You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'This paper studies majorization of high tensor powers of finitely supported probability distributions. Viewing probability distributions as a resource with majorization as a means of transformation corresponds to the resource theory of pure bipartite quantum states under LOCC transformations vis-à-vis Nielsen’s Theorem [@PhysRevLett.83.436]. In [@fritz_2017 Example 8.26] a formula for the asymptotic exchange rate between any two finitely supported probability distributions was conjectured. The main result of the present paper is Theorem \[th1\], which resolves this conjecture.'
author:
- Asger Kjærulff Jensen
bibliography:
- 'all.bib'
title: Asymptotic majorizaiton of finite probability distributions
---

Introduction
============

Majorization of probability distributions is an important notion in the field of information theory. Given probability distributions $P$ and $Q$, we ask whether $P^{\otimes n}\preceq Q^{\otimes n}$ for large $n$, and we ask how large $r\in {\mathbb{R}}$ is allowed to be for $P^{\otimes n}\preceq Q^{\otimes \lfloor n r\rfloor}$ to be true for large $n$. We denote the supremum of such $r$ by $R(P,Q)$. This question is of particular interest to the author as it relates to LOCC transformation of bipartite pure quantum states. In this context $R(P,Q)$ is the optimal rate by which one can extract copies of the bipartite state with Schmidt coefficients $Q$ from copies of the bipartite state with Schmidt coefficients $P$. In Theorem \[th1\] we show that $$R(P,Q) = \min_{\alpha\in [0,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)}.$$ The main tool for obtaining this formula is a description of the growth exponents defined in \[def1\]. This description is found in Proposition \[HayashiLemma\]. The quantum information reader might note the resemblance with the well known entanglement manipulation theorem [@Nielsen:2011:QCQ:1972505 ch. 19.4], which states that the exchange rate, when one allows for non-exact LOCC transformations while demanding that fidelity goes to 1, is given by the ratio of the Shannon entropies (i.e. $\alpha=1$). In the quantum resource theory interpretation of present paper we demand that the transformation is exact.\
\
This paper started out as a section in [@2018arXiv180705130K]. As the section grew larger and we were made aware of the conjecture in [@fritz_2017 Example 8.26], it was decided to write a seperate paper proving this conjecture, while developing the necessary techniques properly.

Asymptotic exponents
====================

Given a probability distribution $P:X\to [0,1]$ with finite support $\left|\operatorname{\text{supp}}(P)\right|=d$, we let $P^\downarrow:[d]=\{1,\ldots ,d\} \to [0,1]$ be $P$ ordered non-increasingly. We may naturally extend $P^\downarrow:{\mathbb{N}}\to [0,1]$ by $P(i)=0$ for $i>d$. In this paper all probability distributions will have finite support.

Given two probability distributions $P,Q$, we say that $Q$ majorizes $P$, written $P\preceq Q$, if $$\sum_{i=1}^N P^\downarrow(i) \le \sum_{i=1}^N Q^\downarrow(i)$$ for all $N\in {\mathbb{N}}$.

Given a probability distribution $P$ with $\operatorname{\text{supp}}(P)=X$ and an $\alpha\in [0,\infty)\backslash \{1\}$, the Rényi $\alpha$-entropy is given by $$H_{\alpha}(P)=\frac{1}{1-\alpha}\log \sum_{x\in X} P(x)^\alpha.$$ For $\alpha\in \{1,\infty\}$, $H_\alpha(P)$ is defined by taking the limit $\lim_{\beta\to \alpha} H_\beta(P)$, that is $$H_1(P)=H(P)=-\sum_{x\in X} P(x)\log P(x)$$ $$H_\infty(P) = -\max_{x\in X} \log P(x).$$

Let $P,Q:X\to [0,1]$ be two probability distributions with $\operatorname{\text{supp}}(Q)\subset \operatorname{\text{supp}}(P)$. The relative entropy, also known as the Kullback-Leibler divergence, is defined as $$D(Q||P) = 
        \sum_{\operatorname{\text{supp}}(Q)} Q(i)\log \left( \frac{Q(i)}{P(i)} \right).$$ Note that the relative entropy is always non-negative.

For $n\in{\mathbb{N}}$, $P^{\otimes n}:X^n\to [0,1]$ is the $n$’th product distribution given by $P^{\otimes n}(I)=\prod_{j=1}^n P(I_j)$. We wish to study majorization of $P^{\otimes n}$ by $Q^{\otimes n}$ for large $n$. To this end, given a value $v$, we are interested in the size of the set of multiindicies $I$, such that $P^{\otimes n}(I)\ge v$ and the sum of these probabilities. In order to asymptotically compare these for different probability distributions, it is useful to let $v$ depend exponentially on $n$ and look at asymptotic growth rates.

