when $t_0 < t \le T$. Hence, $\tilde{y}_{t+1|t}$ is approximately normally distributed with variance $\mathbb{V}_t[\tilde{y}_{t+1}] = (\hat{k}'(\bar{r})\bar{r} + b_{1t})^2\chi_t^2\sigma_\omega^2 \equiv \sigma_{y_{t+1|t}}^2$ when $t_0 < t \le T$. For the tax-standard policy, as $b_1^{std}$ and $\chi^{std}$ do not depend on $t$ (as shown below), $\sigma_{y_{t+1|t}}^2$ is a constant.

The coefficients $a_1, a_2, a_{3t}, b_{1t}, b_2$, and $Z_t$ in (34), and $\chi_t$ and $\rho_t$ in (33), are derived as follows. Using (20), (17), (61), (62) and $\mathbb{E}_t[\tilde{k}_{t+2}] \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t]$, it comes

$$ 
\begin{align*}
\mathbb{E}_t[\tilde{y}_{t+1}] - y_t &\approx q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - \mathbb{E}_t[\tilde{\mu}_{t+2}] + (1+g)\mu_{t+1} \\
&\quad - g\mu_t + p(e_{t+1} - e_t) \\
&= q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t \\
&\quad - (a_1\mu_{t+1} + a_2e_{t+1} + a_{3t} + Z_{t+1} + b_{1t}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + b_2\varepsilon_t) + (1+g)\mu_{t+1} \\
&\quad - \mu_t g + p(e_{t+1} - e_t) \\
&= q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - a_{3t} - Z_{t+1} \\
&\quad - b_{1t}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] - b_2\varepsilon_t + (1+g-a_1)\mu_{t+1} - \mu_t g + (p-a_2)e_{t+1} - pe_t \\
&= q_{t+1} - q_t - \bar{k} - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - a_{3t} - Z_{t+1} \\
&\quad - b_2\varepsilon_t + (1+g-a_1)\mu_{t+1} - \mu_t g + (p-a_2)(\theta e_t + \xi\mu_t - \varphi q_t + \hat{e}) - pe_t \\
&= q_{t+1} - q_t[1 + \varphi(p-a_2)] - \bar{k} - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t \\
&\quad + (1+g-a_1)\mu_{t+1} - \mu_t[g - \xi(p-a_2)] + [\theta(p-a_2) - p]e_t + (p-a_2)\hat{e} - a_{3t} \\
&\quad - Z_{t+1} - b_2\varepsilon_t.
\end{align*}
$$

Using $(1+r_t)z_t = (\bar{h} + \bar{r}\varepsilon_t)z_t$ in (12) and (32), we get

$$ (1-\lambda)(g\mu_t - \mu_{t+1}) + (1-g)(\mu_t - 1) + \varpi\omega_{t+1} + \xi\tau_{t+1} + (\bar{h} + \bar{r}\varepsilon_t)z_t - gz_{t+1} \approx r_t \approx \psi - \gamma^2\sigma_{y_{t+1}}^2/2 + \gamma\{\mathbb{E}_t[\tilde{y}_{t+1}] - y_t\} $$

where $\varpi \equiv (1-\lambda)/(1-\lambda+\sigma_v^2/\sigma_\omega^2)$, which gives, denoting $a_0 \equiv 1/[1-\lambda+\gamma(1+g-a_1)]$,

$$ 
\begin{align*}
\mu_{t+1}/a_0 &\approx [1 - \lambda g + \gamma g - \gamma \xi(p - a_2)]\mu_t - \gamma[\theta(p - a_2) - p]e_t + \varpi\omega_{t+1} + \xi\tau_{t+1} + (\bar{h} + \bar{r}\varepsilon_t)z_t - gz_{t+1} \\
&\quad - (\psi - \gamma^2\sigma_{y_{t+1}}^2/2) - \gamma(p - a_2)\hat{e} + \gamma a_3 + g - 1 + \gamma Z_{t+1} + \gamma(b_1 + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] \\
&\quad + \gamma b_2\varepsilon_t - \gamma q_{t+1} + \gamma q_t[1 + \varphi(p - a_2)] + \gamma\bar{k} - \gamma k_{t+1}(1 + g) + \gamma gk_t.
\end{align*}
$$

Using $k_{t+1} \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\varepsilon_t$ and $q_{t+1} \approx \bar{q} + \hat{q}'(\bar{r})\bar{r}\varepsilon_t$ where $\hat{q}'(\bar{r}) = q'(\bar{k})\hat{k}'(\bar{r})$ for $t$ and $t-1$, the last terms can be approximated by

$$ -\gamma q_{t+1} + \gamma q_t[1 + \varphi(p - a_2)] \approx -\gamma \hat{q}'(\bar{r})\bar{r}\varepsilon_t + \gamma \bar{q}\varphi(p - a_2) + \gamma \hat{q}'(\bar{r})\bar{r}\varepsilon_{t-1}[1 + \varphi(p - a_2)] $$
As $P(1) = \xi p - \theta g - 1 + \theta + g$, this equation simplifies to

$$ \begin{aligned} \left[ (1 - \theta)(1 - g)\nu_1 \right] / \left( \xi p \bar{h}^{(T-t_0-1)} \right) = & (\theta + g) \left[ (1 - \theta)(1 - g)\nu_1 \right] / \left( \xi p \bar{h}^{(T-t_0)} \right) \\ & - (\xi p - \theta g) (\phi \bar{h}^{t_0-1} + g^{\alpha(t_0-1)} \nu_0). \end{aligned} $$

