \dots$} where $s(j)$ is the number of species with $j$ individuals. Show that:

$$S_a = S - \sum_j s(j) \left(1 - \frac{a}{A}\right)^j$$

Application: Calculate $S_a$ for the log-series distribution:

$$s(j) = \alpha \frac{x^j}{j}$$

where $\alpha$ is a parameter related to the total diversity and $x \in (0, 1)$ is a parameter determining the relative abundance.

2. We say that the species is locally endemic to subarea $a \subseteq A$ if it is found exclusively in subarea $a$ and not in the complementary area $A - a$.

(a) Calculate the probability $\tilde{p}_i$ of endemism of species $i$.

(b) What is the expected number $E_a$ of locally endemic species?

(c) Calculate $S_{A-a}$ and $E_{A-a}$ for the complementary area $A - a$. Deduce that:

$$S_{A-a} + E_a = S$$

(d) Show that:

$$0 \le S_a + E_a \le 2S$$

## A.4 Valuation of risks to life and health

*Remark: This exercise was inspired by the research article by Hammitt (2023).*

Let $\tau$ be a survival time$^{170}$, whose survival function is $S(t) = \Pr\{\tau > t\}$ and density function is $f(t) = -\partial S(t)$. We have $S(0) = 1$ and $S(\infty) = 0$.

1. Assume that the individual is alive at time $t$. What is the conditional survival function $S(u|t) = \Pr\{\tau > u | \tau > t\}$?

2. Show that the average length of life (or life expectancy) for an individual who is alive at time $t$ is given by:

$$LE(t) = \mathbb{E}[\tau | \tau > t] = \frac{1}{S(t)} \int_{t}^{\infty} S(u) \, du$$

3. Let $\rho$ be the discount rate and $X(t)$ a payoff function. Following the seminal paper of Yaari (1965), we define the expected present value of the payoff, taking into account the future lifetime, as:

$$\mathbb{E}[X; t, \rho] = \int_{t}^{\infty} e^{-\rho(u-t)} S(u|t) X(u) \, du$$

What is the rationale behind this formula?

<footnote n="170">¹⁷⁰In mortality analysis, this is referred to as time-to-event or time-to-death.</footnote>
4. The discounted remaining life expectancy is defined as the expected present value of the future lifetime:

$$LE(t; \varrho) = \mathbb{E}[1; t, \varrho] = \frac{1}{S(t)} \int_{t}^{\infty} e^{-\varrho(u-t)} S(u) \, du$$

Shows that:

$$LE(t; \varrho) \geq DLE(t; \varrho) = \int_{t}^{\infty} e^{-\varrho(u-t)} (u-t) f(u|t) \, du$$

What is the interpretation of DLE (t; ϱ)? Under which condition do we have equality?

5. Assume that the survival time follows an exponential distribution: $\tau \sim \mathcal{E}(\lambda)$. Derive the formulas for LE (t; ϱ) and DLE (t; ϱ)? Comment on these results.

6. We assume that life expectancy is 70 years ($\mathbb{E}[\tau] = 70$). We consider two survival functions: $\tau \sim \mathcal{E}(\lambda)$ and $\tau \sim W_{\text{eibull}}(a, b)$, where the survival function is defined as $S(t) = \exp\left(-\left(\frac{x}{a}\right)^b\right)$.

(a) Calibrate the parameters $\lambda$ and $a$ so that $b = 5$. Plot the survival functions $S(t)$ and $S(u | t = 50)$. Comment on these results.

(b) We assume that $\varrho = 3\%$. Calculate LE(0; ϱ) and DLE(0; ϱ) if the survival time is exponential.

(c) Assume that the survival time is Weibull distributed. Using numerical integration, plot the discounted life expectancy LE(t; ϱ) for $t \in [0, 100]$ and for different discount rates ϱ (0%, 1%, 2%, 3% and 10%). Comment on these results.

7. We assume that the quality life weight $\hat{Q}(t)$ is a random variable between 0 and 1, and we note $Q(t) = \mathbb{E}[\hat{Q}(t)]$ its expected value. The quality-adjusted life expectancy (QALE) is defined as $\text{QALE}(t; \varrho) = \mathbb{E}[Q(t); t, \varphi]$.

(a) Give the formula of QALE (t; ϱ) when the function $Q(t)$ is constant.

(b) Show that:

$$\text{QALE}(t; \varrho) \leq \text{LE}(t; \varrho)$$

(c) Consider the previously calibrated Weibull distribution. We assume that:

$$Q_1(t) = \begin{cases} 1 & \text{if } t \leq t_1 \\ 1 - \kappa_1 \frac{(t - t_1)}{t_2 - t_1} & \text{if } t_1 \leq t \leq t_2 \\ 1 - \kappa_1 - \kappa_2 \frac{(t - t_2)}{t_3 - t_2} & \text{if } t_2 \leq t \leq t_3 \\ 1 - \kappa_1 - \kappa_2 & \text{if } t \geq t_3 \end{cases}$$

What is the rationale for this specification? We consider a second HRQL function:

$$Q_2(t) = \exp(-\kappa t)$$

Compare QALE (t; ϱ) and LE (t; ϱ) for the following set of parameters: $t_1 = 50$, $\kappa_1 = 10\%$, $t_2 = 70$, $\kappa_2 = 20\%$, $t_3 = 90$, and $\kappa = 2\%$.
8. Let $\mathcal{R}(t)$ be the impact on mortality risk. We use the convention that $\mathcal{R}(t)$ is positive if the impact is a risk reduction and negative if the impact is a risk increase. We define $\Delta L(t) = \mathcal{R}(t) \cdot \Delta t$ as the expected number of lives saved (or the decrease in the number of deaths) during the short time interval $\Delta t$. Hammitt (2023) also defines the increase in life expectancy as $\Delta LE(t) = \Delta L(t) \cdot LE(t; \varrho)$ and the increase in quality-adjusted life expectancy as $\Delta QALE(t) = \Delta L(t) \cdot QALE(t; \varrho)$. The monetary value $v$ of the risk reduction $\mathcal{R}(t)$ is the product of the value of a statistical life (VSL) and the expected number of lives saved. Similarly, $v$ can be expressed as the product of the value per statistical life year (VSLY) and the increase in life expectancy or as the product of the value per quality-adjusted life year (VQALY) and the increase in quality-adjusted life expectancy. Following Hammitt (2023), we have:

$$v = \text{VSL}(t) \cdot \Delta L(t) = \text{VSLY}(t) \cdot \Delta LE(t) = \text{VQALY}(t) \cdot \Delta QALE(t)$$

(a) Find $\text{VSL}(t)$ and express $\text{VSLY}(t)$ and $\text{VQALY}(t)$ as functions of $\text{VSL}(t)$.

(b) Hammitt (2023) considers persistent risk impact $\mathcal{I}(t)$ and defines the economic gain from risk reduction as:

$$\mathcal{G}(t; \varrho) = \frac{1}{\mathbf{S}(t)} \int_{t}^{\infty} e^{-\varrho(u-t)} \mathbf{S}(u) \text{VSL}(u) \mathcal{I}(u) \, du$$

What is the interpretation of $\mathcal{G}(t; \varrho)$? Assume that $\mathcal{I}(t) = \mathcal{R}(t)$. What is the associated payoff function. Give the expression for the payoff function using $\text{VSLY}(t)$ or $\text{VQALY}(t)$.

(c) What is the value of $\mathcal{G}(t; \varrho)$ if $\text{VSL}(t)$ and $\mathcal{I}(t)$ are assumed to be constant?

(d) In the case of air pollution, the economic cost is sometimes estimated using the following formula:

$$C = \text{VSLY} \cdot \text{YLL}$$

What is the rationale for this formula?

(e) Consider a numerical application using the previously calibrated Weibull distribution and the HRQL function $Q_1(t)$. Compute the values of $LE(t; \varrho)$ and $QALE(t; \varrho)$ for $t \in \{0, 40, 80\}$ and $\varrho \in \{0, 3\%\}$. Workers in an industry are paid an additional $1000 per year to face a 1 in 10000 increased risk of death. Compute the value of a statistical life. Deduce the values of $\text{VSLY}(t)$ and $\text{VQALY}(t)$. What is the drawback of assuming a constant VSL regardless the worker's age. Therefore, we prefer to assume a constant $\text{VSLY}(t)$, equal to $150000$. Deduce the values of $\text{VSL}(t)$ and $\text{VQALY}(t)$. Calculate the economic gain $\mathcal{G}(t; \varrho)$ of a policy that avoids 5 deaths among 100000 people. Comment on these results.
# B Solutions to the Tutorial Exercises

## B.1 Calculating the prevalence of undernourishment

1. (a) We have:

$$ \ln D = \ln X - \ln R \sim \mathcal{N}(\mu_d, \sigma_d^2) $$

where:

$$ \mu_d = \mathbb{E}[\ln X - \ln R] = \mathbb{E}[\ln X] - \mathbb{E}[\ln R] = \mu_x - \mu_r $$

and:

$$ \begin{aligned} \sigma_d^2 &= \operatorname{var}(\ln X - \ln R) \\ &= \operatorname{var}(\ln X) + \operatorname{var}(\ln R) - 2 \operatorname{cov}(\ln X, \ln R) \\ &= \sigma_x^2 + \sigma_r^2 - 2\rho\sigma_x\sigma_r \end{aligned} $$

Thus, we deduce that $D = X/R$ is a log-normal random variable: $D \sim \mathcal{LN}(\mu_d, \sigma_d^2)$. It follows that:

$$ \begin{aligned} \mathrm{PoU}^* &= \mathrm{Pr}\{X < R\} \\ &= \mathrm{Pr}\{D < 1\} \\ &= \mathrm{Pr}\{\ln D < 0\} \\ &= \Phi\left(-\frac{\mu_d}{\sigma_d}\right) \\ &= \Phi\left(-\frac{\mu_x - \mu_r}{\sqrt{\sigma_x^2 + \sigma_r^2 - 2\rho\sigma_x\sigma_r}}\right) \end{aligned} $$

(b) The probability density functions of $X$ and $R$ are given in Figure 83.

