# Barro's (2009) rare disaster model in the AER: How to derive equation (10)?

In Barro (2009) Rare disasters, asset prices and welfare costs Barro develops a Lucas tree model with Epstein-Zin preferences.

My question concerns the paper's equation (10). In this equation Barro states that under the optimal solution utility $U_t$ is proportional to consumption $C_t$ rased to the power of $1-\gamma$, where $\gamma$ is the coefficient of relative risk aversion, i.e.

$U_t=\Phi C_t^{1-\gamma}$

While I understand the logic of this result, I do not understand how he derives the constant $\Phi$, which is shown in footnote 7 of the mentioned paper:

Alberto Giovannini and Philippe Weil (1989, appendix) show that, with the utility function in equation (9), attained utility, $U_t$ , is proportional to wealth raised to the power $1-\gamma$. The form in equation (10) follows because $C_t$ is optimally chosen as a constant ratio to wealth in the i.i.d. case. The formula for $\Phi$ is, if $\gamma \neq 1$ $\theta \neq 1$, $$\Phi = (\frac{1}{1-\gamma})\{\rho+(\theta-1)g^* - (1/2)\gamma(\theta -1)\sigma^2 - (\frac{\theta-1}{\gamma-1})p[E(1-b)^{1-\gamma} - 1 - (\gamma - 1)Eb] \}^{(\gamma-1)/(1-\theta)}$$

Barro quotes the 1989 NBER paper by Giovannini and Weil. In this paper I can derive the constant. However, it looks completely different than Barro's version, because I end up with an expression that includes $E[R_t^{1-\gamma}]$, where $R_t$ is the return on equity. I believe Barro has replaced $E[R_t^{1-\gamma}]$ with the equilibrium solution of $R_t$. However, his expression does not include any logs or exp expressions.

I would be grateful for a solution or any hints to the solution.

• This looks great! Thanks for your effort. It will take me a couple of days to review part 2 and 3 of your answer, but it looks very intuitive. – drcms02 Oct 24 '17 at 22:03

I think Barro means in the footnote that Giovanni and Weil find the same equation, $U_t=\Phi C^{1-\gamma}$, but using the optimal path of $C_t$. In Barro's paper, the approach is different given that the dynamics of $C_t$ is exogenous: $C_t=Y_t$ by assumption.

Barro uses the limit case when the length of a period gets close to 0. Maybe what may bother the reader is that the model is defined as discrete.

## Rewrite the model

First, we can rewrite the model with a length of period $\delta$ and then use $\delta\to 0$. The GDP dynamics write $$\log(Y_{t+\delta})=\log(Y_t)+g\delta+u_{t+\delta}+v_ {t+\delta}$$ with $u_{t+\delta}\sim \mathcal{N}(0,\delta\sigma^2)$, and $v_{t+\delta}=0$ with probability $1-p\delta$ and $\log(1-b)$ with probability $p\delta$. The utility satisfies $$U_t=\frac{1}{1-\gamma}\left\lbrace C_t^{1-\theta}+\frac{1}{1+\rho\delta}\left[(1-\gamma)E_tU_{t+\delta}\right]^\frac{1-\theta}{1-\gamma}\right\rbrace^\frac{1-\gamma}{1-\theta}.$$

## 1) Find $\Phi$ as a function of $E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]$

From now suppose there is a $\Phi$ such that $U_t=\Phi C^{1-\gamma}$ (note that $\Phi$ depends on $\delta$ a priori). Define $H(U)=[(1-\gamma)U]^\frac{1-\theta}{1-\gamma}$, the utility satisfies \begin{align} H(U_t)= C_t^{1-\theta}+\frac{1}{1+\rho\delta}H(E_tU_{t+\delta}). \end{align} We substitute $U_t$: \begin{align} H(\Phi)C_t^{1-\theta}= C_t^{1-\theta}+\frac{1}{1+\rho\delta}H(\Phi)\left(E_t\left[C_{t+\delta}^{1-\gamma}\right]\right)^\frac{1-\theta}{1-\gamma}. \end{align} Hence, we obtain for $C_t\neq 0$, \begin{align} \frac{1}{H(\Phi)}= 1-\frac{1}{1+\rho\delta}\left(E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]\right)^\frac{1-\theta}{1-\gamma}. \end{align}

## 2) Find $E_t\left[\left(\frac{C_{t+\delta}}{C_t}\right)^{1-\gamma}\right]$ fromp the GDP dynamics

The trick is to find the expectation in the right-hand side from the GDP dynamics. \begin{align} \left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).\exp\left((1-\gamma)u_{t+\delta}\right).\exp\left((1-\gamma)v_ {t+\delta}\right). \end{align} Taking the expectation and using the independence between $u_{t+1}$ and $v_{t+1}$, it follows \begin{align} E_t\left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).E_t\exp\left((1-\gamma)u_{t+\delta}\right).E_t\exp\left((1-\gamma)v_ {t+\delta}\right). \end{align} The expectation of $\exp(X)$ where $X$ follows $\mathcal{N}(0,\sigma^2)$ is $\exp(\sigma^2/2)$. $\exp\left((1-\gamma)v_ {t+\delta}\right)$ is a random variable equal to $1$ with probability $1-p\delta$ and $(1-b)^{1-\gamma}$ with probability $p\delta$. We substitute the expectation operator: \begin{align} E_t\left(\frac{Y_{t+\delta}}{Y_t}\right)^{1-\gamma}= \exp\left((1-\gamma)g\delta\right).\exp\left(\frac{(1-\gamma)^2\sigma^2\delta}{2}\right).\left(1-p\delta+pE[(1-b)^{1-\gamma}]\delta\right). \end{align} Finally, we use $C_t=Y_t$ to compute an equation for $\Phi$: \begin{align} \frac{1}{H(\Phi)}&= 1-\frac{1}{1+\rho\delta}\left\lbrace\exp\left((1-\theta)g\delta\right).\exp\left(\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\right)\right.\\ &\left. .\left(1-p\delta+pE[(1-b)^{1-\gamma}]\delta\right)^\frac{1-\theta}{1-\gamma}\right\rbrace. \end{align}

## 3) Take the approximation $\delta\to 0$

The last step consists in taking a first-order approximation (I abusively keep the equal symbol): \begin{align} \frac{1}{H(\Phi)}&= 1-(1-\rho\delta). \left(1+(1-\theta)g\delta\right).\left(1+\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\right)\\ & .\left(1-\frac{1-\theta}{1-\gamma}p\delta+\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta\right). \end{align} Pursuing the first-order apprixmation (all the $\delta^i$ with $i>1$ can be neglected), we have \begin{align} \frac{1}{H(\Phi)}&= \rho\delta -(1-\theta)g\delta-\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\\ & +\frac{1-\theta}{1-\gamma}p\delta-\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta. \end{align} Substitute $g$ using $g^*=g+\frac{\sigma^2}{2}-pEb$, \begin{align} \frac{1}{H(\Phi)}&= \rho\delta -(1-\theta)g^*\delta+(1-\theta)\frac{\sigma^2}{2}\delta -(1-\theta)pEb\delta -\frac{(1-\gamma)(1-\theta)\sigma^2\delta}{2}\\ & +\frac{1-\theta}{1-\gamma}p\delta-\frac{1-\theta}{1-\gamma}pE[(1-b)^{1-\gamma}]\delta. \end{align} We take $\delta=1$ and invert function $H$ to find the solution in the footnote 7 of the paper. The right-hand side of this equation "simplifies" to the within braces in the formula.