Multiplying by $\xi p \bar{h}^{T-t_0+1}$ yields

$$ [(1 - \theta)(1 - g)\nu_1]\bar{h}^2 = (\theta + g)[(1 - \theta)(1 - g)\nu_1]\bar{h} - (\xi p - \theta g)(\phi \bar{h}^{t_0-1} + g^{\alpha(t_0-1)}\nu_0)\xi p \bar{h}^{T-t_0+1}, $$

and re-organizing terms,

$$ [(1 - \theta)(1 - g)\nu_1][\bar{h}^2 - (\theta + g)\bar{h}] = -(\xi p - \theta g)(\phi \bar{h}^{t_0-1} + g^{\alpha(t_0-1)}\nu_0)\xi p \bar{h}^{T-t_0+1}, $$

where, as $P(\bar{h}) = 0$, $p\xi - \theta g = \bar{h}^2 - (\theta + g)\bar{h}$. Simplifying, we get

$$ (1 - \theta)(1 - g)\nu_1 = -[\phi + (g^{\alpha}/\bar{h})^{t_0-1}\nu_0]\xi p \bar{h}^T, $$

which gives

$$ \phi = - \left( \frac{g^{\alpha}}{\bar{h}} \right)^{t_0-1} \nu_0 - \frac{(1 - \theta)(1 - g)\nu_1}{\xi p \bar{h}^T}. $$

Substituting into (58) yields

$$ m_t = \nu_0 g^{\alpha t} [1 - (g^{\alpha}/\bar{h})^{t_0-1-t}] + \nu_1 - [(1 - \theta)(1 - g)\nu_1] / (\xi p \bar{h}^{T-t}) $$

for all $0 < t < t_0$. As $m_t > 0$, $e_{t+1} > e_t$ for all $0 < t < t_0$. (30) is deduced from

$$ \begin{aligned} \sum_{\tau=1}^{t-1} m_{\tau} &= e_t - e_1 = \sum_{\tau=1}^{t-1} (\phi \bar{h}^{\tau} + g^{\alpha\tau} \nu_0 + \nu_1) = \phi \frac{\bar{h}^t - \bar{h}}{\bar{h}-1} + \nu_0 \frac{g^\alpha - g^{\alpha t}}{1-g^\alpha} + (t-1)\nu_1 \\ &= \nu_0 \left( \frac{g^\alpha - g^{\alpha t}}{1-g^\alpha} - \left( \frac{g^\alpha}{\bar{h}} \right)^{t_0-1} \frac{\bar{h}^t - \bar{h}}{\bar{h}-1} \right) + \nu_1(t-1) - \frac{(1-\theta)(1-g)\nu_1 \bar{h}^t - \bar{h}}{\xi p \bar{h}^T (\bar{h}-1)} \end{aligned} $$

for all $1 < t < t_0$, where $e_1 = \theta e_0 + \xi \mu_0 - \varphi q_0 + \hat{e}$, and from

$$ \begin{aligned} \sum_{\tau=t_0-1}^{t-1} m_{\tau} &= e_t - e_{t_0-1} = \phi \sum_{\tau=t_0-1}^{t-1} \bar{h}^{\tau} + (t-t_0)\nu_1 = \phi \frac{\bar{h}^t - \bar{h}^{t_0-1}}{\bar{h}-1} + (t-t_0+1)\nu_1 \\ &= \nu_1(t-t_0+1) - \frac{(1-\theta)(1-g)\nu_1 \bar{h}^t - \bar{h}^{t_0-1}}{\xi p \bar{h}^T (\bar{h}-1)}, \end{aligned} $$
and

$$ \gamma \bar{k} - \gamma k_{t+1}(1+g) + \gamma g k_t \approx -\gamma \hat{k}'(\bar{r}) \bar{r} \varepsilon_t (1+g) + \gamma g \hat{k}'(\bar{r}) \bar{r} \varepsilon_{t-1}. $$

Substituting and, from (33), using $\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] = \rho_t \varepsilon_t$, we arrive at

$$ \begin{aligned} \mu_{t+1} \approx & a_0[1 - \lambda g + \gamma g - \gamma \xi(p - a_2)]\mu_t - a_0\gamma[\theta(p - a_2) - p]e_t + a_0\varpi\omega_{t+1} \\ & - a_0[\psi - \gamma^2\sigma^2_{y_{t+1|t}}/2 + \gamma(p - a_2)(\hat{e} - \bar{q}\varphi) - \gamma a_{3t} + 1 - g] + a_0[\xi\tau_{t+1} + \bar{h}z_t - gz_{t+1} + \gamma Z_{t+1}] \\ & + a_0\gamma(\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r}))\bar{r}\varepsilon_{t-1} \\ & - a_0\gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1 + g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t \end{aligned} \quad (63) $$

Identifying the non-stochastic terms with those of (34) and simplifying gives

$$ a_1 = \frac{1 - \lambda g + \gamma [g + \xi(a_2 - p)]}{1 - \lambda + \gamma(1 + g - a_1)}, \quad a_2 = \frac{\gamma p(1 - \theta)}{1 - \lambda + \gamma(1 + g - a_1 - \theta)}, \quad (64) $$

$$ a_{3t} = - \frac{\psi - \gamma^2 \sigma^2_{y_{t+1|t}} / 2 - \gamma(p - a_2)(\varphi\bar{q} - \hat{e}) + 1 - g}{1 - \lambda + \gamma(g - a_1)}, $$

and $Z_t = a_0(\xi\tau_{t+1} + \bar{h}z_t - gz_{t+1} + \gamma Z_{t+1})$. Solving the recursion gives (35).