(c) Figure 84 illustrates the relationship between the correlation and the prevalence of undernourishment. We observe a decreasing function between $\rho$ and $\mathrm{PoU}^*$, because the standard deviation of $D$ is a decreasing function of the correlation between $X$ and $R$ and $\mu_x > \mu_r$. Conversely, if $\mu_x < \mu_r$, the relationship between $\rho$ and $\mathrm{PoU}^*$ becomes increasing.

(d) We have:

$$ \mathrm{PoU} = \mathrm{Pr}\{X < r_L\} = \mathrm{Pr}\{\ln X < \ln r_L\} = \Phi\left(\frac{\ln r_L - \mu_x}{\sigma_x}\right) $$

The relationship between $r_L$ and $\mathrm{PoU}$ is shown in Figure 85.

(e) We deduce that:

$$ \begin{aligned} \mathrm{PoU} = \mathrm{PoU}^* &\Leftrightarrow \Phi\left(\frac{\ln r_L^* - \mu_x}{\sigma_x}\right) = \Phi\left(-\frac{\mu_x - \mu_r}{\sqrt{\sigma_x^2 + \sigma_r^2 - 2\rho_{x,r}\sigma_x\sigma_r}}\right) \\ &\Leftrightarrow \frac{\ln r_L^* - \mu_x}{\sigma_x} = -\frac{\mu_x - \mu_r}{\sqrt{\sigma_x^2 + \sigma_r^2 - 2\rho_{x,r}\sigma_x\sigma_r}} \end{aligned} $$

which implies:

$$ r_L^* = \exp\left(\mu_x - \frac{(\mu_x - \mu_r)\sigma_x}{\sqrt{\sigma_x^2 + \sigma_r^2 - 2\rho_{x,r}\sigma_x\sigma_r}}\right) $$

The calibration of $r_L^*$ is shown in Figure 86.
<figure><img src="image_119.png" /><figcaption>Figure 84: Relationship between the correlation $\rho$ and the prevalence of undernourishment PoU*</figcaption></figure>

<figure><img src="image_120.png" /></figure>
<figure><img src="image_121.png" /><figcaption>Figure 86: Calibration of $r_L^*$</figcaption></figure>

<figure><img src="image_122.png" /></figure>
2. (a) We have:

$$ \mu(X) = e^{\mu_x + \frac{1}{2}\sigma_x^2} $$

and:

$$ \sigma^2(X) = e^{2\mu_x + \sigma_x^2} (e^{\sigma_x^2} - 1) = \mu^2(X) (e^{\sigma_x^2} - 1) $$

(b) We deduce that:

$$ CV(X) = \frac{\sigma(X)}{\mu(X)} = \sqrt{e^{\sigma_x^2} - 1} $$

(c) We have:

$$ CV(X) = \sqrt{e^{\sigma_x^2} - 1} \Leftrightarrow e^{\sigma_x^2} = CV^2(X) + 1 \\ \Leftrightarrow \sigma_x = \sqrt{\ln(CV^2(X) + 1)} $$

and:

$$ \begin{align*} \mu(X) = e^{\mu_x + \frac{1}{2}\sigma_x^2} &\Leftrightarrow \mu_x = \ln \mu(X) - \frac{1}{2}\sigma_x^2 \\ &\Leftrightarrow \mu_x = \ln \mu(X) - \frac{1}{2}\ln(CV^2(X) + 1) \\ &\Leftrightarrow \mu_x = \ln \mu(X) - \ln \sqrt{CV^2(X) + 1} \\ &\Leftrightarrow \mu_x = \ln \frac{\mu(X)}{\sqrt{CV^2(X) + 1}} \end{align*} $$

(d) The food available for human consumption is equal to:

$$ Q_1 = 349\,000 + 1\,025 - 40\,000 - (-1\,000) - 45\,000 - 9\,000 = 257\,025 \text{ tonnes} $$

$$ Q_2 = 50\,000 + 9\,010 - 1\,000 - 6\,000 - 500 - 500 = 51\,010 \text{ tonnes} $$

We deduce that the ADEC value is:

$$ \mu(X) = \frac{257\,025 \times 10^3 \times 3\,500 + 51\,010 \times 10^3 \times 500}{365 \times 10^6} = 2\,534.50 \text{kcal/capita/day} $$

(e) Let $x_j^y$ be the average dietary energy consumption for the jth household expenditure decile. We have:

$$ \sigma (X | Y) = \sqrt{\sum_{j=1}^{10} f_j (x_i^y - \bar{x}^y)^2} = 533.4906 $$

where $f_j = 10\%$ is the frequency of each decile group and $\bar{x}^y = \sum_{j=1}^{10} f_j x_i^y$ is the mean. We verify that $\bar{x}^y = 2\,534.50 = \mu(X)$. It follows that:

$$ CV(X | Y) = \frac{\sigma (X | Y)}{\mu (X | Y)} = \frac{533.4906}{2\,534.50} = 0.2105 $$

We deduce that:

$$ CV(X) = \sqrt{CV^2(X | Y) + CV^2(X | R)} = \sqrt{0.2105^2 + 0.20^2} = 0.2904 $$
(f) We have:

$$ \sigma_x = \sqrt{\ln(0.2904^2 + 1)} = 0.2845 $$

and:

$$ \mu_x = \ln \frac{2534.50}{\sqrt{0.2904^2 + 1}} = 7.7973 $$

The prevalence of undernourishment is equal to:

$$ \text{PoU} = \Phi \left( \frac{1850 - 7.7973}{0.2845} \right) = 16.75\% $$

Finally, the number of undernourished people is around 167 500:

$$ \text{NoU} = N \cdot \text{PoU} = 10^6 \times 16.75\% = 167\,451 $$

3. (a) We have:

$$ \text{weight}_{\text{age,sex}}^{*} = \text{BMI}_{\text{age,sex}}(p) \cdot \text{height}_{\text{age,sex}}^{2} $$

For example, we have:

$$ \begin{aligned} \text{weight}_{-3\text{yr,female}}^{*} &= \text{BMI}_{-3\text{yr,female}} (5\%) \cdot \text{height}_{-3\text{yr,female}}^{2} \\ &= 15.5 \times 0.80^{2} \\ &= 9.92 \text{ kg} \end{aligned} $$

We obtain the following results:

<table><thead><tr><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><td>Female</td><td>9.92</td><td>22.32</td><td>40.84</td><td>44.80</td><td>44.80</td><td>44.80</td></tr><tr><td>Male</td><td>12.00</td><td>23.83</td><td>42.44</td><td>54.73</td><td>54.73</td><td>54.73</td></tr></tbody></table>

This means that a 40-year-old woman is considered undernourished if she weighs less than 44.8 kg.

(b) The basal metabolic rate (BMR) is given by the Schofield equation:

$$ \text{BMR}_{\text{age,sex}} = \alpha_{\text{age,sex}} + \beta_{\text{age,sex}} \cdot \text{weight}_{\text{age,sex}}^{*} $$

For example, we have:

$$ \begin{aligned} \text{BMR}_{-3\text{yr,female}} &= \alpha_{-3\text{yr,female}} + \beta_{-3\text{yr,female}} \cdot \text{weight}_{-3\text{yr,female}}^{*} \\ &= -31.1 + 58.317 \times 9.92 \\ &= 547.40 \, \text{kcal/capita/day} \end{aligned} $$

We obtain the following results:

<table><thead><tr><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><td>Female</td><td>547</td><td>939</td><td>1 239</td><td>1 150</td><td>1 210</td><td>1 065</td></tr><tr><td>Male</td><td>684</td><td>1 045</td><td>1 409</td><td>1 516</td><td>1 501</td><td>1 229</td></tr></tbody></table>
(c) The minimum dietary energy requirement is equal to the physical activity level times the basal metabolic rate:

$$MDER_{age,sex} = PAL_{age,sex} \cdot BMR_{age,sex}$$

We obtain the following results:

<table><thead><tr><th>Age</th><th>-3</th><th>3-10</th><th>10-18</th><th>18-30</th><th>30-60</th><th>60+</th></tr></thead><tbody><tr><td>Female</td><td>848</td><td>1 456</td><td>1 921</td><td>1 783</td><td>1 875</td><td>1 651</td></tr><tr><td>Male</td><td>1 060</td><td>1 620</td><td>2 184</td><td>2 350</td><td>2 326</td><td>1 904</td></tr></tbody></table>

This means that a 40-year-old woman is considered undernourished if her dietary energy consumption is less than 1 875 Calories per day.

