<p>This might only be a partial answer.</p>
<p>Let <span class="math-container">$z \sim N(\mu, \sigma^2)$</span>. Then it is known that <a href="https://math.stackexchange.com/questions/176196/calculate-the-expected-value-of-y-ex-where-x-sim-n-mu-sigma2">see here</a>:
<span class="math-container">$$
\mathbb{E}(e^z) = e^{\mu + \frac{1}{2}\sigma^2}
$$</span></p>
<p>Assume that <span class="math-container">$y$</span> is lognormal distributed, so <span class="math-container">$\ln(y) \sim N(\mu, \sigma^2)$</span> then,
<span class="math-container">$$
(1-\gamma) \ln(y) \sim N((1-\gamma) \mu, (1-\gamma)^2 \sigma^2).
$$</span></p>
<p>Then,
<span class="math-container">$$
\begin{align*}
\mathbb{E}(u(y)) &amp;= \mathbb{E}\left(\frac{y^{1 - \gamma}}{1 - \gamma}\right),\\
&amp;=\mathbb{E}\left(\frac{e^{(1 - \gamma)\ln(y)}}{1 - \gamma}\right),\\
&amp;= \frac{1}{1 - \gamma} \mathbb{E}\left(e^z\right) \text{ where } z = (1-\gamma)\ln(y) \sim N((1-\gamma) \mu, (1-\gamma)^2 \sigma^2),\\
&amp;= \frac{1}{1 - \gamma} e^{(1 - \gamma)\mu + \frac{1}{2}(1-\gamma)^2 \sigma^2},\\
&amp;= \frac{\exp\left((1 - \gamma)\left[\mu - (\gamma - 1)\sigma^2/2\right]\right)}{1 - \gamma},
\end{align*}
$$</span>
This is as far as I got.</p>
<p>Note that,
<span class="math-container">$
\mu - (\gamma - 1) \sigma^2/2 = E(\ln(y)) - 0.5 (\gamma - 1) V(\ln(y))$</span>.</p>

CC BY-SA 4.0
2024-11-25T11:09:37.033Z
2024-11-25T11:09:37.033Z
tdm
<p>I am reading a paper that assumes that individuals just consume their income and that welfare is <span class="math-container">$y^{1-\gamma} / 1-\gamma$</span>, and they focus on the case where <span class="math-container">$\gamma&gt;1$</span>. If income is log-normally distributed, the we can write
<span class="math-container">$$
\frac{1}{1-\gamma} e^{\ln \left(y^{1-\gamma}\right)}=\frac{1}{1-\gamma} e^{-(\gamma-1) \ln (y)}
$$</span>
And they say that &quot;the standard constant absolute risk aversion calculations follow to derive a linear mean-variance frontier&quot;, so they conclude that then expected welfare is equivalent to <span class="math-container">$E(\ln y)-0.5(\gamma-1) V(\ln y)$</span>. I'm not sure how the expression for expected welfare is retrieved. Thank you very much.</p>

<macroeconomics><welfare>
Expected welfare log-normal
nini
# Expected welfare log-normal

(Asked by: username_0 on 2024-08-31)

I am reading a paper that assumes that individuals just consume their income and that welfare is $y^{1-\gamma} / 1-\gamma$, and they focus on the case where $\gamma>1$. If income is log-normally distributed, the we can write
$$
\frac{1}{1-\gamma} e^{\ln \left(y^{1-\gamma}\right)}=\frac{1}{1-\gamma} e^{-(\gamma-1) \ln (y)}
$$
And they say that "the standard constant absolute risk aversion calculations follow to derive a linear mean-variance frontier", so they conclude that then expected welfare is equivalent to $E(\ln y)-0.5(\gamma-1) V(\ln y)$. I'm not sure how the expression for expected welfare is retrieved. Thank you very much.

**Tags:** macroeconomics, welfare

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## 1 answers

### Answer by username_1 on 2024-11-25. Score: 1

This might only be a partial answer.

Let $z \sim N(\mu, \sigma^2)$. Then it is known that [see here](https://math.stackexchange.com/questions/176196/calculate-the-expected-value-of-y-ex-where-x-sim-n-mu-sigma2):
$$
\mathbb{E}(e^z) = e^{\mu + \frac{1}{2}\sigma^2}
$$

Assume that $y$ is lognormal distributed, so $\ln(y) \sim N(\mu, \sigma^2)$ then,
$$
(1-\gamma) \ln(y) \sim N((1-\gamma) \mu, (1-\gamma)^2 \sigma^2).
$$

Then,
$$
\begin{align\*}
\mathbb{E}(u(y)) &= \mathbb{E}\left(\frac{y^{1 - \gamma}}{1 - \gamma}\right),\\
&=\mathbb{E}\left(\frac{e^{(1 - \gamma)\ln(y)}}{1 - \gamma}\right),\\
&= \frac{1}{1 - \gamma} \mathbb{E}\left(e^z\right) \text{ where } z = (1-\gamma)\ln(y) \sim N((1-\gamma) \mu, (1-\gamma)^2 \sigma^2),\\
&= \frac{1}{1 - \gamma} e^{(1 - \gamma)\mu + \frac{1}{2}(1-\gamma)^2 \sigma^2},\\
&= \frac{\exp\left((1 - \gamma)\left[\mu - (\gamma - 1)\sigma^2/2\right]\right)}{1 - \gamma},
\end{align\*}
$$
This is as far as I got.

Note that,
$
\mu - (\gamma - 1) \sigma^2/2 = E(\ln(y)) - 0.5 (\gamma - 1) V(\ln(y))$.


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