For the standard policy, using (14) and (32) to get

$$ (1 - \lambda)(g\mu_t - \mu_{t+1}) + (1 - g)(\underline{x}_t - 1) + \varpi\omega_{t+1} + \xi\tau_{t+1} = r_t \approx \psi - \gamma^2\sigma^2_{y_{t+1}}/2 + \gamma\{\mathbb{E}_t[\tilde{y}_{t+1}] - y_t\}, $$

it comes following the same steps that $a_2^{std}$ and $a_3^{std}$ are given also by (64) (but where $\sigma^2_{y_{t+1|t}}$ is a constant in the latter) while we have

$$ a_1^{std} = \frac{(1 - \lambda)g + \gamma[g + \xi(a_2^{std} - p)]}{1 - \lambda + \gamma(1 + g - a_1^{std})} $$

and $Z_t^{std} = a_0^{std}(\xi\tau_{t+1} + (1 - g)\underline{x}_t + \gamma Z_{t+1}^{std})$, with $a_0^{std} = 1/[1 - \lambda + \gamma(1 + g - a_1^{std})]$. Solving the recursion gives (37).

The parameters of the stochastic terms are derived as follows. When $\varepsilon_{t-1} = \omega_{t+1} = 0$, we have $\varepsilon_t = 0$ implying $r_t = \bar{r}$. Denoting by $\mu_{t+1|r_t=\bar{r}}$ the value of $\mu_{t+1}$ in such a case, we get from (12),

$$ \mu_{t+1} - \mu_{t+1|r_t=\bar{r}} \approx \frac{\varpi\omega_{t+1} + (1 + r_t - \bar{h})z_t + \bar{r} - r_t}{1 - \lambda} = \frac{\varpi\omega_{t+1} - (1 - z_t)\bar{r}\varepsilon_t}{1 - \lambda}, $$
for all $t_0 \le t \le T$. Using $m_T = \ln(\beta\bar{h})^{1/(p\gamma)}$ and $m_T = e_{T+1} - e_T = \hat{e} - (1-\theta)e_T$, it comes that $T$ is solution of $e_T = e_N - \ln(\beta\bar{h})^{1/(1-\theta)p\gamma}$. As

$$ m_t - m_{t-1} = \nu_0(g^{\alpha t} - g^{\alpha(t_0-1)}/\bar{h}^{t_0-1-t} - g^{\alpha(t-1)} + g^{\alpha(t_0-1)}/\bar{h}^{t_0-t}) - \frac{\nu_1(1-\theta)(1-g)}{\xi p \bar{h}^{T-t+1}}(\bar{h}-1) \\ = -\nu_0[(1/g^{\alpha}-1)g^{\alpha t} + (\bar{h}-1)g^{\alpha(t_0-1)}/\bar{h}^{t_0-t}] - \frac{\nu_1(1-\theta)(1-g)}{\xi p \bar{h}^{T-t+1}}(\bar{h}-1), $$

where $\nu_0 > 0$ if $t < t_0$ and $\nu_0 = 0$ if $t_0 \le t \le T$, $m_t - m_{t-1} < 0$ for all $t \le T$: the sequence $\{e_t\}_{2\le t \le T}$ increases at a decreasing rate. From (11), the AGTI index is deduced from these expressions using $\mu_t = (e_{t+1} - \theta e_t + \varphi q_t - \hat{e})/\xi$. From (55),

$$ c_{t+1} - c_t = \ln(\beta\bar{h})^{1/\gamma} - pm_t $$

for all $t \le T$ where

$$ m_t = \nu_0 g^{\alpha t} [1 - (g^{\alpha}/\bar{h})^{t_0-1-t}] + \nu_1 [1 - (1-\theta)(1-g)/(\xi p \bar{h}^{T-t})] \\ = \nu_0 g^{\alpha t} [1 - (g^{\alpha}/\bar{h})^{t_0-1-t}] + \ln(\beta\bar{h})^{1/\gamma} [\xi p - (1-\theta)(1-g)/\bar{h}^{T-t}]/[pP(1)] $$

for all $1 < t \le T$. We thus have

$$ c_{t+1} - c_t = -\nu_0 g^{\alpha t} [1 - (g^{\alpha}/\bar{h})^{t_0-1-t}] + \ln(\beta\bar{h})^{1/\gamma} [P(1) - (\xi p - (1-\theta)(1-g)/\bar{h}^{T-t})]/P(1) \\ = -\nu_0 g^{\alpha t} [1 - (g^{\alpha}/\bar{h})^{t_0-1-t}] - \ln(\beta\bar{h})^{1/\gamma} (1-\theta)(1-g)(1-1/\bar{h}^{T-t})/P(1) $$

which is negative and increasing, with a maximum equal to 0 at $t = T$. As $c_{t+1} - c_t - (c_t - c_{t-1}) = -p(m_t - m_{t-1}) > 0$, the consumption sequence decreases at a decreasing rate.

# I Proof of Proposition 7

For ease of notation, superscript $m$ is omitted and results are derived in the tax-subsidy policy case ($m = sub$). Variations from these results in the tax-standard policy case are specified in the text. Also, subscript $t$ is used as a shorthand for coefficients function of $z_t$, e.g. $a_{3t} \equiv a_3^{sub}(z_t)$. I first characterize the distribution of $\tilde{y}_{t+1}$ given the information available in period $t$, $t_0 < t \le T$, a random variable denoted by $\tilde{y}_{t+1|t}$. Define

$$ \hat{k}(r) = q'^{-1}((1+r-g)/(1-(\bar{h}-g)\varphi/\xi)), $$
using $(1+r_t)z_t \approx (\bar{h} + \bar{r}\varepsilon_t)z_t$, and from (63),

$$ \mu_{t+1} - \mu_{t+1|r_t=\bar{r}} \approx a_0 \{ \varpi\omega_{t+1} + \gamma(\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r}))\bar{r}\varepsilon_{t-1} \\ - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1+g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t \}. \quad (65) $$

Equalizing gives

$$ \begin{aligned} \varpi\omega_{t+1} = & \frac{\bar{r}(1-z_t) - (1-\lambda)a_0\gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1+g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t}{1 - (1-\lambda)a_0} \\ & + (1-\lambda)\gamma a_0 \frac{\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r})}{1 - (1-\lambda)a_0} \bar{r}\varepsilon_{t-1}. \end{aligned} \quad (66) $$