(d) The minimum dietary energy requirement of the population is the weighted average of the different MDER values:

$$MDER = \sum_{sex} \sum_{age} f_{age,sex} \cdot MDER_{age,sex}$$

where $f_{age,sex}$ is the frequency of the group in the population. Finally, we obtain:

$$MDER = 1 849.90 \text{ kcal/capita/day}$$

4. (a) FD indicates how many calories would be needed to ensure that undernourished would be eliminated if properly distributed.

(b) We have:

$$ 
\begin{aligned}
\text{FD} &= \int_{x<r_L} (\bar{r}-x) f_x(x) \, dx \\
&= \int_0^{r_L} \bar{r} f_x(x) \, dx - \int_0^{r_L} x f_x(x) \, dx \\
&= \bar{r} \int_0^{r_L} f_x(x) \, dx - \frac{\int_0^{r_L} x f_x(x) \, dx}{\int_0^{r_L} f_x(x) \, dx} \int_0^{r_L} f_x(x) \, dx \\
&= (\bar{r} - \mathbb{E}[X | X < r_L]) \cdot \text{Pr}\{X < r_L\} \\
&= \text{PoU} \cdot (\bar{r} - \mathbb{E}[X | X < r_L])
\end{aligned}
$$

The depth of the food deficit is the product of the prevalence of undernourishment and the difference between the average dietary energy requirement and the average dietary energy consumption, conditional on consumption being below the minimum requirement. Another expression is:

$$\text{FD} = \text{PoU} \cdot \mathbb{E}[(\bar{r} - X)_+ | X < r_L]$$

because $\bar{r} \ge r_L$. $\mathbb{E}[(\bar{r} - X)_+ | X < r_L]$ is the expected shortfall of food security.

(c) Following Roncalli (2020a, page 319), we introduce the notation $\Phi_c(x) = \Phi((x - \mu_x)/\sigma_x)$, and we calculate the conditional moment $\mu'_m(X) = \mathbb{E}[X^m | X < r_L]$ for $m \ge 1$ by using the change of variable $y = \ln x$:

$$ 
\begin{aligned}
\mu'_{m}(X) &= \frac{1}{\Phi_c(\ln r_L)} \int_{0}^{r_L} \frac{x^m}{x\sigma_x\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{\ln x - \mu_x}{\sigma_x}\right)^2\right) dx \\
&= \frac{1}{\Phi_c(\ln r_L)} \int_{-\infty}^{\ln r_L} \frac{1}{\sigma_x\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{y - \mu_x}{\sigma_x}\right)^2 + my\right) dy
\end{aligned}
$$
We have:

$$ 
\begin{aligned}
-\frac{1}{2} \left( \frac{y - \mu_x}{\sigma_x} \right)^2 + my &= -\frac{1}{2} \left( \frac{y^2 - 2y(\mu_x + m\sigma_x^2) + \mu_x^2}{\sigma_x^2} \right) \\
&= -\frac{1}{2} \left( \frac{y - (\mu_x + m\sigma_x^2)}{\sigma_x} \right)^2 + \left( m\mu_x + \frac{1}{2}m^2\sigma_x^2 \right)
\end{aligned}
$$

We deduce that:

$$ \mu'_{m}(X) = \frac{\exp(m\mu_{x} + m^{2}\sigma_{x}^{2}/2)}{\Phi_{c}(\ln r_{L})} \int_{-\infty}^{\ln r_{L}} \frac{1}{\sigma_{x}\sqrt{2\pi}} \exp\left(-\frac{1}{2}\left(\frac{y - (\mu_{x} + m\sigma_{x}^{2})}{\sigma_{x}}\right)^{2}\right) dy $$

Using the change of variable $z = \frac{y - (\mu_x + m\sigma_x^2)}{\sigma_x}$, it follows that:

$$ \mu'_{m}(X) = \frac{\exp(m\mu_{x} + m^{2}\sigma_{x}^{2}/2)}{\Phi_{c}(\ln r_{L})} \int_{-\infty}^{z_{L}} \frac{1}{\sqrt{2\pi}} \exp\left(-\frac{1}{2}z^{2}\right) dz $$

where:

$$ z_L = \frac{\ln r_L - (\mu_x + m\sigma_x^2)}{\sigma_x} $$

Finally, we obtain:

$$ \mu'_{m}(X) = \frac{\Phi_{c}(\ln r_{L} - m\sigma_{x}^{2})}{\Phi_{c}(\ln r_{L})} \exp\left(m\mu_{x} + \frac{1}{2}m^{2}\sigma_{x}^{2}\right) $$

and:

$$ \mathbb{E}[X | X < r_L] = \mu'_1(X) = \frac{\Phi\left(\frac{\ln r_L - \mu_x - \sigma_x^2}{\sigma_x}\right)}{\Phi\left(\frac{\ln r_L - \mu_x}{\sigma_x}\right)} \exp\left(\mu_x + \frac{1}{2}\sigma_x^2\right) $$

The analytical expression of the depth of the food deficit is:

$$ FD = \bar{r}\Phi\left(\frac{\ln r_L - \mu_x}{\sigma_x}\right) - e^{\mu_x + \frac{1}{2}\sigma_x^2}\Phi\left(\frac{\ln r_L - \mu_x - \sigma_x^2}{\sigma_x}\right) $$

because:

$$ \text{PoU} = \text{Pr}\{X < r_L\} = \Phi\left(\frac{\ln r_L - \mu_x}{\sigma_x}\right) $$

(d) The depth of the food deficit is equal to:

$$ 
\begin{aligned}
FD &= 2500 \times \Phi\left(\frac{\ln 1850 - 7.7973}{0.2845}\right) - \\
&\quad \exp\left(7.7973 + \frac{1}{2}0.2845^2\right) \times \Phi\left(\frac{\ln 1850 - 7.7973 - 0.2845^2}{0.2845}\right) \\
&= 150.2968 \text{ kcal/capita/year}
\end{aligned}
$$
## B.2 Calculating the species-area relationship using the theory of island biogeography

1. (a) We have:

$$ 
\begin{aligned}
f'(x) &= ae^{b(x/c)^d}bd\left(\frac{x}{c}\right)^{d-1}\frac{1}{c} \\
&= \frac{abd}{c}\left(\frac{x}{c}\right)^{d-1}e^{b(x/c)^d}
\end{aligned}
$$

and:

$$ 
\begin{aligned}
f''(x) &= \frac{abd}{c}(d-1)\left(\frac{x}{c}\right)^{d-2}\frac{1}{c}e^{b(x/c)^d} + \frac{abd}{c}\left(\frac{x}{c}\right)^{d-1}e^{b(x/c)^d}bd\left(\frac{x}{c}\right)^{d-1}\frac{1}{c} \\
&= \frac{abd(d-1)}{c^2}\left(\frac{x}{c}\right)^{d-2}e^{b(x/c)^d} + \frac{ab^2d^2}{c^2}\left(\frac{x}{c}\right)^{2d-2}e^{b(x/c)^d} \\
&= \frac{abd}{c^2}\left(\frac{x}{c}\right)^{d-2}\left(d-1+bd\left(\frac{x}{c}\right)^d\right)e^{b(x/c)^d}
\end{aligned}
$$

Since $a, c$ and $d$ are positive, we deduce that $\frac{x}{c} \ge 0$, $e^{b(x/c)^d} \ge 0$ and:

$$ f'(x) \ge 0 \Leftrightarrow b \ge 0 $$

$f(x)$ is an increasing function of $x$ if the parameter $b$ is positive, otherwise $f(x)$ is a decreasing function of $x$. We have:

$$ f''(x) \ge 0 \Leftrightarrow g(x; b) = \underbrace{b(d-1)}_{-/+} + \underbrace{b^2d\left(\frac{x}{c}\right)^d}_{+} \ge 0 $$

because we have:

$$ 0 \le b^2 d \left(\frac{x}{c}\right)^d \le b^2 d $$

We consider two cases:

i. $b = -1$

If $d \le 1$, then $g(x; -1) = 1 - d + d\left(\frac{x}{c}\right)^d$, $f''(x) \ge 0$ and the function is convex. If $d > 1$, the function is neither convex nor concave.

ii. $b = 1$

If $d \ge 1$, then $g(x; 1) = (d-1) + d\left(\frac{x}{c}\right)^d$, $f''(x) \ge 0$ and the function is convex. If $d \le \frac{1}{2}$, then $g(0; 1) = d-1 < 0$, $g(c; 1) = 2d-1 \le 0$, $f''(x) \le 0$ and the function is concave. If $d \in (\frac{1}{2}, 1)$, the function is neither convex nor concave.