\[def1\] For $V\in \left[\log P(d),\log P(1)\right]$ let $$m_n^P(V)=\sum_{\stackrel{I\in [d]^n}{P^{\otimes n}(I)\ge 2^{nV}}} P^{\otimes n}(I),$$ $$m_{n*}^P(V)=\sum_{\stackrel{I\in [d]^n}{P^{\otimes n}(I)\le 2^{nV}}} P^{\otimes n}(I),$$ $$s_n^P(V)=\left| \{I\in [d]^n{\lvert}P^{\otimes n}(I)\ge 2^{nV} \}\right|,$$ $$s_{n*}^P(V)=\left| \{I\in [d]^n{\lvert}P^{\otimes n}(I)\le 2^{nV} \}\right|.$$ We define asymptotic exponents of these functions as follows: $$\label{eq10}
    M^P(V)=\lim_{n\to \infty}\frac{1}{n}\log m_n^P(V),$$ $$\label{eq11}
    M_*^P(V)=\lim_{n\to \infty}\frac{1}{n}\log m_{n*}^P(V),$$ $$\label{eq12}
    S^P(V)=\lim_{n\to \infty}\frac{1}{n}\log s_n^P(V),$$ $$\label{eq13}
    S_*^P(V)=\lim_{n\to \infty}\frac{1}{n}\log s_{n*}^P(V).$$

It is not immediately clear that the limits describing $M^P,M^P_*,S^P$ and $S_*^P$ are well defined. This will follow from Proposition \[HayashiLemma\]. The letters chosen, stand for value, mass and size. $V,M^P$ and $S^P$ might be called the value, mass and size exponents, respectively. $M_*^P$ and $S_*^P$ might then be called the converse mass and size exponents. For the purpose of proving Proposition \[HayashiLemma\], we need Lemma \[concavityLemma\]. It should be said that Proposition \[HayashiLemma\] has been extracted from [@MR1960075], and should merely be viewed as a concise summary and slight extension of some of the tools presented in that paper.

\[concavityLemma\] Let $X\subset {\mathbb{R}}^n$ be a compact, convex set. Let $g:X\to {\mathbb{R}}$ be continuous and $h:X\to {\mathbb{R}}$ be continuous and strictly concave. Suppose $h$ takes its maximum value at $x_2\in X$.\
If $g$ takes its minimum value at $x_1\in X$, then $$y\mapsto \max_{x:g(x)=y}h(x)\quad y\in [g(x_1),g(x_2)]$$ is strictly monotone increasing.\
If $g$ takes its maximum value at $x_1\in X$. Then $$y\mapsto \max_{x:g(x)=y}h(x)\quad y\in [g(x_2),g(x_1)]$$ is strictly monotone decreasing.

Assume that $g$ takes its minimum value at $x_1$. Let $g(x_1)\le y'<y''\le g(x_2)$. Let $x'\in g^{-1}\left(y'\right)$ such that $\max_{x:g(x)=y'}h(x)=h(x')$. By continuity of $g$ we may find $x''$ on the line segment between $x'$ and $x_2$, such that $g(x'')=y''$. That is $$x''=\lambda x_2+(1-\lambda)x'$$ for some $\lambda\in (0,1]$. Since $h$ is strictly concave $$h(x'') \ge \lambda h(x_2)+(1-\lambda)h(x')>h(x').$$ So $$\max_{x:g(x)=y'}h(x)=h(x') < h(x'') \le \max_{x:g(x)=y''}h(x).$$ The second part of the lemma follows from the first by replacing $g$ with $-g$.

Given a probability distribution $P$ with support $[d]$, we let $$F_P(\alpha) = \log \sum P(i)^\alpha.$$ In order to make things simpler, we shall only consider $F_P$ for probability distributions that are non-uniform (such that $F_P$ is strictly convex) and ordered non-increasingly (such that we may simply write $P(1)$ instead of $\max_{x\in X} P(x)$ and $P(d)$ instead of $\min_{x\in X} P(x)$).\
\
The function $F_P$ will be central to the rest of the paper. Note that $$F_P'(\alpha) = \frac{\sum P(i)^\alpha \log P(i)}{\sum P(i)^\alpha}$$ is negative and monotone increasing $F_P':{\mathbb{R}}\to \left( \log P(d), \log P(1) \right)$. We shall define $$F_P'(\infty)=\lim_{\alpha\to \infty} F_P'(\alpha)=\log P(1)$$ and $$F_P'(-\infty)=\lim_{\alpha\to -\infty} F_P'(\alpha)=\log P(d).$$ $F_P$ is decreasing and strictly convex. Two important values to keep in mind are $$\begin{split}
    F(0)&= H_0(P) = \log d,
    \\
    F(1)&= 0.
\end{split}$$ Also note the following bijections $$F_P'
    \begin{cases}
    [-\infty,0]\quad & \longleftrightarrow \quad\left[\log P(d), \frac{\sum_i\log P(i)}{d} \right], \\
    \left[0,1\right]\quad & \longleftrightarrow \quad\left[\frac{\sum_i\log P(i)}{d}, -H(P) \right], \\
    \left[1,\infty \right]\quad & \longleftrightarrow \quad\left[-H(P), \log P(1) \right].
    \end{cases}$$ We are now ready to give explicit formulas for the exponent functions (\[eq10\]),(\[eq11\]),(\[eq12\]),(\[eq13\]).