Substituting in (65) yields

$$ \mu_{t+1} - \mu_{t+1|r_t=\bar{r}} \approx \gamma a_0 \frac{\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r})}{1 - (1-\lambda)a_0} \bar{r}\varepsilon_{t-1} \\ + a_0 \frac{\bar{r} - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1+g) - b_2 - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t}{1 - (1-\lambda)a_0}. $$

Identifying with the stochastic terms of (34) gives, using $1 - (1 - \lambda)a_0 = \gamma a_0(1 + g - a_1)$,

$$ b_2 = \frac{\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r})}{1 + g - a_1} \bar{r} \quad (67) $$

$$ b_{1t} = \frac{\bar{r} - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1 + g - \rho_t) - b_2 - \rho b_{1t}]}{\gamma(1 + g - a_1)}. \quad (68) $$

As (66) can be written as $\varpi\omega_{t+1} = [(1 - \lambda)b_1 + \bar{r}(1 - z_t)]\varepsilon_t + b_2(1 - \lambda)\varepsilon_{t-1}$, we get from (33) that $\chi_t = \varpi / [(1 - \lambda)b_1 + \bar{r}(1 - z_t)]$ and

$$ \rho_t = \frac{-b_2(1 - \lambda)}{(1 - \lambda)b_{1t} + \bar{r}(1 - z_t)}. \quad (69) $$

Reorganizing terms to get $b_2 = -\rho_t[b_{1t} + \bar{r}(1 - z_t)/(1 - \lambda)]$ and substituting in (68), we arrive at

$$ b_{1t} = \frac{1 - \gamma[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1 + g - \rho_t) + \rho_t(1 - z_t)/(1 - \lambda)]}{\gamma(1 + g - a_1)} \bar{r}. $$
and $\hat{q}(r) \equiv q(\hat{k}(r))$. We have $\tilde{k} = \hat{k}(\bar{r})$, and $\bar{q} \equiv \hat{q}(\bar{r})$ denotes the corresponding GDP.

Given the information available in period $t$, the prevailing interest rate $r_t$ (and thus the realization $\varepsilon_t$ of $\tilde{\varepsilon}_t$) is known, and both $k_{t+1} = \hat{k}(r_t)$ and $q_{t+1} = \hat{q}(r_{t+1})$ are known. From (11), $e_{t+1}$ is also known since it depends on variables of the previous period, that are function of $\varepsilon_{t-1}$. Hence, from (8), we get

$$ \tilde{y}_{t+1|t} = \tilde{c}_{t+1|t} + pe_{t+1} = q_{t+1} - \tilde{k}_{t+2|t} + gk_{t+1} - (\tilde{\mu}_{t+2|t} - g\mu_{t+1}) + pe_{t+1}, \quad (60) $$

where $\tilde{k}_{t+2}$ and $\tilde{\mu}_{t+2}$ are unknown since they depend on the realization of $\tilde{\varepsilon}_{t+1}$: we have $\tilde{k}_{t+2} = \hat{k}(\tilde{r}_{t+1})$ and, from (33),

$$ \tilde{\mu}_{t+2|t} = a_1\mu_{t+1} + a_2e_{t+1} + a_{3t} + Z_{t+1} + b_{1t}\tilde{\varepsilon}_{t+1} + b_2\varepsilon_t. \quad (61) $$

Replacing, we get

$$ \tilde{y}_{t+1|t} = q_{t+1} - \tilde{k}_{t+2|t} + gk_{t+1} + (g - a_1)\mu_{t+1} + (p - a_2)e_{t+1} - a_{3t} - Z_{t+1} - b_{1t}\tilde{\varepsilon}_{t+1} - b_2\varepsilon_t. $$

Using the linear approximation $k_t \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\varepsilon_{t-1}$ where $\hat{k}'(\bar{r}) = 1/[(1 - \tau\varphi)q''(\bar{k})]$, we get using $\mathbb{E}_t[\tilde{k}_{t+2}] \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t]$ and (32),

$$ \begin{aligned} \tilde{y}_{t+1|t} - \mathbb{E}_t[\tilde{y}_{t+1}] &\approx -(\tilde{k}_{t+2|t} - \mathbb{E}[\tilde{k}_{t+2}|\varepsilon_t]) - b_{1t}(\tilde{\varepsilon}_{t+1} - \mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t]) \\ &= -(\hat{k}'(\bar{r})\bar{r} + b_{1t})(\tilde{\varepsilon}_{t+1} - \mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t]) \\ &= -(\hat{k}'(\bar{r})\bar{r} + b_{1t})\chi_t\tilde{\omega}_{t+2} \end{aligned} $$

when $t_0 < t \le T$. Hence, $\tilde{y}_{t+1|t}$ is approximately normally distributed with variance $\mathbb{V}_t[\tilde{y}_{t+1}] = (\hat{k}'(\bar{r})\bar{r} + b_{1t})^2\chi_t^2\sigma_\omega^2 \equiv \sigma_{y_{t+1|t}}^2$ when $t_0 < t \le T$. For the tax-standard policy, as $b_1^{std}$ and $\chi^{std}$ do not depend on $t$ (as shown below), $\sigma_{y_{t+1|t}}^2$ is a constant.

The coefficients $a_1, a_2, a_{3t}, b_{1t}, b_2$, and $Z_t$ in (33), and $\chi_t$ and $\rho_t$ in (32), are derived
From (67) and (69), we obtain that $\rho_t$ solves

$$ \rho_t = \frac{-(1-\lambda)\gamma a_0 \{\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r})\}}{1 - \gamma a_0 \{(1-\lambda)[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1+g)] + \rho_t [1 - z_t - \hat{k}'(\bar{r})(1-\lambda)]\}} \quad (70) $$

For the standard policy, the same steps give identical formulae but with $z_t$ replaced by 0. As a result, parameters $b_1^{std}$, $\rho^{std}$ and $\chi^{std}$ are constant.