(b) We have$^{171}$:

$$ \frac{\partial \lambda(t)}{\partial S(t)} = -\frac{\beta_1 \lambda_0^2}{(1 - e^{-1}) S_s} \left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0 - 1} e^{-\left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0}} \le 0 $$

with:

$$ \lambda(t_0) = \lambda_0 \frac{e^{-0} - e^{-1}}{1 - e^{-1}} = \lambda_0 $$

<footnote n="171">171 We set $a = \frac{\lambda_0}{1 - e^{-1}}$, $b = -1$, $c = S_s$ and $d = \beta_1 \lambda_0$.</footnote>
and:

$$ \lambda(t_s) = \lambda_0 \frac{e^{-1} - e^{-1}}{1 - e^{-1}} = 0 $$

The second-order derivative is:

$$ \frac{\partial^2 \lambda(t)}{\partial S(t)^2} = - \frac{\beta_1 \lambda_0^2}{(1 - e^{-1}) S_s^2} \left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0 - 2} \left(\beta_1 \lambda_0 - 1 - \beta_1 \lambda_0 \left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0}\right) e^{-\left(\frac{S(t)}{S_s}\right)^{\beta_1 \lambda_0}} $$

$\lambda(t)$ is convex if and only if $\beta_1\lambda_0 \le 1$. The function $\lambda(t)$ is plotted in Figure 87. We verify that it is decreasing and convex for the first three sets of parameters because $\beta_1\lambda_0 \le 1$. For the last set of parameters, it is decreasing but not convex.

<figure><img src="image_123.png" /><figcaption>Figure 87: Immigrate rate function $\lambda(t)$</figcaption></figure>

(c) We have$^{172}$:

$$ \frac{\partial \mu^{\text{long}}(t)}{\partial S(t)} = \frac{\beta_2 \mu_s^2}{(e^{-1} - 1) S_s} \left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s - 1} e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} \geq 0 $$

with:

$$ \mu^{\text{long}}(t_0) = \mu_s \frac{e^0 - 1}{e^1 - 1} = 0 $$

and:

$$ \mu^{\text{long}}(t_s) = \mu_s \frac{e^1 - 1}{e^1 - 1} = \mu_s $$

<footnote n="172">172 We set $a = \frac{\mu_s}{e^{-1} - 1}$, $b = 1$, $c = S_s$ and $d = \beta_2\mu_s$.</footnote>
The second-order derivative is:

$$ \frac{\partial^2 \mu^{\text{long}}(t)}{\partial S(t)^2} = \frac{\beta_2 \mu_s^2}{(e^{-1} - 1) S_s^2} \left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s - 2} \left(\beta_2 \mu_s - 1 + \beta_2 \mu_s \left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}\right) e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} $$

$\mu^{\text{long}}(t)$ is convex if and only if $\beta_2\mu_s \ge 1$ and concave if and only if $\beta_2\mu_s \le \frac{1}{2}$. The function $\mu^{\text{long}}(t)$ is plotted in Figure 88. We verify that it is increasing and concave for the first two sets of parameters because $\beta_2\mu_s \le \frac{1}{2}$. For the last two sets of parameters, it is increasing and convex.

<figure><img src="image_124.png" /><figcaption>Figure 88: Long-term extinction rate function $\mu^{\text{long}}(t)$</figcaption></figure>

(d) $\mu^{\text{short}}(t)$ is the product of two positive and decreasing functions: $\lambda(t)$ and $\beta_3\lambda(t) e^{-\beta_4 S(t)}$. We deduce that it is decreasing. We also have:

$$ \mu^{\text{short}}(t_0) = \beta_3 \lambda_0 e^{-\beta_4 S_0} \le \beta_3 \lambda_0 $$

and:

$$ \mu^{\text{short}}(t_s) = \beta_3 \lambda_s e^{-\beta_4 S_s} = 0 $$

The function $\mu^{\text{short}}(t)$ is plotted in Figure 89.

(e) We have:

$$ \mu(t_0) = \mu^{\text{short}}(t_0) + \mu^{\text{long}}(t_0) = \mu^{\text{short}}(t_0) + 0 = \mu^{\text{short}}(t_0) $$

and:

$$ \mu(t_s) = \mu^{\text{short}}(t_s) + \mu^{\text{long}}(t_s) = 0 + \mu_s = \mu_s $$
<figure><img src="image_125.png" /><figcaption>Figure 90: Extinction rate function $\mu(t)$</figcaption></figure>

<figure><img src="image_126.png" /></figure>
Figure 90 illustrates the function $\mu(t)$. Since the extinction rate is the sum of two functions, $\mu^{short}(t)$ which is decreasing, and $\mu^{long}(t)$ which is increasing, $\mu(t)$ is not monotonie and can exhibit several shapes. However, for reasonable values of the parameters, we observe that $\mu(t)$ first decreases and then increases. This behavior arises because $\mu^{short}(t_0) > \mu^{long}(t_0)$ and $\mu^{short}(t_s) < \mu^{long}(t_s)$. Consequently, when $S(t)$ is low and close to zero, the short-term extinction rate dominates the long-term extinction rate, leading to a general decrease in $\mu(t)$. In contrast, when $S(t)$ is high and approaches the saturation state $S_s$, the long-term extinction rate becomes dominant, implying that $\mu(t)$ generally increases for high values of species richness. In summary, the extinction rate $\mu(t)$ has a U-shaped relationship with $S(t)$.

(f) An equilibrium occurs if and only if $\delta(t) = 0$ or $\lambda(t) = \mu(t)$. We deduce that such an equilibrium $S^*$ is achieved if the following equation has at least one solution:

$$ \lambda_0 \frac{e^{-\left(\frac{S^*}{S_s}\right)^{\beta_1 \lambda_0}} - e^{-1}}{1 - e^{-1}} \left(1 - \beta_3 e^{-\beta_4 S^*}\right) = \mu_s \frac{e^{\left(\frac{S^*}{S_s}\right)^{\beta_2 \mu_s}} - 1}{e^1 - 1} \quad (38) $$

Figure 91 shows the functions $\lambda(t)$ and $\mu(t)$. Solving Equation (38) gives $S^* = 96.472$. This equilibrium is stable, because $\frac{dS(t)}{dt} < 0$ when $S(t) > S^*$ and $\frac{dS(t)}{dt} > 0$ when $S(t) < S^*$. This implies that any perturbation from $S^*$ will return to the steady state $S^*$.

(g) In Table 61, we give the direction of variation of the functions $\lambda(t)$, $\mu^{short}(t)$, $\mu^{long}(t)$ and $\mu(t)$. We can then derive the table of variation of the net diversification rate $\delta(t) = \lambda(t) - \mu(t)$. When $\beta_3 = 0$, $\mu^{short}(t) = 0$ and $\delta(t)$ is a decreasing function with $\delta(t_0) = \delta_0 = \lambda_0 > 0$ and $\delta(t_s) = -\mu_s < 0$. We conclude that there is only one equilibrium. When $0 < \beta_3 \le 1$, $\delta(t)$ increases up to a threshold $S^*$ and then decreases. We have $\delta_0 = \lambda_0(1 - \beta_3) > 0$ and $\delta(t_s) = -\mu_s < 0$. We obtain the table of variation of the net diversification rate given in Table 62. When $\beta_3 > 1$, $\delta(t)$ has the same behavior as the previous case, but $\delta_0 = \lambda_0(1 - \beta_3) < 0$ and we obtain the table of variation of the net diversification rate given in Table 63. We deduce that there are two equilibria $S_1^*$ and $S_2^*$, with $S_1^* \le S_2^*$. The equilibrium $S_1^*$ is unstable because $\frac{dS(t)}{dt} < 0$ when $S(t) < S_1^*$ and $\frac{dS(t)}{dt} > 0$ when $S(t) > S_1^*$. Only $S_2^*$ is a stable steady state. Furthermore, the net diversification rate is negative when $S(t) < S_1^*$, which means that $S(t + dt) < S(t)$ and $S(t)$ tends to the equilibrium $S_0^* = 0$.

(h) The simulations of the species richness process are presented in Figures 93 and 94. In the one-equilibrium case, the process $S(t)$ converges to the equilibrium $S^* = 96.47$, regardless of the initial state $S_0$. In the three-equilibrium case, the process $S(t)$ converges to either $S_0^* = 0$ or $S_2^* = 97.35$, depending on the initial state $S_0$. The equilibrium $S_1^* = 27.65$ is reached only if $S_0 = S_1^*$. Otherwise, we get the equilibrium $S_0^* = 0$ if $S_0 < S_1^*$, and the equilibrium $S_2^* = 97.35$ if $S_0 > S_1^*$.