\[HayashiLemma\] Let $P$ be a non-uniform probability distribution with $\operatorname{\text{supp}}(P)=[d]$ which is ordered non-increasingly. For $V\in \left[ -\log P(d),\log P(1)\right]$ let $\alpha_V\in [-\infty,\infty]$ be the unique solution to $F_P'(\alpha)=V$, then $$\label{equation16}
    M^P(V) =
    \left\{
    \begin{array}{rl}
    0 & \text{if } V\in \left[-\log P(d),-H(P)\right],\\
    F_P(\alpha_V)+(1-\alpha_V)F_P'(\alpha_V) & \text{if } V \in \left[-H(P),\log P(1)\right].
    \end{array} \right.$$ $$\label{equation17}
    M_*^P(V) =
    \left\{
    \begin{array}{rl}
    F_P(\alpha_V)+(1-\alpha_V)F_P'(\alpha_V) & \text{if } V\in \left[-\log P(d),-H(P)\right],\\
    0 & \text{if } V \in \left[-H(P),\log P(1)\right].
    \end{array} \right.$$ $$\label{equation18}
    S^P(V) =
    \left\{
    \begin{array}{rl}
    \log d & \text{if } V\in \left[-\log P(d),\frac{\sum \log P(i)}{d}\right],\\
    F_P(\alpha_V)-\alpha_V F_P'(\alpha_V) & \text{if } V \in \left[\frac{\sum \log P(i)}{d},\log P(1)\right].
    \end{array} \right.$$ $$\label{equation19}
    S_*^P(V) =
    \left\{
    \begin{array}{rl}
    F_P(\alpha_V)-\alpha_V F_P'(\alpha_V) & \text{if } V\in \left[-\log P(d),\frac{\sum \log P(i)}{d}\right],\\
    \log d & \text{if } V \in \left[\frac{\sum \log P(i)}{d},\log P(1)\right].
    \end{array} \right.$$ Whenever $\alpha_V=\pm\infty$ the above formulas are to be interpreted as the limit $\alpha\to \pm\infty$.

Let ${\mathcal{P}}([d])$ be the set of probability distributions on $[d]$. The map $h:Q\mapsto H(Q)$ is concave on ${\mathcal{P}}([d])$ and takes its maximum value at the uniform distribution, where $-H(Q)-D(Q||P)=\frac{\sum\log P(i)}{d}$. The map $g:Q\to -H(Q)-D(Q||P)$ has maximim value $\log P(1)$ and minimum value $\log P(d)$. According to Lemma \[concavityLemma\], $$\label{equation13}
    V\mapsto \max_{-H(Q)-D(Q||P) = V} H(Q)$$ is strictly monotone decreasing on $\left[\frac{\sum\log P(i)}{d},\log P(1)\right]$ and strictly monotone increasing on $\left[ \log P(d) ,\frac{\sum\log P(i)}{d}\right]$.\
\
Similarly, according to Lemma \[concavityLemma\], $$\label{equation14}
    V\mapsto \max_{-H(Q)-D(Q||P) = V} -D(Q||P)$$ is strictly monotone increasing on $\left[\log P(d),-H(P)\right]$ and strictly monotone decreasing on $\left[-H(P),\log P(1)\right]$. For each $V\in \left[ \log P(d),\log P(1) \right]$ we wish to find the probability distribution, $Q$, that solves the maximization problems in (\[equation13\]) and (\[equation14\]).\
\
Given $V\in \left( \log P(d), \log P(1) \right) $, let $\alpha$ be the solution to $F_P'(\alpha)=V$ and consider the distribution $P_\alpha(i)=\frac{P(i)^\alpha}{\sum_j P(j)^\alpha}$. Note that $-H(P_\alpha)-D(P_\alpha||P)=F_P'(\alpha)=V$. We prove that $P_\alpha$ solves the above optimization problems. Let $V\in (\log P(d),\log P(1))$ be given and choose $\alpha\in (-\infty,\infty)$ such that $-H(P_\alpha)-D(P_\alpha||P)=V$. Let $Q\in {\mathcal{P}}([d])$ be such that also $-H(Q)-D(Q||P)=V$. We need to show that $H(P_\alpha)\ge H(Q)$. $$\begin{split}