The first two equations of (64) form a system involving only coefficients $a_1$ and $a_2$ that can be solved separately from the others. More precisely, defining $a_1^0 \equiv g + (1 - \lambda)/\gamma$, we can express (64) as

$$ \begin{gathered} a_1 = \frac{a_1^0 + \lambda(1-g)/\gamma + \xi(a_2-p)}{a_1^0 - a_1 + 1}, \quad a_2 = \frac{p(1-\theta)}{a_1^0 - a_1 + 1 - \theta}, \quad (71) \\ a_{3t} = -\frac{\psi - \gamma^2 \sigma_{y_{t+1|t}}^2 / 2 - \gamma(p-a_2)(\varphi\bar{q} - \hat{e}) + 1 - g}{\gamma(a_1^0 - a_1)}, \quad a_0 = \frac{1}{\gamma(a_1^0 - a_1 + 1)} \end{gathered} $$

From the expression of $a_2$, we get

$$ a_2 - p = -p \frac{a_1^0 - a_1}{a_1^0 - a_1 + 1 - \theta}, \quad (72) $$

which, plugged into the expression of $a_1^0 - a_1$, gives

$$ (a_1^0 - a_1) \left( a_1 - 1 + \frac{\xi p}{a_1^0 - a_1 + 1 - \theta} \right) = \frac{\lambda(1-g)}{\gamma} $$

that can be expressed as $Q(a_1^0 - a_1) = 0$ where

$$ Q(x) = x^3 + (2 - \theta - a_1^0)x^2 + [(1 - \theta)(1 - a_1^0) - \xi p + \lambda(1 - g)/\gamma]x + \lambda(1 - \theta)(1 - g)/\gamma $$

is a third degree polynomial. As $Z_t$ given by (35) converges if $\gamma a_0 < 1$, i.e. if $a_1 < a_1^0$, only positive roots are relevant. From (71) and (72), $a_1 > 0$ if $a_1^0 + \lambda(1 - g)/\gamma + \xi(a_2 - p) = g + (1 - \lambda g)/\gamma - p\xi(a_1^0 - a_1)/(a_1^0 - a_1 + 1 - \theta) > 0$. Since $a_1^0 > a_1$, it suffices to have $g + (1 - \lambda g)/\gamma \ge p\xi$. Under the tax-standard policy, $a_1^{std} > 0$ if $a_1^0 - (1 - g)(1 - \lambda)/\gamma + \xi(a_2 - p)g = g(1 + (1 - \lambda)/\gamma) - p\xi(a_1^0 - a_1)/(a_1^0 - a_1 + 1 - \theta) > 0$. A sufficient condition is thus $g(1 + (1 - \lambda)/\gamma) \ge p\xi$ which is more stringent than the tax-subsidy one. Both are satisfied if $g \ge p\xi$.

Equation (70) can be written as $\zeta_t(\rho) = -A_t\rho^2 + B\rho - C = 0$, where $A_t \equiv [1 - z_t - \hat{k}'(\bar{r})(1-\lambda)]\gamma a_0$, $B \equiv 1 - (1-\lambda)\gamma a_0[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1+g)]$, and $C \equiv -(1-\lambda)\gamma a_0\{\hat{q}'(\bar{r})[1+$
as follows. Using (11), (8), (60), (61) and $\mathbb{E}_t[\tilde{k}_{t+2}] \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t]$, it comes

$$
\begin{align*}
\mathbb{E}_t[\tilde{y}_{t+1}] - y_t &\approx q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - \mathbb{E}_t[\tilde{\mu}_{t+2}] + (1+g)\mu_{t+1} \\
&\quad - g\mu_t + p(e_{t+1} - e_t) \\
&= q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t \\
&\quad - (a_1\mu_{t+1} + a_2e_{t+1} + a_{3t} + Z_{t+1} + b_{1t}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + b_2\varepsilon_t) + (1+g)\mu_{t+1} \\
&\quad - \mu_t g + p(e_{t+1} - e_t) \\
&= q_{t+1} - q_t - \bar{k} - \hat{k}'(\bar{r})\bar{r}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - a_{3t} - Z_{t+1} \\
&\quad - b_{1t}\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] - b_2\varepsilon_t + (1+g-a_1)\mu_{t+1} - \mu_t g + (p-a_2)e_{t+1} - pe_t \\
&= q_{t+1} - q_t - \bar{k} - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t - a_{3t} - Z_{t+1} \\
&\quad - b_2\varepsilon_t + (1+g-a_1)\mu_{t+1} - \mu_t g + (p-a_2)(\theta e_t + \xi\mu_t - \varphi q_t + \hat{e}) - pe_t \\
&= q_{t+1} - q_t[1 + \varphi(p-a_2)] - \bar{k} - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] + k_{t+1}(1+g) - gk_t \\
&\quad + (1+g-a_1)\mu_{t+1} - \mu_t[g - \xi(p-a_2)] + [\theta(p-a_2) - p]e_t + (p-a_2)\hat{e} - a_{3t} \\
&\quad - Z_{t+1} - b_2\varepsilon_t.
\end{align*}
$$