2. (a) We have:

$$ \frac{\partial \mu(t)}{\partial \beta_2} = \frac{\mu_s^2}{e^1 - 1} e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} \left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s} \ln\left(\frac{S(t)}{S_s}\right) \le 0 $$

This means that the relationship $\mu(t)$ shifts downward as $\beta_2$ increases. Figure 95 illustrates the equilibrium for the three values of $\beta_2$. From this, we deduce that if $\beta'_2 \ge \beta_2$, then $S^*(\beta'_2) \ge S^*(\beta_2)$.
<table><caption>Table 61: Table of variation for the immigration and extinction rates</caption><thead><tr><th>S</th><th>0</th><th>S*</th><th>S<sub>s</sub></th></tr></thead><tbody><tr><td>λ(t)</td><td>λ<sub>0</sub></td><td></td><td>0</td></tr><tr><td>μ<sup>short</sup>(t)</td><td>β<sub>3</sub>λ<sub>0</sub></td><td></td><td>0</td></tr><tr><td>μ<sup>long</sup>(t)</td><td>0</td><td>μ<sub>s</sub></td><td></td></tr><tr><td>μ(t)<br>(β<sub>3</sub> = 0)</td><td>0</td><td>μ<sub>s</sub></td><td></td></tr><tr><td>μ(t)<br>(β<sub>3</sub> > 0)</td><td>β<sub>3</sub>λ<sub>0</sub></td><td>μ<sup>*</sup></td><td>μ<sub>s</sub></td></tr></tbody></table>

<table><caption>Table 62: Table of variation for the net diversification rate $\delta(t)$</caption><thead><tr><th>S</th><th>0</th><th>S<sup>*</sup></th><th>S<sup>*</sup></th><th>S<sub>s</sub></th></tr></thead><tbody><tr><td>λ(t) - μ(t)<br>(0 < β<sub>3</sub> ≤ 1)</td><td>δ<sub>0</sub></td><td>δ<sup>*</sup></td><td>0</td><td>-μ<sub>s</sub></td></tr></tbody></table>

<table><caption>Table 63: Table of variation for the net diversification rate $\delta(t)$</caption><thead><tr><th>S</th><th>0</th><th>S<sub>1</sub><sup>*</sup></th><th>S<sup>*</sup></th><th>S<sub>2</sub><sup>*</sup></th><th>S<sub>s</sub></th></tr></thead><tbody><tr><td>λ(t) - μ(t)<br>(β<sub>3</sub> > 1)</td><td>δ<sub>0</sub></td><td>0</td><td>δ<sup>*</sup></td><td>0</td><td>-μ<sub>s</sub></td></tr></tbody></table>
<figure><img src="image_127.png" /><figcaption>Figure 92: The three-equilibrium case</figcaption></figure>

<figure><img src="image_128.png" /></figure>
<figure><img src="image_129.png" /><figcaption>Figure 94: Simulation of the species richness process under a three-equilibrium scenario</figcaption></figure>

<figure><img src="image_130.png" /></figure>
<figure><img src="image_131.png" /><figcaption>Figure 95: How the equilibrium shifts with the parameter $\beta_2$</figcaption></figure>

(b) The relationship between $\beta_2$ and $S^*$ is shown in Figure 96.

(c) We have:

$$ \frac{\partial \mu(t)}{\partial \mu_s} = \frac{e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} - 1}{e^1 - 1} + \frac{\beta_2 \mu_s}{e^1 - 1} e^{\left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s}} \left(\frac{S(t)}{S_s}\right)^{\beta_2 \mu_s} \ln\left(\frac{S(t)}{S_s}\right) $$

Using the default parameter values and $S(t) = 100$, we find that $\partial_{\mu_s}\mu(t) = 0.045$ when $\mu_S = 1$ and $\partial_{\mu_s}\mu(t) = -0.047$ when $\mu_S = 5$. Since the derivative can be either positive or negative, we cannot determine the effect of $\mu_s$ on the equilibrium $S^*$.

(d) The relationship between $\mu_s$ and $S^*$ is shown in Figure 97.

(e) We can assume that the relationship between $A$ and $\beta_2$ is increasing. In this case, we obtain the species-area relationship. We can also assume that the relationship between $A$ and $\mu_s$ is increasing. In this case, we retrieve the species-area relationship only if $\mu_s$ is greater than a threshold.

(f) We obtain the following estimated values for the parameters:

<table><thead><tr><td>Parameter</td><td>Power</td><td>Exponential</td><td>Kobayashi</td></tr></thead><tbody><tr><td>c</td><td>74.513</td><td>82.098</td><td>73.539</td></tr><tr><td>z</td><td>0.562</td><td>45.199</td><td>1.797</td></tr></tbody></table>

The fitted models are displayed in Figure 98. We observe that the fitted curves closely align with the equilibrium curve predicted by the theory of island biogeography.
<figure><img src="image_132.png" /><figcaption>Figure 97: Relationship between $\mu_s$ and $S^*$</figcaption></figure>

<figure><img src="image_133.png" /></figure>
<figure><img src="image_134.png" /><figcaption>Figure 98: Species-area relationship (TIB, power, exponential and Kobayashi models)</figcaption></figure>

## B.3 Species abundance models

1. (a) We have:

$$ \tilde{S}_a = \sum_{i=1}^{S} \tilde{S}_i $$

Assuming that the random variables $\tilde{S}_1, \dots, \tilde{S}_S$ are independent, $\tilde{S}_a$ follows a Poisson binomial distribution:

$$ \tilde{S}_a \sim \mathcal{P}\mathcal{B}(p_1, \dots, p_S) $$

The probability mass function is given by:

$$ \mathrm{Pr} \{ \tilde{S}_a = k \} = \sum_{\mathcal{E} \in \mathcal{F}_k} \prod_{i \in \mathcal{E}} p_i \prod_{j \in \mathcal{E}^c} (1 - p_j) $$

where $\mathcal{F}_k$ is the set of all subsets of $k$ integers that can be selected from $\{1, \dots, n\}$ and $\mathcal{E}^c$ denotes the complement of the subset $\mathcal{E}$. From this, we deduce the expected value:

$$ S_a = \mathbb{E}[\tilde{S}_a] = \mathbb{E}\left[\sum_{i=1}^{S} \tilde{S}_i\right] = \sum_{i=1}^{S} p_i $$

Given $p = (20\%, 30\%, 10\%, 65\%, 10\%)$, we obtain the following probability mass function for $\tilde{S}_a$:

<table><thead><tr><th>k</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead><tbody><tr><td>Pr{S&#x0303;<sub>a</sub> = k}</td><td>15.88%</td><td>43.79%</td><td>30.85%</td><td>8.48%</td><td>0.97%</td><td>0.04%</td></tr></tbody></table>

The expected number of species is $S_a = 1.35$.
(b) Let $\pi$ be the probability that an individual occupies the subarea $a$. We have:

$$ \pi = \frac{a}{A} $$

For example, if $A = 10 \text{ km}^2$, the occupancy probability is 20% when $a = 2 \text{ km}^2$. Let $\tilde{N}_i$ be the random variable indicating the number of individuals of species $i$ present in the subarea $a$. $\tilde{N}_i$ follows a binomial distribution with parameters $n_i$ (the total number of individuals of species $i$) and $\pi$. Thus, we have:

$$ \Pr\{\tilde{N}_i = k\} = C_{n_i}^k \pi^k (1-\pi)^{n_i-k} $$

The probability $p_i$ of observing the species $i$ on the subarea is then:

$$ \begin{aligned} p_i = \Pr\{\tilde{S}_i = 1\} &= \Pr\{\tilde{N}_i > 0\} \\ &= 1 - \Pr\{N_i = 0\} \\ &= 1 - (1 - \pi)^{n_i} \\ &= 1 - \left(1 - \frac{a}{A}\right)^{n_i} \end{aligned} $$

Consequently, $\tilde{S}_a$ follows then a Poisson binomial distribution:

$$ \tilde{S}_a \sim \mathcal{P}\mathcal{B}\left(p_i = 1 - \left(1 - \frac{a}{A}\right)^{n_i}\right) $$

It follows that:

$$ \begin{aligned} S_a &= \sum_{i=1}^{S} p_i \\ &= \sum_{i=1}^{S} \left(1 - \left(1 - \frac{a}{A}\right)^{n_i}\right) \\ &= S - \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i} \end{aligned} $$

Given $(n_1, \dots, n_5) = (10, 30, 25, 66, 8)$, $A = 10 \text{ km}^2$, and $a = 2 \text{ km}^2$, $\pi = 20\%$ and we obtain the following results:

<table><thead><tr><th>k or i</th><th>0</th><th>1</th><th>2</th><th>3</th><th>4</th><th>5</th></tr></thead><tbody><tr><td>p<sub>i</sub></td><td></td><td>89.26%</td><td>59.04%</td><td>99.62%</td><td>73.79%</td><td>83.22%</td></tr><tr><td>Pr{S&#x0303;<sub>a</sub> = k}</td><td>0.00%</td><td>0.21%</td><td>3.45%</td><td>19.78%</td><td>44.32%</td><td>32.34%</td></tr></tbody></table>

The expected number of species is $S_a = 4.05$.