    &\ \frac{1}{1-\alpha} \bigg[ D(Q||P_\alpha) - H(P_\alpha)+ H(Q)\bigg]
    \\
    =&\ \frac{1}{1-\alpha}\bigg[D(Q||P_\alpha) + D(P_\alpha|| P) - D(Q||P)\bigg]
    \\
    =&\
    \frac{1}{1-\alpha}\bigg[\sum_i -Q(i)\log\frac{P(i)^\alpha}{\sum_j P(j)^\alpha}+\frac{P(i)^\alpha}{\sum_j P(j)^\alpha}\log \frac{P(i)^\alpha}{\sum_j P(j)^\alpha}-\frac{P(i)^\alpha}{\sum_j P(j)^\alpha}\log P(i)+Q(i)\log P(i) \bigg]
    \\
    =&\
    \frac{1}{1-\alpha}\bigg[(1-\alpha)\sum_i \Big( Q(i) - \frac{P(i)^\alpha}{\sum_j P(j)^\alpha}\Big)\log P(i) \bigg]
    \\ =&\ 
    \sum_i \Big( Q(i) - \frac{P(i)^\alpha}{\sum_j P(j)^\alpha}\Big)\log P(i) = H(P_\alpha) + D(P_\alpha || P)-H(Q) - D(Q || P) = 0.
    \end{split}$$ So $$H(P_\alpha) - H(Q) = D(Q||P_\alpha) \ge 0,$$ which proves that $P_\alpha$ solves the optimization problems with the values $$\begin{split}
    H(P_\alpha) &= -\sum_i \frac{P(i)^\alpha}{\sum_j P(j)^\alpha}\log \frac{P(i)^\alpha}{\sum_j P(j)^\alpha} = \log \sum P(i)^\alpha - \alpha\frac{\sum P(i)^\alpha \log P(i)}{\sum P(i)^\alpha}   
    \\
    &= F_P(\alpha) -\alpha F_P'(\alpha),
    \\\\
    -D(P_\alpha|| P)&=H(P_\alpha)-\Big(H(P_\alpha)+D(P_\alpha||P)\Big) 
    = H(P_\alpha) + F_P'(\alpha)
    \\
    &=
    F_P(\alpha) +(1-\alpha) F_P'(\alpha).
    \end{split}$$ By standard type class arguments (approximating $P_\alpha$ by type classes while using [@MR1960075 Lemma 4 and Lemma 5] and the fact that the number of type classes grows only polynomially) and monotonicity of the maps (\[equation13\]) and (\[equation14\]) we get the following:\
When $V\in \left[ \frac{\sum \log P(i)}{d},\log P(1) \right)$ $$\begin{split}
    S(V)&=\lim_{n\to \infty}\frac{1}{n}\log\Big| \{ I\in [d]^n | p_I\ge 2^{nV} \}\Big| = \max_{-H(Q)-D(Q|| P)\ge V} H(Q)
    \\
    &= \max_{-H(Q)-D(Q|| P)= V} H(Q)=H(P_\alpha)= F_P(\alpha)-\alpha F_P'(\alpha).
    \end{split}$$ When $V\in \left( \log P(d), \frac{\sum \log P(i)}{d}\right]$ $$\begin{split}
    S_*(V)&=\lim_{n\to \infty}\frac{1}{n}\log\Big| \{ I\in [d]^n | p_I\le 2^{nV} \}\Big| = \max_{-H(Q)-D(Q|| P)\le V} H(Q)
    \\
    &= \max_{-H(Q)-D(Q|| P)= V} H(Q)=H(P_\alpha)= F_P(\alpha)-\alpha F_P'(\alpha).
    \end{split}$$ When $V\in \left(-\log P(d),-H(P)\right]$ $$\begin{split}
    M_*(V)&= \lim_{n\to \infty}\frac{1}{n}\log \sum_{\stackrel{I\in [d]^n}{p_I\le 2^{nV}}} p_I= \max_{-H(Q)-D(Q|| P)\le V} -D(Q|| P)
    \\
    &= \max_{-H(Q)-D(Q|| P)= V} -D(Q|| P)=-D(P_\alpha||P)= F_P(\alpha)+(1-\alpha) F_P'(\alpha).
    \end{split}$$ When $V\in \left[-H(P),\log P(1)\right)$ $$\begin{split}
    M(V)&= \lim_{n\to \infty}\frac{1}{n}\log \sum_{\stackrel{I\in [d]^n}{p_I\ge 2^{nV}}} p_I= \max_{-H(Q)-D(Q|| P)\ge V} -D(Q|| P)
    \\
    &= \max_{-H(Q)-D(Q|| P)= V} -D(Q|| P)=-D(P_\alpha||P)= F_P(\alpha)+(1-\alpha) F_P'(\alpha).
    \end{split}$$ We may take $\alpha$ to $-\infty$ or $\infty$ and get the results at the boundary.