Using $(1+r_t)z_t = (\bar{h} + \bar{r}\varepsilon_t)z_t$ in (23) and (31), we get

$$(1-\lambda)(g\mu_t - \mu_{t+1}) + (1-g)(\mu_t - 1) + \varpi\omega_{t+1} + \xi\tau_{t+1} + (\bar{h} + \bar{r}\varepsilon_t)z_t - gz_{t+1} \approx r_t \approx \psi - \gamma^2\sigma_{y+1}^2/2 + \gamma\{\mathbb{E}_t[\tilde{y}_{t+1}] - y_t\}$$

where $\varpi \equiv (1-\lambda)/(1-\lambda + \sigma_v^2/\sigma_\omega^2)$, which gives, denoting $a_0 \equiv 1/[1-\lambda+\gamma(1+g-a_1)]$,

$$\mu_{t+1}/a_0 \approx [1 - \lambda g + \gamma g - \gamma \xi(p - a_2)]\mu_t - \gamma[\theta(p - a_2) - p]e_t + \varpi\omega_{t+1} + \xi\tau_{t+1} + (\bar{h} + \bar{r}\varepsilon_t)z_t - gz_{t+1} \\ - (\psi - \gamma^2\sigma_{y+1}^2/2) - \gamma(p - a_2)\hat{e} + \gamma a_3 + g - 1 + \gamma Z_{t+1} + \gamma(b_1 + \hat{k}'(\bar{r})\bar{r})\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] \\ + \gamma b_2\varepsilon_t - \gamma q_{t+1} + \gamma q_t[1 + \varphi(p - a_2)] + \gamma \bar{k} - \gamma k_{t+1}(1 + g) + \gamma gk_t.$$

Using $k_{t+1} \approx \bar{k} + \hat{k}'(\bar{r})\bar{r}\varepsilon_t$ and $q_{t+1} \approx \bar{q} + \hat{q}'(\bar{r})\bar{r}\varepsilon_t$ where $\hat{q}'(\bar{r}) = q'(\bar{k})\hat{k}'(\bar{r})$ for $t$ and $t-1$, the last terms can be approximated by

$$-\gamma q_{t+1} + \gamma q_t[1 + \varphi(p - a_2)] \approx -\gamma \hat{q}'(\bar{r})\bar{r}\varepsilon_t + \gamma \bar{q}\varphi(p - a_2) + \gamma \hat{q}'(\bar{r})\bar{r}\varepsilon_{t-1}[1 + \varphi(p - a_2)]$$

and

$$\gamma \bar{k} - \gamma k_{t+1}(1 + g) + \gamma gk_t \approx -\gamma \hat{k}'(\bar{r})\bar{r}\varepsilon_t(1 + g) + \gamma g\hat{k}'(\bar{r})\bar{r}\varepsilon_{t-1}.$$
$\varphi(p - a_2)] + g\hat{k}'(\bar{r})\}$ are positive coefficients. As $\zeta(0) < 0$ and $\zeta'(0) > 0$, $\zeta$ admits two positive roots provided that $\Delta_t \equiv B^2 - 4A_tC > 0$, the smallest root being given by $(B - \sqrt{\Delta_t})/(2A_t)$. In the tax-standard policy case, formulae are identical but with $z_t$ replace by 0, implying that $\rho^{std}$ is a constant.

Solving the recursion of (33), it comes

$$ \tilde{\varepsilon}_t = \chi_t \tilde{\omega}_{t+1} + \rho_t \tilde{\varepsilon}_{t-1} = \sum_{k=0}^{t-t_0} \chi_{t-k} \tilde{\omega}_{t+1-k} \prod_{i=0}^{k-1} \rho_{t-i} + \varepsilon_{t_0} \prod_{k=0}^{t_0} \rho_{t-k}, $$

hence $\mathbb{E}[\tilde{\varepsilon}_t | \varepsilon_{t_0}] = \varepsilon_{t_0} \prod_{k=0}^{t_0} \rho_{t-k}$ and

$$ \mathbb{V}[\tilde{\varepsilon}_t | \varepsilon_{t_0}] = \mathbb{E}[(\tilde{\varepsilon}_t - \mathbb{E}[\tilde{\varepsilon}_t | \varepsilon_{t_0}])^2 | \varepsilon_{t_0}] = \mathbb{E} \left[ \left( \sum_{k=0}^{t-t_0} \chi_{t-k} \prod_{i=0}^{k-1} \rho_{t-i} \right)^2 \right] \mathbb{E}[\tilde{\omega}]^2 $$

using independence. In the tax-standard policy case, as $\rho_t = \rho^{std}$ and $\chi_t = \chi^{std}$, it comes $\mathbb{E}[\tilde{\varepsilon}_t | \varepsilon_{t_0}] = \varepsilon_{t_0} (\rho^{std})^{t-t_0}$ and $\mathbb{V}[\tilde{\varepsilon}_t | \varepsilon_{t_0}] = (\chi^{std} \sigma_\omega)^2 (1 - (\rho^{std})^{2(t-t_0)+1})/(1 - (\rho^{std})^2)$.
Substituting and, from (32), using $\mathbb{E}[\tilde{\varepsilon}_{t+1}|\varepsilon_t] = \rho_t \varepsilon_t$, we arrive at

$$ 
\begin{aligned}
\mu_{t+1} & \approx a_0[1 - \lambda g + \gamma g - \gamma \xi(p - a_2)]\mu_t - a_0\gamma[\theta(p - a_2) - p]e_t + a_0\varpi\omega_{t+1} \\
& - a_0[\psi - \gamma^2\sigma_{y_{t+1}|t}^2/2 + \gamma(p - a_2)(\hat{e} - \bar{q}\varphi) - \gamma a_{3t} + 1 - g] + a_0[\xi\tau_{t+1} + \bar{h}z_t - gz_{t+1} + \gamma Z_{t+1}] \\
& + a_0\gamma(\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r}))\bar{r}\varepsilon_{t-1} \\
& - a_0\gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1 + g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t
\end{aligned} 
\quad (62) $$

Identifying the non-stochastic terms with those of (33) and simplifying gives

$$ 
\begin{aligned}
a_1 &= \frac{1 - \lambda g + \gamma[g + \xi(a_2 - p)]}{1 - \lambda + \gamma(1 + g - a_1)}, \quad a_2 = \frac{\gamma p(1 - \theta)}{1 - \lambda + \gamma(1 + g - a_1 - \theta)}, \quad (63) \\
a_{3t} &= -\frac{\psi - \gamma^2\sigma_{y_{t+1}|t}^2/2 - \gamma(p - a_2)(\varphi\bar{q} - \hat{e}) + 1 - g}{1 - \lambda + \gamma(g - a_1)},
\end{aligned} $$

and $Z_t = a_0(\xi\tau_{t+1} + \bar{h}z_t - gz_{t+1} + \gamma Z_{t+1})$. Solving the recursion gives (34).