(c) i. If the distribution is homogenous across species ($n_i = \frac{n}{S}$), we have:

$$ p_i = 1 - \left(1 - \frac{a}{A}\right)^{\frac{n}{S}} $$

$\tilde{S}_a$ follows then a binomial distribution:

$$ \tilde{S}_a \sim \mathcal{B}\left(S, p = 1 - \left(1 - \frac{a}{A}\right)^{\frac{n}{S}}\right) $$
The expected number of species is then equal to:

$$S_a = Sp = S \left( 1 - \left( 1 - \frac{a}{A} \right)^{\frac{n}{S}} \right)$$

ii. If the distribution is the most uneven ($n_i = 1$ for $i < S$ and $n_S = n - S + 1$), we have:

$$ 
\begin{aligned}
S_a &= S - \sum_{i=1}^{S-1} \left(1 - \frac{a}{A}\right)^{1} - \left(1 - \frac{a}{A}\right)^{n-S+1} \\
&= S - (S-1)\left(1 - \frac{a}{A}\right) - \left(1 - \frac{a}{A}\right)^{n-S+1} \\
&= 1 + (S-1)\frac{a}{A} - \left(1 - \frac{a}{A}\right)^{n-S+1}
\end{aligned}
$$

iii. We have:

$$ 
\begin{aligned}
S_a &= S - \sum_{i=1}^{s} \left(1 - \frac{a}{A}\right)^{1} - \sum_{i=s+1}^{s} \left(1 - \frac{a}{A}\right)^{n_{s+1}} \\
&= S - s\left(1 - \frac{a}{A}\right) - (S - s)\left(1 - \frac{a}{A}\right)^{n_{s+1}} \\
&= s\frac{a}{A} + (S - s)\left(1 - \left(1 - \frac{a}{A}\right)^{n_{s+1}}\right) \\
&= s\frac{a}{A} + (S - s)\left(1 - \left(1 - \frac{a}{A}\right)^{\frac{n-s}{S-s}}\right)
\end{aligned}
$$

If $s = 0$, we retrieve the most even model:

$$S_a = S \left( 1 - \left( 1 - \frac{a}{A} \right)^{\frac{n}{S}} \right)$$

If $s = S - 1$, we retrieve the most uneven model:

$$S_a = (S-1) \frac{a}{A} + \left(1 - \left(1 - \frac{a}{A}\right)^{\frac{n-S+1}{S-S+1}}\right) = 1 + (S-1) \frac{a}{A} - \left(1 - \frac{a}{A}\right)^{n-S+1}$$

iv. We have:

$$ 
\begin{aligned}
S_a &= \sum_{i=1}^{S} \left( 1 - \left(1 - \frac{a}{A}\right) \left(1 + \frac{n_i a}{\kappa_i A}\right)^{-\kappa_i} \right) \\
&= S - \left(1 - \frac{a}{A}\right) \sum_{i=1}^{S} \left(1 + \frac{n_i a}{\kappa_i A}\right)^{-\kappa_i}
\end{aligned}
$$

v. We can approximate the harmonic sum by:

$$\sum_{k=i}^{S} \frac{1}{k} \approx \int_{i}^{S} \frac{1}{x} dx = \ln \left( \frac{S}{i} \right)$$

We deduce that:

$$n_i = \frac{n}{S} \sum_{k=i}^{S} \frac{1}{k} \approx \frac{n}{S} \ln \left( \frac{S}{i} \right)$$
We have:

$$ \left(1 - \frac{a}{A}\right)^{n_i} = e^{(n_i \ln(1 - \frac{a}{A}))} = \exp\left(\frac{n}{S} \ln\left(\frac{S}{i}\right) \ln\left(1 - \frac{a}{A}\right)\right) = \left(\frac{i}{S}\right)^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right)} $$

and:

$$ \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i} = \sum_{i=1}^{S} \left(\frac{i}{S}\right)^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right)} \approx \int_{0}^{S} \left(\frac{x}{S}\right)^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right)} dx $$

Using the change of variable $u = \frac{x}{S}$, we obtain:

$$ \begin{aligned} \int_{0}^{S} \left(\frac{x}{S}\right)^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right)} dx &= S \int_{0}^{1} u^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right)} du \\ &= S \left[ \frac{u^{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right) + 1}}{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right) + 1} \right]_{0}^{1} \\ &= \frac{S}{-\frac{n}{S} \ln\left(1 - \frac{a}{A}\right) + 1} \\ &= \frac{S^2}{S - n \ln\left(1 - \frac{a}{A}\right)} \end{aligned} $$

Since we have:

$$ S_a = S - \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i} \approx S - \frac{S^2}{S - n \ln\left(1 - \frac{a}{A}\right)} = \frac{-Sn \ln\left(1 - \frac{a}{A}\right)}{S - n \ln\left(1 - \frac{a}{A}\right)} $$

we conclude that:

$$ S_a = \frac{S \ln\left(1 - \frac{a}{A}\right)}{\ln\left(1 - \frac{a}{A}\right) - \frac{S}{n}} $$

(d) We have:

$$ S_a = \sum_{i=1}^{S} p_i = \sum_{j} \sum_{i \in j} p_i = \sum_{j} s(j) p_j $$

where $p_j$ is the probability that a species with abundance $j$ is present in the subarea $a$. Since $p_j = 1 - \left(1 - \frac{a}{A}\right)^j$, we get:

$$ S_a = \sum_{j} s(j) \left(1 - \left(1 - \frac{a}{A}\right)^j\right) = S - \sum_{j} s(j) \left(1 - \frac{a}{A}\right)^j $$

where $S = \sum_j s(j)$ is the total number of species. To calculate $S_a$ for the log-series distribution, we use a preliminary result: $\sum_{j=1}^{\infty} x^j/j$ is the series expansion for $-\ln(1-x)$ when $|x| < 1$. We have:

$$ S = \sum_{j=1}^{\infty} s(j) = \alpha \sum_{j=1}^{\infty} \frac{x^j}{j} = -\alpha \ln(1-x) $$
and:

$$ \sum_{j=1}^{\infty} s(j) \left(1 - \frac{a}{A}\right)^j = \alpha \sum_{j=1}^{\infty} \frac{x^j}{j} \left(1 - \frac{a}{A}\right)^j = -\alpha \ln \left(1 - x \left(1 - \frac{a}{A}\right)\right) $$

We deduce that:

$$ \begin{aligned} S_a &= -\alpha \ln(1-x) + \alpha \ln\left(1-x\left(1-\frac{a}{A}\right)\right) \\ &= \alpha \ln\left(\frac{1-x\left(1-\frac{a}{A}\right)}{1-x}\right) \\ &= \alpha \ln\left(1+\frac{x}{1-x}\frac{a}{A}\right) \end{aligned} $$

2. (a) Species $i$ is locally endemic to subarea $a \subseteq A$ if all the individuals of this species are found in $a$. Let $E_i \sim \mathcal{B}(\tilde{p}_i)$ be the Bernoulli random variable that takes the value 1 if species $i$ is locally endemic to $a$. We have:

$$ \tilde{p}_i = \Pr\{E_i = 1\} = \prod_{k=1}^{n_i} \left(\frac{a}{A}\right) = \left(\frac{a}{A}\right)^{n_i} $$

(b) The expected number $E_a$ of locally endemic species is equal to:

$$ E_a = \mathbb{E}[E_1 + E_2 + \dots + E_S] = \sum_{i=1}^{S} \mathbb{E}[E_i] = \sum_{i=1}^{S} \tilde{p}_i = \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i} $$

(c) Since we have $S_a = S - \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i}$ and $E_a = \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i}$, we deduce that:

$$ S_{A-a} = S - \sum_{i=1}^{S} \left(1 - \frac{A-a}{A}\right)^{n_i} = S - \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i} $$

and:

$$ E_{A-a} = \sum_{i=1}^{S} \left(\frac{A-a}{A}\right)^{n_i} = \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i} $$

It follows that:

$$ S_{A-a} + E_a = S - \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i} + \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i} = S $$

(d) We have:

$$ T_a = S_a + E_a = S - \sum_{i=1}^{S} \left(1 - \frac{a}{A}\right)^{n_i} + \sum_{i=1}^{S} \left(\frac{a}{A}\right)^{n_i} $$