\[remark1\] Define $\overline{m_n^P}(V)$ on $\left[ \log P(d),\log P(1) \right]$ to be equal to $m_n^P(V)$ at the endpoints, but for $V\in \left( \log P(d),\log P(1) \right)$ we use a strict inequality and define $$\overline{m_n^P}(V)=\sum_{\stackrel{I\in [d]^n}{P^{\otimes n}(I)> 2^{nV}}} P^{\otimes n} (I).$$ Define $\overline{m_{n*}^P},\overline{s_{n}^P}$ and $\overline{s_{n*}^P}$ similarly. By continuity of $M^P,M_*^P,S^P$ and $S_*^P$ one sees that we could replace $m_n^P,m_{n*}^P,s_n^P,s_{n*}^P$ in equations (\[eq10\]),(\[eq11\]),(\[eq12\]),(\[eq13\]) with respectively $\overline{m_{n}^P},\overline{m_{n*}^P},\overline{s_{n}^P},\overline{s_{n*}^P}$, without the limit changing. Furtermore since all functions are monotone and the limit functions are monotone, continuous and bounded, the convergences are all uniform. This will be important later.

A few nice values to keep in mind for $S^P,M^P$ and $M_*^P$ are the following $$\begin{split}
        &M^P(-H(P))=0
        \\
        &M^P(\log P(1)) = \log P(1)+\log \left| \left\{i\in [d]\ |\ P(i)=P(1) \right\}\right|
        \\
        &M_*^P(-H(P))=0
        \\
        &S^P(\log P(1))=\log \left| \left\{i\in [d]\ |\  P(i)=P(1) \right\}\right|
        \\
        &S^P(-H(P)) = H(P)
        \\
        &S^P\left(\sum \frac{\log P(i)}{d} \right) = \log d=H_0(P).
    \end{split}$$

A sufficient and almost necessary condition for asymptotic majorization
=======================================================================

\[convexityLemma\] Let $F_1, F_2:[0,1]\to {\mathbb{R}}$ be differentiable with $F_2$ convex, such that $F_1(x) > F_2(x)$ for all $x\in [0,1]$. Let $\epsilon < \min_{x\in [0,1]} F_1(x)-F_2(x)$. Then for all $x,y\in [0,1]$ $$\label{eq18}
    F_1(x)-xF_1'(x) \le F_2(y)-yF_2'(y)+\epsilon \implies
    F_1(x)+(1-x)F_1'(x) > F_2(y)+(1-y)F_2'(y) +\epsilon .$$

We prove the assertion by contraposition. Fix $x,y\in [0,1]$ and assume that $$F_1(x)+(1-x)F_1'(x) \le F_2(y)+(1-y)F_2'(y)+\epsilon.$$ Consider the function $$g(t)=F_2(y)+(t-y)F_2'(y)-F_1(x)-(t-x)F_1'(x).$$ By assumption $g(1)\ge-\epsilon$ and by convexity of $F_2$ $$g(x) = F_2(y)+(x-y)F_2'(y)-F_1(x)\le F_2(x)-F_1(x)<-\epsilon.$$ Since $g$ is linear with $g(1)\ge -\epsilon$ and $g(x)<-\epsilon$, we must have $g(0)<-\epsilon$ which is equivalent to $$F_1(x)-xF_1'(x) > F_2(y)-yF_2'(y)+\epsilon.$$

\[convexityLemma2\] Let $F_1, F_2:[1,\infty)\to {\mathbb{R}}$ be continuously differentiable, strictly decreasing and convex, such that $F_1(x) > F_2(x)$ for all $x\in [1,\infty)$. Assume further that $\lim_{x\to \infty} F_1'(x) > \lim_{x\to \infty} F_2'(x)$. Then for sufficiently small $\epsilon>0$ and all $x,y\in [1,\infty)$ $$\label{eq7}
    F_1(x)-xF_1'(x) +\epsilon \ge F_2(y)-yF_2'(y)
    \implies 
    F_1(x)+(1-x)F_1'(x) > F_2(y)+(1-y)F_2'(y) +\epsilon.$$

Again we prove the assertion by contraposition. Choosing $$\epsilon< \frac{\lim_{x\to \infty} F_1'(x)-\lim_{x\to \infty} F_2'(x)}{2},$$ we have $F_1'(x)>F_2'(y)+2\epsilon$ for all sufficiently large $x$ and all $y\in [1,\infty)$, which implies (\[eq7\]). For $x=1$, the left hand side of (\[eq7\]) is never true, so (\[eq7\]) holds. We thus only need to check (\[eq7\]) for $x\in (1,R]$, where $R>1$ is some large number.\
\
Choose epsilon such that $$\epsilon + R^2\epsilon < \min_{z\in [1,R]}  \frac{F_1(z)-F_2(z)}{R}.$$ Let $x\in (1,R],y\in [1,\infty)$, and assume that $F_1(x)+(1-x)F_1'(x) \le F_2(y)+(1-y)F_2'(y)+\epsilon$. Consider the function $$g(t)=F_2(y)+(t-y)F_2'(y)-F_1(x)-(t-x)F_1'(x).$$ By assumption $g(1)\ge -\epsilon$ and by convexity of $F_2$ $$g(x) = F_2(y)+(x-y)F_2'(y)-F_1(x)\le F_2(x)-F_1(x) \le \max_{z\in [1,R]} F_2(z)-F_1(z).$$ Since $g$ is linear with $g(1)\ge -\epsilon$ and $g(x)<\max_{z\in [1,R]} F_2(z)-F_1(z)$ we have $$\begin{split}
        g(0)
        &=\frac{xg(1)-g(x)}{x-1} 
        \ge \frac{-x\epsilon-\max_{z\in [1,R]} F_2(z)-F_1(z)}{x-1}
        \ge \frac{-R\epsilon+\min_{z\in [1,R]} F_1(z)-F_2(z)}{x-1} 
        \\
        &= \frac{\min_{z\in [1,R]} \frac{F_1(z)-F_2(z)}{R}-\epsilon}{R(x-1)}
        >
        \frac{\min_{z\in [1,R]} \frac{F_1(z)-F_2(z)}{R}-\epsilon}{R^2}> \epsilon.
    \end{split}$$ $g(0)>\epsilon$ is equivalent to the left-hand-side of (\[eq7\])

We now combine Proposition \[HayashiLemma\] with (\[eq18\]) and (\[eq7\]). Firstly, by considering the formulas (\[equation17\]) and (\[equation18\]), and applying Lemma \[convexityLemma\] to $F_1=F_P$ and $F_2=F_Q$ one gets

\[lemma1\] Let $P$ and $Q$ be non-uniform probobability distributions with $$\min_{\alpha\in [0,1]} \frac{H_\alpha(P)}{H_\alpha(Q)} > 1.$$ For sufficiently small $\epsilon>0$ and all $V\in \left[ \frac{\sum \log P(i)}{d_1},-H(P)\right]$ and $W\in \left[ \frac{\sum \log Q(i)}{d_2},-H(Q)\right]$ $$S^P(V)\le S^Q(W)+\epsilon \implies M_*^P(V)\ge M_*^Q(W)+\epsilon.$$