For the standard policy, using (25) and (31) to get

$$ (1 - \lambda)(g\mu_t - \mu_{t+1}) + (1 - g)(\underline{x}_t - 1) + \varpi\omega_{t+1} + \xi\tau_{t+1} = r_t \approx \psi - \gamma^2\sigma_{y_{t+1}}^2/2 + \gamma\{\mathbb{E}_t[\tilde{y}_{t+1}] - y_t\}, $$

it comes following the same steps that $a_2^{std}$ and $a_3^{std}$ are given also by (63) (but where $\sigma_{y_{t+1}|t}^2$ is a constant in the latter) while we have

$$ a_1^{std} = \frac{(1 - \lambda)g + \gamma[g + \xi(a_2^{std} - p)]}{1 - \lambda + \gamma(1 + g - a_1^{std})} $$

and $Z_t^{std} = a_0^{std}(\xi\tau_{t+1} + (1 - g)\underline{x}_t + \gamma Z_{t+1}^{std})$, with $a_0^{std} = 1/[1 - \lambda + \gamma(1 + g - a_1^{std})]$. Solving the recursion gives (36).

The parameters of the stochastic terms are derived as follows. When $\varepsilon_{t-1} = \omega_{t+1} = 0$, we have $\varepsilon_t = 0$ implying $r_t = \bar{r}$. Denoting by $\mu_{t+1|r_t=\bar{r}}$ the value of $\mu_{t+1}$ in such a case, we get from (23),

$$ \mu_{t+1} - \mu_{t+1|r_t=\bar{r}} \approx \frac{\varpi\omega_{t+1} + (1 + r_t - \bar{h})z_t + \bar{r} - r_t}{1 - \lambda} = \frac{\varpi\omega_{t+1} - (1 - z_t)\bar{r}\varepsilon_t}{1 - \lambda}, $$

using $(1 + r_t)z_t \approx (\bar{h} + \bar{r}\varepsilon_t)z_t$, and from (62),

$$ 
\begin{aligned}
\mu_{t+1} - \mu_{t+1|r_t=\bar{r}} & \approx a_0\{\varpi\omega_{t+1} + \gamma(\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r}))\bar{r}\varepsilon_{t-1} \\
& \quad - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1 + g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]\varepsilon_t\}.
\end{aligned} 
\quad (64) $$
Equalizing gives

$$ \begin{aligned} \varpi\omega_{t+1} = & \frac{\bar{r}(1-z_t) - (1-\lambda)a_0\gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1+g) - b_2 - \bar{r}z_t/\gamma - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]}{1 - (1-\lambda)a_0} \varepsilon_t \\ & + (1-\lambda)\gamma a_0 \frac{\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r})}{1 - (1-\lambda)a_0} \bar{r}\varepsilon_{t-1}. \end{aligned} \tag{65} $$

Substituting in (64) yields

$$ \begin{aligned} \mu_{t+1} - \mu_{t+1|r_t=\bar{r}} & \approx \gamma a_0 \frac{\hat{q}'(\bar{r})[1+\varphi(p-a_2)] + g\hat{k}'(\bar{r})}{1 - (1-\lambda)a_0} \bar{r}\varepsilon_{t-1} \\ & + a_0 \frac{\bar{r} - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1+g) - b_2 - (b_{1t} + \hat{k}'(\bar{r})\bar{r})\rho_t]}{1 - (1-\lambda)a_0} \varepsilon_t. \end{aligned} $$

Identifying with the stochastic terms of (33) gives, using $1 - (1 - \lambda)a_0 = \gamma a_0(1 + g - a_1)$,

$$ b_2 = \frac{\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r})}{1 + g - a_1} \bar{r} \tag{66} $$

$$ b_{1t} = \frac{\bar{r} - \gamma[\hat{q}'(\bar{r})\bar{r} + \hat{k}'(\bar{r})\bar{r}(1 + g - \rho_t) - b_2 - \rho b_{1t}]}{\gamma(1 + g - a_1)}. \tag{67} $$

As (65) can be written as $\varpi\omega_{t+1} = [(1 - \lambda)b_1 + \bar{r}(1 - z_t)]\varepsilon_t + b_2(1 - \lambda)\varepsilon_{t-1}$, we get from (32) that $\chi_t = \varpi / [(1 - \lambda)b_1 + \bar{r}(1 - z_t)]$ and

$$ \rho_t = \frac{-b_2(1 - \lambda)}{(1 - \lambda)b_{1t} + \bar{r}(1 - z_t)}. \tag{68} $$

Reorganizing terms to get $b_2 = -\rho_t[b_{1t} + \bar{r}(1 - z_t)/(1 - \lambda)]$ and substituting in (67), we arrive at

$$ b_{1t} = \frac{1 - \gamma[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1 + g - \rho_t) + \rho_t(1 - z_t)/(1 - \lambda)]}{\gamma(1 + g - a_1)} \bar{r}. $$

From (66) and (68), we obtain that $\rho_t$ solves

$$ \rho_t = \frac{-(1 - \lambda)\gamma a_0 \{\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r})\}}{1 - \gamma a_0 \{(1 - \lambda)[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1 + g)] + \rho_t[1 - z_t - \hat{k}'(\bar{r})(1 - \lambda)]\}}. \tag{69} $$

For the standard policy, the same steps give identical formulae but with $z_t$ replaced
by 0. As a result, parameters $b_1^{std}$, $\rho^{std}$ and $\chi^{std}$ are constant.