It follows that:

$$ \begin{aligned} \frac{\partial T_a}{\partial a} &= -\sum_{i=1}^{S} n_i \left(1 - \frac{a}{A}\right)^{n_i-1} \left(-\frac{1}{A}\right) + \sum_{i=1}^{S} n_i \left(\frac{a}{A}\right)^{n_i-1} \left(\frac{1}{A}\right) \\ &= \frac{1}{A} \left( \sum_{i=1}^{S} n_i \left(1 - \frac{a}{A}\right)^{n_i-1} + \sum_{i=1}^{S} n_i \left(\frac{a}{A}\right)^{n_i-1} \right) \\ &\ge 0 \end{aligned} $$
This implies that $T_a$ is an increasing function of $a$. Moreover, we have:

$$T_{\emptyset} = S_{\emptyset} + E_{\emptyset} = S - \sum_{i=1}^{S} \left(1 - \frac{0}{A}\right)^{n_i} + \sum_{i=1}^{S} \left(\frac{0}{A}\right)^{n_i} = S - S + 0 = 0$$

and:

$$T_A = S_A + E_A = S - \sum_{i=1}^{S} \left(1 - \frac{A}{A}\right)^{n_i} + \sum_{i=1}^{S} \left(\frac{A}{A}\right)^{n_i} = S - 0 + S = 2S$$

We deduce that:

$$0 \le S_a + E_a \le 2S$$

## B.4 Valuation of risks to life and health

1. Using Bayes theorem, we have:

$$\mathbf{S}(u | t) = \Pr\{\tau > u | \tau > t\} = \frac{\Pr\{\tau > u, \tau > t\}}{\Pr\{\tau > t\}} = \frac{\mathbf{S}(u)}{\mathbf{S}(t)}$$

2. Since $d\mathbf{S}(t) = \partial\mathbf{S}(t) dt = -f(t) dt$, we can express the expected survival time as:

$$\mathbb{E}[\tau] = \int_{0}^{\infty} tf(t) dt = - \int_{0}^{\infty} t d\mathbf{S}(t)$$

Using integration by parts, we get:

$$\mathbb{E}[\tau] = -[t\mathbf{S}(t)]_{0}^{\infty} + \int_{0}^{\infty} \mathbf{S}(t) dt = 0 + \int_{0}^{\infty} \mathbf{S}(t) dt = \int_{0}^{\infty} \mathbf{S}(t) dt$$

We deduce the conditional life expectancy for an individual alive at time $t$:

$$\mathrm{LE}(t) = \mathbb{E}[\tau | \tau > t] = \frac{\mathbb{E}[\mathbb{1}\{\tau > t\} \cdot \tau]}{\mathrm{Pr}\{\tau > t\}} = \frac{\int_{t}^{\infty} \mathbf{S}(u) \, du}{\mathbf{S}(t)} = \int_{t}^{\infty} \mathbf{S}(u | t) \, du$$

3. The expected present value of a payoff, taking into account the future lifetime, is given by:

$$\mathbb{E}[X; t, \varrho] = \int_{t}^{\infty} e^{-\varrho(u-t)} \mathbf{S}(u | t) X(u) \, du = \mathbb{E}[\delta(u) X(u)] \quad (39)$$

where $\delta(u) = B(t, u) \mathbf{S}(u | t)$ is the discount factor under uncertain lifetime and $B(t, u) = e^{-\varrho(u-t)}$ is the standard discount factor at time $u$. Therefore, we have:

$$\delta(u) = e^{-\varrho(u-t)} \mathrm{Pr}\{\tau > u | \tau > t\}$$

Equation (39) is the classical formula for the present value when the discount rate accounts for uncertainty about the individual's lifetime.

4. DLE($t; \varrho$) is the mathematical expectation of the discounted survival time, given that the survival time exceeds $t$:

$$\mathrm{DLE}(t; \varrho) = \mathbb{E}[e^{-\varrho t} \tau | \tau > t]$$
Using integration by parts with $u = e^{-\varrho t}t$ and $v' = f(t)$, we can express the discounted life expectancy at $t = 0$ as:

$$ 
\begin{aligned}
\text{DLE}(0; \varrho) &= \int_{0}^{\infty} e^{-\varrho t} tf(t) \, dt \\
&= [-e^{-\varrho t} t \mathbf{S}(t)]_{0}^{\infty} + \int_{0}^{\infty} (-\varrho e^{-\varrho t} t + e^{-\varrho t}) \mathbf{S}(t) \, dt \\
&= \int_{0}^{\infty} e^{-\varrho t} \mathbf{S}(t) \, dt - \varrho \int_{0}^{\infty} e^{-\varrho t} t \mathbf{S}(t) \, dt
\end{aligned}
$$

because $u' = -\varrho e^{-\varrho t}t + e^{-\varrho t}$ and $v = -\mathbf{S}(t)$. We deduce that:

$$ \text{DLE}(0; \varrho) = \text{LE}(0; \varrho) - \underbrace{\varrho \int_{0}^{\infty} e^{-\varrho t} t \mathbf{S}(t) \, dt}_{\ge 0} $$

and:

$$ \text{LE}(0; \varrho) \ge \text{DLE}(0; \varrho) = \int_{0}^{\infty} e^{-\varrho t} tf(t) \, dt $$

The generalization to the case $t \ne 0$ is straightforward. Using the change of variable $x = u - t$, we obtain $\text{DLE}(t; \varrho) = \text{LE}(t; \varrho) - \varrho \int_{0}^{\infty} e^{-\varrho x} x \mathbf{S}(t+x) \, dx$. Equality is achieved if and only if $\varrho = 0$.

5. Since $\mathbf{S}(t) = e^{-\lambda t}$, we have:

$$ 
\begin{aligned}
\text{LE}(t; \varrho) &= \frac{1}{e^{-\lambda t}} \int_{t}^{\infty} e^{-\varrho(u-t)} e^{-\lambda u} \, du \\
&= \int_{t}^{\infty} e^{-(\varrho+\lambda)(u-t)} \, du \\
&= \left[ \frac{e^{-(\varrho+\lambda)(u-t)}}{(\varrho+\lambda)} \right]_{t}^{\infty} \\
&= \frac{1}{\varrho+\lambda}
\end{aligned}
$$

and:

$$ 
\begin{aligned}
\text{DLE}(t; \varrho) &= \int_{t}^{\infty} e^{-\varrho(u-t)} (u-t) f(u \mid t) \, du \\
&= \int_{t}^{\infty} e^{-\varrho(u-t)} (u-t) \lambda e^{-\lambda(u-t)} \, du \\
&= \int_{0}^{\infty} \lambda e^{-(\varrho+\lambda)s} \, ds \\
&= \left[ -\lambda s \frac{e^{-(\varrho+\lambda)s}}{(\varrho+\lambda)} \right]_{0}^{\infty} + \lambda \int_{0}^{\infty} \frac{e^{-(\varrho+\lambda)s}}{(\varrho+\lambda)} \, ds \\
&= \lambda \left[ -\frac{e^{-(\varrho+\lambda)s}}{(\varrho+\lambda)^2} \right]_{0}^{\infty} \\
&= \frac{\lambda}{(\varrho+\lambda)^2}
\end{aligned}
$$
We verify that:

$$ \mathrm{LE}(t; \varrho) = \left(1 + \frac{\varrho}{\lambda}\right) \mathrm{DLE}(t; \varrho) \geq \mathrm{DLE}(t; \varrho) $$

Moreover, we note that $LE(t; \varrho)$ does not depend on time $t$ because exponential survival times satisfy the property of lack of memory.

6. (a) In the case of exponential survival time, we have:

$$ \mathbb{E}[\tau] = \frac{1}{\lambda} $$

It follows that:

$$ \lambda = \frac{1}{\mathbb{E}[\tau]} = \frac{1}{70} = 0.01429 $$

For the Weibull distribution, we have:

$$ \mathbb{E}[\tau] = a\Gamma\left(1 + \frac{1}{b}\right) $$

We deduce that:

$$ a = \frac{\mathbb{E}[\tau]}{\Gamma\left(1 + \frac{1}{b}\right)} = \frac{70}{\Gamma\left(1 + \frac{1}{5}\right)} = 76.23871 $$

The survival functions $S(t)$ and $S(u | t = 50)$ are shown in Figure 99. It is evident that the exponential distribution is not suitable for modeling human lifetimes, as the probability of dying before reaching 50 years of age is approximately 50%. In contrast, the Weibull distribution provides a more realistic model for human lifetime.

(b) We have:

$$ \mathrm{LE}(0; 3\%) = \frac{1}{\varrho + \lambda} = \frac{1}{3\% + 1.43\%} = 22.58 \text{ years} $$

and:

$$ \mathrm{DLE}(0; 3\%) = \frac{\lambda}{(\varrho + \lambda)^2} = \frac{1.43\%}{(3\% + 1.43\%)^2} = 7.28 \text{ years} $$

The function $DLE(t; \varrho)$ discounts lifetime too rapidly and is not a suitable approach for calculating present values while accounting for future lifetime.

(c) Figure 100 shows the results. We observe that $LE(t; \varrho)$ is a decreasing function with respect to both time $t$ and the discount rate $\varrho$. Using a standard discount rate of 3%, a life expectancy of 70 years corresponds to a discounted life expectancy of approximately 30 years at birth and 20 years at age 40.