Secondly, by considering the formulas (\[equation16\]) and (\[equation18\]), and applying Lemma \[convexityLemma2\] to $F_1=F_Q$ and $F_2=F_P$ one gets

\[lemma2\] Let $P$ and $Q$ be non-uniform probobability distributions with $$\min_{\alpha\in [1,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)} > 1.$$ For sufficiently small $\epsilon>0$ and all $V\in \left[ -H(P), \log P(1)\right]$ and $W\in \left[ -H(Q),-\log Q(1) \right]$ $$S^P(V)\le S^Q(W)+\epsilon \implies M_*^P(V)+\epsilon \le M_*^Q(W).$$

\[prop1\] Let $P=P^\downarrow:[d_1]\to [0,1]$ and $Q=Q^\downarrow:[d_2]\to [0,1]$ be non-uniform probobability distributions with $$\min_{\alpha\in [0,1]} \frac{H_\alpha(P)}{H_\alpha(Q)} > 1.$$ Let $V^*$ be such that $S^P(V^*)\in \left(H(Q),H(P)\right)$. Then for all sufficiently large $n$, and all $N$ such that $V=\frac{1}{n}\log(P^{\otimes n\downarrow}(N))\in  \left[\log P(d_1),V^*\right]$ $$\label{eq6}
        \sum_{i=1}^{N-1} P^{\otimes n\downarrow}(i)\le \sum_{i=1}^{N-1} Q^{\otimes n\downarrow}(i).$$

Let $\epsilon>0$ be small enough that Lemma \[lemma1\] applies. Assuming $\epsilon< H_0(P)-H_0(Q)$, we may let $V_1$ be such that $S_P(V_1)=H_0(Q)+\epsilon$. Now let $n$ be large enough that for both $P$ and $Q$ and all $V$ and $W$ (\[eq10\]),(\[eq11\]),(\[eq12\]),(\[eq13\]) are good approximations, and also good approximations when replaced by the alternative versions in Remark \[remark1\] (this may be done since the convergences are uniform). Note that $P^{\otimes n\downarrow}(N)=2^{nV}$ such that $N\ge \overline{s_n^P}(V)$. We split into three cases.\
\
First assume that $V\in \left[ \log P(d_1),V_1 \right]$. Then $$\frac{1}{n} \log  N \ge \frac{1}{n}\log \overline{s_n^P}(V) \ge \frac{1}{n}\log \overline{s_n^P}(V_1) \simeq S^P(V_1) = H_0(Q)+\epsilon>H_0(Q),$$ which implies $N> 2^{nH_0(Q)}$, so $$\sum_{i=1}^{N-1} Q^{\otimes n\downarrow}(i) = 1$$ and (\[eq6\]) holds trivially.\
\
Assume now that $V\in \left[ V_1,-H(P) \right]$. Let $W\in \left[ \frac{\sum \log Q(i)}{d_2},-H(Q) \right]$ be such that $S^Q(W)+\epsilon=S^P(V)$. This is possible by the definition of $V_1$ and by assuming that $\epsilon <H(P)-H(Q)$. Then $$S^P(V)\le S^Q(W)+\epsilon,$$ which by Lemma \[lemma1\] implies $$M_*^P(V) \ge M_*^Q(W)+\epsilon.$$ And $$\label{eq14}
    \frac{1}{n} \log  N \ge \frac{1}{n} \log \overline{s_n^P}(V) \simeq
    S^P(V) = S^Q(W)+\epsilon > S^Q(W) \simeq \frac{1}{n}\log {s_n^Q}(W),$$ which implies $N>{s_n^Q}(W)$. Therefore $$\label{eq15}
    \begin{split}
        &\frac{1}{n} \log\sum_{i=N}^{\infty} P^{\otimes n\downarrow}(i)
        \ge     
        \frac{1}{n} \log \overline{m_{n*}^P}(V)
        \simeq 
        M_*^P(V)
        \ge M_*^Q(W)+\epsilon 
        \\
        &>
        M_*^Q(W)
        \simeq 
        \frac{1}{n} \log m_{n*}^Q(W)
        =
        \frac{1}{n}\log \sum_{\stackrel{I\in [d_2]^n}{Q^{\otimes n}(I)\le 2^{nV}}} Q^{\otimes n}(I)
        \\
        &\ge
        \frac{1}{n} \log \sum_{i={s_n^Q}(W)}^{\infty} Q^{\otimes n\downarrow}(i)
        \ge
        \frac{1}{n} \log \sum_{i=N}^{\infty} Q^{\otimes n\downarrow}(i),
    \end{split}$$ which implies (\[eq6\]).\
\
Finally, assume that $V\in [-H(P),V^*]$. Let $W$ be such that $S^Q(W)\in \left( H(Q),S^P(V^*) \right)$. Again, as in (\[eq14\]), $N>{s_n^Q}(W)$, and since $M_*^P(V)=0>M_*^Q(W)$, we conclude as in (\[eq15\]) that $$\frac{1}{n} \log\sum_{i=N}^{\infty} P^{\otimes n\downarrow}(i)\ge \sum_{i=N}^{\infty} Q^{\otimes n\downarrow}(i).$$

\[prop2\] Let $P$ and $Q$ be non-uniform probobability distributions with $$\min_{\alpha\in [1,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)} > 1.$$ Let $V^*$ be such that $S^P(V^*)\in \left(H(Q),H(P)\right)$. Then for all sufficiently large $n$, and all $N$ such that $V=\frac{1}{n}\log(P^{\otimes n\downarrow}(N))\in  \left[ V^*, \log P(1) \right]$ $$\label{eq16}
    \sum_{i=1}^{N} P^{\otimes n\downarrow}(i)\le \sum_{i=1}^{N} Q^{\otimes n\downarrow}(i)$$