The first two equations of (63) form a system involving only coefficients $a_1$ and $a_2$ that can be solved separately from the others. More precisely, defining $a_1^0 \equiv g + (1 - \lambda)/\gamma$, we can express (63) as

$$ 
\begin{aligned}
a_1 &= \frac{a_1^0 + \lambda(1-g)/\gamma + \xi(a_2-p)}{a_1^0 - a_1 + 1}, & a_2 &= \frac{p(1-\theta)}{a_1^0 - a_1 + 1 - \theta}, \\
a_{3t} &= -\frac{\psi - \gamma^2\sigma_{y_{t+1|t}}^2/2 - \gamma(p-a_2)(\varphi\bar{q} - \hat{e}) + 1 - g}{\gamma(a_1^0 - a_1)}, & a_0 &= \frac{1}{\gamma(a_1^0 - a_1 + 1)}
\end{aligned} 
\quad (70) $$

From the expression of $a_2$, we get

$$ a_2 - p = -p \frac{a_1^0 - a_1}{a_1^0 - a_1 + 1 - \theta}, \quad (71) $$

which, plugged into the expression of $a_1^0 - a_1$, gives

$$ (a_1^0 - a_1) \left( a_1 - 1 + \frac{\xi p}{a_1^0 - a_1 + 1 - \theta} \right) = \frac{\lambda(1-g)}{\gamma} $$

that can be expressed as $Q(a_1^0 - a_1) = 0$ where

$$ Q(x) \equiv x^3 + (2 - \theta - a_1^0)x^2 + [(1 - \theta)(1 - a_1^0) - \xi p + \lambda(1 - g)/\gamma]x + \lambda(1 - \theta)(1 - g)/\gamma $$

is a third degree polynomial. As $Z_t$ given by (34) converges if $\gamma a_0 < 1$, i.e. if $a_1 < a_1^0$, only positive roots are relevant. From (70) and (71), $a_1 > 0$ if $a_1^0 + \lambda(1 - g)/\gamma + \xi(a_2 - p) = g + (1 - \lambda g)/\gamma - p\xi(a_1^0 - a_1)/(a_1^0 - a_1 + 1 - \theta) > 0$. Since $a_1^0 > a_1$, it suffices to have $g + (1 - \lambda g)/\gamma \ge p\xi$. Under the tax-standard policy, $a_1^{std} > 0$ if $a_1^0 - (1 - g)(1 - \lambda)/\gamma + \xi(a_2 - p)g = g(1 + (1 - \lambda)/\gamma) - p\xi(a_1^0 - a_1)/(a_1^0 - a_1 + 1 - \theta) > 0$. A sufficient condition is thus $g(1 + (1 - \lambda)/\gamma) \ge p\xi$ which is more stringent than the tax-subsidy one. Both are satisfied if $g \ge p\xi$.

Equation (69) can be written as $\zeta_t(\rho) = -A_t\rho^2 + B\rho - C = 0$, where $A_t \equiv [1 - z_t - \hat{k}'(\bar{r})(1 - \lambda)]\gamma a_0$, $B \equiv 1 - (1 - \lambda)\gamma a_0[\hat{q}'(\bar{r}) + \hat{k}'(\bar{r})(1 + g)]$, and $C \equiv -(1 - \lambda)\gamma a_0\{\hat{q}'(\bar{r})[1 + \varphi(p - a_2)] + g\hat{k}'(\bar{r})\}$ are positive coefficients. As $\zeta(0) < 0$ and $\zeta'(0) > 0$, $\zeta$ admits two positive roots provided that $\Delta_t \equiv B^2 - 4A_tC > 0$, the smallest root being given by $(B - \sqrt{\Delta_t})/(2A_t)$. In the tax-standard policy case, formulae are identical but with $z_t$ replace by 0, implying that $\rho^{std}$ is a constant.
Solving the recursion of (32), it comes

$$ \tilde{\epsilon}_t = \chi_t \tilde{\omega}_{t+1} + \rho_t \tilde{\epsilon}_{t-1} = \sum_{k=0}^{t-t_0} \chi_{t-k} \tilde{\omega}_{t+1-k} \prod_{i=0}^{k-1} \rho_{t-i} + \epsilon_{t_0} \prod_{k=0}^{t_0} \rho_{t-k}, $$

hence $\mathbb{E}[\tilde{\epsilon}_t | \epsilon_{t_0}] = \epsilon_{t_0} \prod_{k=0}^{t_0} \rho_{t-k}$ and

$$ \mathbb{V}[\tilde{\epsilon}_t | \epsilon_{t_0}] = \mathbb{E}[(\tilde{\epsilon}_t - \mathbb{E}[\tilde{\epsilon}_t | \epsilon_{t_0}])^2 | \epsilon_{t_0}] = \mathbb{E} \left[ \left( \sum_{k=0}^{t-t_0} \chi_{t-k} \prod_{i=0}^{k-1} \rho_{t-i} \right)^2 \right] \mathbb{E}[\tilde{\omega}]^2 $$

using independence. In the tax-standard policy case, as $\rho_t = \rho^{std}$ and $\chi_t = \chi^{std}$, it comes $\mathbb{E}[\tilde{\epsilon}_t | \epsilon_{t_0}] = \epsilon_{t_0} (\rho^{std})^{t-t_0}$ and $\mathbb{V}[\tilde{\epsilon}_t | \epsilon_{t_0}] = (\chi^{std} \sigma_\omega)^2 (1 - (\rho^{std})^{2(t-t_0)+1}) / (1 - (\rho^{std})^2)$.