7. (a) We have:

$$ \begin{align*} \mathrm{QALE}(t; \varrho) &= \frac{1}{\mathbf{S}(t)} \int_t^\infty e^{-\varrho(u-t)} \mathbf{S}(u) Q(t) \, du \\ &= \left( \frac{1}{\mathbf{S}(t)} \int_t^\infty e^{-\varrho(u-t)} \mathbf{S}(u) \, du \right) \cdot Q(t) \\ &= \mathrm{LE}(t; \varrho) \cdot Q(t) \end{align*} $$

The quality-adjusted life expectancy is the product of the discounted life expectancy and the average quality of life weight.
<figure><img src="image_135.png" /><figcaption>Figure 100: Discounted lifetime expectancy $LE(t; \rho)$</figcaption></figure>

<figure><img src="image_136.png" /></figure>
(b) Since $Q(t) \le 1$, we deduce that:

$$0 \le e^{-\varrho(u-t)} \frac{\mathbf{S}(u)}{\mathbf{S}(t)} Q(u) \le e^{-\varrho(u-t)} \frac{\mathbf{S}(u)}{\mathbf{S}(t)}$$

and:

$$\int_t^\infty e^{-\varrho(u-t)} \frac{\mathbf{S}(u)}{\mathbf{S}(t)} Q(u) \, du \le \int_t^\infty e^{-\varrho(u-t)} \frac{\mathbf{S}(u)}{\mathbf{S}(t)} du$$

We conclude that $\text{QALE}(t; \varrho) \le \text{LE}(t; \varrho)$.

(c) $Q_1(t)$ is a piecewise linear function representing three phases of life. Before $t_1$, the quality of life is equal to 1. Between $t_1$ and $t_2$, the quality of life decreases linearly, reaching $1 - \kappa_1$ at $t = t_2$. The rate of decrease per year in this phase is $\kappa_1/(t_2 - t_1)$. In the third phase, between $t_2$ and $t_3$, the quality of life continues to decline linearly at a rate of $\kappa_2/(t_3 - t_2)$. Figure 101 shows the functions $\text{LE}(t; \varrho)$, $\text{QALE}_1(t; \varrho)$ calculated with $Q_1(t)$, and $\text{QALE}_2(t; \varrho)$ calculated with $Q_2(t)$. The difference between $\text{LE}(t; \varrho)$ and $\text{QALE}_1(t; \varrho)$ is small, especially when $t \le t_1$. It is more pronounced for large values of $t$ because quality of life is strongly affected when people are old. In contrast, $\text{QALE}_2(t; \varrho)$ is much lower than $\text{LE}(t; \varrho)$, because the quality of life decreases exponentially 2% per year.

<figure><img src="image_137.png" /><figcaption>Figure 101: Discounted lifetime expectancy $\text{LE}(t; \varrho)$</figcaption></figure>

8. (a) We have:

$$\text{VSL}(t) = \frac{v}{\Delta L(t)} = \frac{v}{\mathcal{R}(t) \cdot \Delta t}$$

We deduce that:

$$\text{VSLY}(t) = \frac{\text{VSL}(t) \cdot \Delta L}{\Delta \text{LE}(t)} = \frac{\text{VSL}(t)}{\text{LE}(t; \varrho)}$$
and:

$$VQALY(t) = \frac{VSL(t) \cdot \Delta L}{\Delta QALE(t)} = \frac{VSL(t)}{QALE(t; \varrho)}$$

(b) Since $\Delta L(t) = \mathcal{R}(t) = \mathcal{I}(t)$ because $\Delta t = 1$ year, the associated payoff is:

$$X(t) = VSL(t) \cdot \mathcal{R}(t) = VSL(t) \cdot \Delta L(t)$$

It follows that:

$$X(t) = VSLY(t) \cdot \Delta LE(t) = VQALY(t) \cdot \Delta QALE(t)$$

(c) If $VSL(t)$ and $\mathcal{I}(t)$ are constant, we get:

$$\begin{aligned} \mathcal{G}(t; \varrho) &= VSL \cdot \left( \frac{1}{\mathbf{S}(t)} \int_t^\infty e^{-\varrho(u-t)} \mathbf{S}(u) \, du \right) \cdot \mathcal{I}(t) \\ &= VSL \cdot \mathcal{L}E(t; \varrho) \cdot \Delta L \end{aligned} \quad (40)$$

The economic gain is the value of a statistical life multiplied by the discounted life expectancy and the expected number of lives saved.

(d) In the case of air pollution, $\mathcal{I}(t)$ is negative and the economic cost becomes:

$$C(t; \varrho) = -\mathcal{G}(t; \varrho) = VSL \cdot \mathcal{L}E(t; \varrho) \cdot \Delta L$$

where $\Delta L$ is the expected number of deaths due to air pollution. If we compare with the formula:

$$C = VSLY \cdot YLL$$

we can deduce that the years of life lost (YLL) is equal to:

$$YLL = \Delta L \cdot \mathcal{L}E(t; \varrho)$$

In this model, years of life lost is the product of the expected number of deaths and the discounted life expectancy. The traditional formula is:

$$YLL = \Delta L \cdot \Delta \tau$$

where $\Delta \tau$ is the difference between life expectancy without air pollution and life expectancy with air pollution. Therefore, we assume that:

$$\Delta \tau = \mathcal{L}E(t; \varrho) = \frac{1}{\mathbf{S}(t)} \int_t^\infty e^{-\varrho(u-t)} \mathbf{S}(u) \, du$$

This means that we take the discount rate into account when calculating years of life lost.

(e) Results are reported in Table 64. By construction, the life expectancy of an individual at age 0 is 70 years. Using a discount rate of 3%, the discounted life expectancy is 28.73 years. At age 40, the remaining life expectancy is 31.49 years, so the average lifetime of an individual alive at age 40 is 71.49 years. This corresponds to a discounted life expectancy of 19.11 years. At age 80, the remaining life expectancy is 8.74 years, so the average lifetime of an individual alive at age 80 is 88.74 years. The impact of the quality-of-life weight is shown in the second row. The estimated value of a statistical life is calculated as:

$$VSL(t) = \frac{\$1\,000}{1/10\,000} = \$10\,mn$$
<table><caption>Table 64: Calculation of the economic gain $G(t; \varrho)$</caption><thead><tr><th rowspan="2">Discount rate<br>Age</th><th colspan="3">&varrho; = 0%</th><th colspan="3">&varrho; = 3%</th><th rowspan="2">Unit</th></tr><tr><th>Birth</th><th>40 years</th><th>80 years</th><th>0</th><th>40 years</th><th>80 years</th></tr></thead><tbody><tr><td>LE (t; &varrho;)</td><td>70.00</td><td>31.49</td><td>8.74</td><td>28.73</td><td>19.11</td><td>7.19</td><td rowspan="2">years</td></tr><tr><td>QALE (t; &varrho;)</td><td>68.13</td><td>29.54</td><td>6.50</td><td>28.52</td><td>18.39</td><td>5.38</td></tr><tr><td>VSL (t)</td><td colspan="6">10 000</td><td></td></tr><tr><td>VSLY (t)</td><td>143</td><td>318</td><td>1 144</td><td>348</td><td>523</td><td>1 390</td><td>$1 000</td></tr><tr><td>VQALY (t)</td><td>147</td><td>339</td><td>1 538</td><td>351</td><td>544</td><td>1 858</td><td></td></tr><tr><td>VSL (t)</td><td>10 500</td><td>4 723</td><td>1 312</td><td>4 310</td><td>2 867</td><td>1 079</td><td></td></tr><tr><td>VSLY (t)</td><td colspan="6">150</td><td>$ 1000</td></tr><tr><td>VQALY (t)</td><td>154</td><td>160</td><td>202</td><td>151</td><td>156</td><td>200</td><td></td></tr><tr><td>G (t; &varrho;)</td><td>36 750</td><td>7 437</td><td>573</td><td>6 191</td><td>2 740</td><td>388</td><td>$</td></tr></tbody></table>

We can then compute the value of a statistical life year. Assuming an age of 40 years and a discount rate of 3%, we obtain:

$$VSLY (40) = \frac{VSL (40)}{LE (40; 3\%)} = \frac{\$10 mn}{19.11} = \$523 206$$

and:

$$VQALY (40) = \frac{VSL (40)}{QALE (40; 3\%)} = \frac{\$10 mn}{18.39} = \$543 817$$

The assumption that VSL(t) is constant is not realistic. For instance, the value per statistical life year for a newborn is $142 857, while for an 80-year-old individual it is $1 390 448. On the other hand, the assumption that VSLY(t) is constant is more realistic, even if it does not fully match reality. In fact, we can expect VSLY(40) ≥ VSLY(0) and VSLY(40) ≥ VSLY(80). Finally, we calculate the economic gain using Equation (40). For example, assuming an age of 40 years and a discount rate of 3%, the economic gain is:

$$G (40; 3\%) = \$2 866 938 \times 19.11 \times \frac{5}{100 000} = \$2 740$$

We observe that the economic gain is greater for newborns than for older individuals.