Like in the proof of Proposition \[prop1\] we let $\epsilon>0$ be small enough that Lemma \[lemma2\] applies. We split into three cases. Letting $\epsilon>0$ be sufficiently small we may let $W^*\in \left( \log P(1),\log Q(1) \right)$ be the solution to $S^Q(W^*)=S^Q(\log Q(1))+\epsilon$.\
\
Firstly we assume that $S^P(V) \le S^Q(W^*)$, then $$\frac{1}{n}\log N \le \frac{1}{n} \log s_n^P(V)\simeq S^P(V) \le S^Q(W^*)<S^Q(\log P(1))\sim \frac{1}{n}\log s_n^Q(\log P(1))$$ showing that $N\le s_n^Q(\log P(1))$ which implies that $Q^{\otimes N\downarrow}(i)\ge \log P(1)$ for all $i\in [N]$. So $$\begin{split}
        \sum_{i=1}^{N} Q^{\otimes n\downarrow}(i)\ge N\log P(1) \ge  \sum_{i=1}^{N} P^{\otimes n\downarrow}(i)
    \end{split}$$ Secondly we assume that $S^P(V)\in \left[ S^Q(W^*) , H(Q) \right]$. Let $W\in \left[ -H(Q),\log Q(1) \right]$ be such that $S^Q(W)+\epsilon = S^P(V)$, which is possible by the choice of $W^*$. By Lemma \[lemma2\] $$M^P(V)+\epsilon \le M^Q(W).$$ And $$\frac{1}{n} \log N \ge \frac{1}{n} \log \overline{s_n^P}(V)\simeq S^P(V) = S^Q(W)+\epsilon/2 > S^Q(W) \simeq \frac{1}{n} \log s_n^Q(W),$$ showing that $N> s_n^Q(W)$. $$\begin{split}
        &\frac{1}{n}\log \sum_{i=1}^N P^{\otimes n\downarrow}(i)
        \le 
        \frac{1}{n}\log m_n^P(V) \simeq M^P(V) 
        \\
        &< 
        M^P(V)+\epsilon
        \le M^Q(W)\simeq \frac{1}{n} \log m_n^Q(W)
        \\
        &
        =
        \frac{1}{n} \log \sum_{i=1}^{s_n^Q(W)} Q^{\otimes n\downarrow}(i)
        \le
        \frac{1}{n}\log \sum_{i=1}^{N} Q^{\otimes n\downarrow}(i).
    \end{split}$$ Finally assume that $S^P(V)\in\left[ H(Q),S^P(V^*) \right]$. Let $W > -H(Q)$ be such that $M^Q(W)>M^P(V^*)$. Then $$\frac{1}{n} \log N \ge \frac{1}{n}\log \overline{s_n^P}(V) \simeq S^P(V) \ge H(Q) > S^Q(W) \simeq \frac{1}{n}\log s_n^Q(W),$$ showing that $N>s_n^Q(W)$. $$\begin{split}
    &\frac{1}{n} \log\sum_{i=1}^{N} P^{\otimes n\downarrow}(i)
    \le     
    \frac{1}{n} \log {m_{n}^P}(V)
    \simeq 
    M^P(V)
    \le M^P(V^*)< M^Q(W) 
    \\
    &\simeq 
    \frac{1}{n} \log m_{n}^Q(W)
    =
    \frac{1}{n}\log \sum_{\stackrel{I\in [d_2]^n}{Q^{\otimes n}(I)\ge 2^{nW}}} Q^{\otimes n}(I)
    \\
    &=
    \frac{1}{n} \log \sum_{i=1}^{s_n^Q(W)} Q^{\otimes n\downarrow}(i)
    \le
    \frac{1}{n} \log \sum_{i=1}^{N} Q^{\otimes n\downarrow}(i).
    \end{split}$$’

So far we have assumed that all probability distributions are non-uniform. This was mainly a matter of convenience. In the following we no longer make this assumption. If $Q$ is the trivial probability distribution (i.e. $\left|\operatorname{\text{supp}}(Q)\right|=1$), then $P^{\otimes n} \preceq Q^{\otimes n}$ holds for any $P$ and $n$, so this case is rather uninsteresting.

\[prop3\] Let $P=P^\downarrow:[d_1]\to {\mathbb{R}}$ and $Q=Q^\downarrow:[d_2]\to {\mathbb{R}}$ be two probability distributions with $d_2>1$ and assume that $$\min_{\alpha\in [0,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)} > 1.$$ For sufficiently large $n$ $$P^{\otimes n} \preceq Q^{\otimes n}$$

If $d_1=1$ then ${H_\alpha(P)}=0$ for all $\alpha$, so we may assume that $d_1>1$. For small $\delta>0$, let $$P_\delta(i) =
        \begin{cases}
            P(1)+\delta & \text{if } i=1,\\
            P(i) & \text{if } 1<i<d_1,\\
            P(d_1)-\delta & \text{if } i=d_1.
        \end{cases}$$ $$Q_\delta(i) =
    \begin{cases}
        Q(1)-\delta & \text{if } i=1,\\
        Q(i) & \text{if } 1<i<d_2,\\
        Q(d_1)+\delta & \text{if } i=d_2.
    \end{cases}$$ When $\delta$ is sufficiently small $$\min_{\alpha\in [0,\infty]} \frac{H_\alpha(P_\delta)}{H_\alpha(Q_\delta)} > 1.$$ By applying Propositions \[prop1\] and \[prop2\] to $P_\delta$ and $Q_\delta$, we get for large $n$ $$P^{\otimes n}\preceq P_\delta^{\otimes n} \preceq Q_\delta^{\otimes n} \preceq Q^{\otimes n}.$$

We have now established a sufficient condition for asymptotic majorization. In fact this condition is almost necesarry. It is well known that for $\alpha\in (0,\infty)$ the $\alpha$-Rényi entropy is strictly Schur-concave (this is a consequence of the fact that $p\mapsto \frac{1}{1-\alpha}p^\alpha$ is strictly concave). In other words:

\[prop5\] Let $P$ and $Q$ be two probability distribution with $P\preceq Q$. Then either $$P^\downarrow=Q^\downarrow$$ or $$H_\alpha(P)>H_\alpha(Q)\quad \text{for all } \alpha\in (0,\infty)$$

Using the fact that $H_\alpha(P^{\otimes n})=nH_\alpha (P)$, we may sum up the contents of Propositions \[prop3\] and \[prop5\] as follows: When $P^\downarrow\not = Q^\downarrow$; $$\label{eq24}
    \begin{split}
    \forall \alpha\in \left[0,\infty\right]&:H_{\alpha}(P)>H_{\alpha}(Q)
    \\
    &\Downarrow \ref{prop3}
    \\
      \exists n\in {\mathbb{N}}&: P^{\otimes n} \preceq Q^{\otimes n}
     \\
     &\Downarrow \ref{prop5}
     \\
     \forall \alpha\in (0,\infty)&:H_{\alpha}(P)>H_{\alpha}(Q).
    \end{split}$$

It is natural to ask if we can make requirements at $0$ and $\infty$ in order to get a biimplication, that is, if we can determine $\exists n\in {\mathbb{N}}: P^{\otimes n} \preceq Q^{\otimes n}$ entirely from comparing Rényi entropies. It seems that in order to do so, we would have to be more careful with our estimations. The author cautiously conjectures that requiring a weak inequality at $\infty$ is sufficient, and that the requirement of a sharp inequality at $0$ could be replaced by a similar condition regarding the $\alpha$-Rényi entropies for negative $\alpha$.

When $P$ and $Q$ are probability distributions with finite support, we let $$R(P,Q) = \sup \left\{r\in {\mathbb{R}}_{\ge 0} \big| \text{ for large } n\ P^{\otimes n} \preceq Q^{\otimes \lfloor nr\rfloor} \right\}.$$

When $Q$ is the trivial probability distribution $R(P,Q) = \infty$.

\[th1\] Given finitely supported probability distributions $P$ and $Q$, with $Q$ non-trivial, $$R(P,Q) = \min_{\alpha\in [0,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)}.$$

Let $r< \min_{\alpha\in [0,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)}$. Then for large $n$ $$\min_{\alpha\in [0,\infty]}\frac{H_\alpha(P^{\otimes n})}{H_\alpha(Q^{\otimes \lfloor nr\rfloor})}
            =
            \min_{\alpha\in [0,\infty]}
            \frac{n}{\lfloor nr\rfloor}\frac{H_{\alpha}(P)}{H_{\alpha}(Q^)}
            >1.$$ By Proposition \[prop3\], $P^{\otimes n}\preceq Q^{\otimes \lfloor nr\rfloor} $.\
\
Let $r> \min_{\alpha\in [0,\infty]} \frac{H_\alpha(P)}{H_\alpha(Q)}$ and choose some ${\alpha_r}$ such that $r>\frac{H_{\alpha_r}(P)}{H_{\alpha_r}(Q)}$. Then for large $n$ $$\frac{H_{\alpha_r}(P^{\otimes n})}{H_{\alpha_r}(Q^{\otimes \lfloor nr\rfloor})}
    =
    \frac{n}{\lfloor nr\rfloor}\frac{H_{\alpha_r}(P)}{H_{\alpha_r}(Q^)}
    <1.$$ By Proposition \[prop5\] $P^{\otimes n}\npreceq Q^{\otimes \lfloor nr\rfloor} $.

#### Acknowledgements.

I acknowledge financial support from the European Research Council (ERC Grant Agreement no. 337603) and VILLUM FONDEN via the QMATH Centre of Excellence (Grant no. 10059).\
Furthermore I would like to thank Péter Vrana for stimulating discussions.