<p>The normal distribution with mean <span class="math-container">$\mu$</span> and standard deviation <span class="math-container">$s$</span> has density:
<span class="math-container">$$
f(\delta) = \frac{1}{\sqrt{2\pi}s} \exp(-\frac{1}{2}\left(\frac{\delta - \mu}{s}\right)^2)
$$</span>
Now, let <span class="math-container">$\delta = \ln(\varepsilon)$</span> so:
<span class="math-container">$$
\varepsilon^{1-1/\sigma} = \exp\left(\delta(1-1/\sigma)\right)
$$</span>
Then we can write:
<span class="math-container">$$
\begin{align*}
&amp;\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon),\\
&amp;= \frac{1}{\sqrt{2\pi} \sqrt{v}}\int \exp(\delta(1 - 1/\sigma)) \exp\left(-\frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2\right) d \delta,\\
&amp;= \frac{1}{\sqrt{2\pi} \sqrt{v}}\int \exp\left(\delta(1 - 1/\sigma) -\frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2\right) d \delta,\\
\end{align*}
$$</span>
Now we manipulate the term in the exponent:
<span class="math-container">$$
\begin{align*}
&amp;\delta(1 - 1/\sigma) - \frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2,\\
&amp;= \frac{1}{2v}\left[2 \delta v(1 - 1/\sigma) - \delta^2 - \delta v- \frac{v^2}{4}\right]\\
&amp;= \frac{1}{2v}\left[-\delta^2 + 2 \delta[v - v/\sigma - v/2] - \frac{v^2}{4}\right]\\
&amp;= \frac{1}{2v}\left[-\delta^2 + 2 \delta[v/2 - v/\sigma] \underbrace{-[v/2 - v/\sigma]^2 + [v/2 - v/\sigma]^2}_{=0}- \frac{v^2}{4}\right],\\
&amp;= -\frac{1}{2v}\left(\delta-[v/2 - v/\sigma]\right)^2 +\frac{1}{2v}[v/2 - v/\sigma]^2 - \frac{v^2}{8v}
\end{align*}
$$</span>
The last two terms give:
<span class="math-container">$$
\begin{align*}
&amp;\frac{v}{2}[1/2 - 1/\sigma]^2 - \frac{v}{8},\\
&amp;=\frac{v}{2}[1/4 - 1/\sigma + 1/\sigma^2 - 1/4],\\
&amp;= \frac{v}{2}\frac{1 - \sigma}{\sigma^2}
\end{align*}
$$</span>
So:
<span class="math-container">$$
\begin{align*}
&amp;\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon),\\
&amp;= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right) \frac{1}{\sqrt{2\pi}\sqrt{v}} \int \exp\left(-\frac{1}{2}\left(\frac{\delta - [v/2 - v/\sigma]}{\sqrt{v}}\right)^2\right) d \delta,\\
&amp;= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right)\int g(\delta) d \delta,\\
&amp;= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right)
\end{align*}
$$</span>
Here <span class="math-container">$g(\delta)$</span> is the distribution of a random normal variable with variance <span class="math-container">$v$</span> and mean <span class="math-container">$v/2 - v/\sigma$</span>, so it integrates to 1.</p>
<p>Then:
<span class="math-container">$$
\Phi = \left(\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon) \right)^\sigma = \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma}\right).
$$</span></p>

CC BY-SA 4.0
2021-11-04T16:53:00.630Z
Sorry I cannot get around why $\int_{I}\left(\varepsilon_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}$ is equivalent  to $\int \varepsilon^{1-1 / \sigma} f(\ln (\varepsilon)) d \varepsilon$? Whatever the measure $\mu$ would be, it will always be the probability measure of the random variable?  So even if there are heterogeneous agents, any aggregate term of their shocks would be irrelevant to the distribution?
Alalalalaki
CC BY-SA 4.0
2021-11-04T17:02:53.133Z
The integration should be over $\ln(\varepsilon)$. I corrected it.
tdm
CC BY-SA 4.0
2021-11-04T15:54:41.113Z
2021-11-04T17:02:02.207Z
2021-11-04T17:02:02.207Z
tdm
<p>I am reading some lecture notes on heterogenous agents and Negishi approach. The question I have is a very simple math problem under this background. The setting is following: (I omit the irrelevant contents)</p>
<p>&quot;&quot;&quot;
The economy is inhabited by a continuum of infinitely lived agents, indexed by <span class="math-container">$i \in I \equiv[0,1]$</span>. Denote by <span class="math-container">$\mu^{i}$</span> the measure of agents <span class="math-container">$i$</span> in the set <span class="math-container">$I$</span> and normalize the total number of agents to one, <span class="math-container">$\int_{I} d \mu^{i}=1$</span>. Agents are subject to idiosyncratic productivity shocks to skills. Let <span class="math-container">$\varepsilon_{t}^{i}$</span> be the shock of agent <span class="math-container">$i$</span>, and suppose shocks are iid, with mean 1, and defined over the set <span class="math-container">$E$</span>.</p>
<p>Using Negishi approach, we can found a fictitious RA with utility <span class="math-container">$$\frac{C_{t}^{1-\gamma}-1}{1-\gamma}+\phi \Phi \frac{\left(1-N_{t}\right)^{1-\sigma}-1}{1-\sigma}$$</span> where <span class="math-container">$\Phi=\frac{\left[\int_{I}\left(\alpha^{i}\right)^{\frac{1}{\sigma}}\left(\varepsilon_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}\right]^{\sigma}}{\left[\int_{I}\left(\alpha^{i}\right)^{\frac{1}{\gamma}} d \mu^{i}\right]^{\gamma}}$</span>.</p>
<p>In the case of no initial wealth heterogeneity <span class="math-container">$\alpha^{i}=1$</span> for all <span class="math-container">$i$</span>, <span class="math-container">$\Phi=\left[\int_{I}\left(\varepsilon_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}\right]^{\sigma}$</span>. Suppose that <span class="math-container">$\log \varepsilon \sim N\left(-v_{\varepsilon} / 2, v_{\varepsilon}\right)$</span>, then <span class="math-container">$\Phi=\exp \left(\sigma\left(\frac{\sigma-1}{\sigma}\right)\left(\frac{\sigma-1}{\sigma}-1\right) \frac{v_{\varepsilon}}{2}\right)=\exp \left(\frac{1-\sigma}{\sigma} \frac{v_{\varepsilon}}{2}\right)$</span>, which shows how the variance of the shocks affects the taste for leisure of the fictitious representative agent.
&quot;&quot;&quot;</p>
<p>I guess the assumption of the log normal distribution is to make the result simple but I don't understand how to derive the last equation.</p>

CC BY-SA 4.0
2021-11-04T21:44:35.560Z
The mathematics of idiosyncratic shocks is not that simple if you care about measure-theoretic details.
Michael Greinecker
CC BY-SA 4.0
2021-11-05T16:17:25.020Z
@Alalalalaki. I honestly think this post could do with a title change.
EB3112
<macroeconomics>
The Simple Math of Aggregation on Idiosyncratic Shocks
Alalalalaki
# The Simple Math of Aggregation on Idiosyncratic Shocks

(Asked by: username_0 on 2021-10-30)

I am reading some lecture notes on heterogenous agents and Negishi approach. The question I have is a very simple math problem under this background. The setting is following: (I omit the irrelevant contents)

"""
The economy is inhabited by a continuum of infinitely lived agents, indexed by $i \in I \equiv[0,1]$. Denote by $\mu^{i}$ the measure of agents $i$ in the set $I$ and normalize the total number of agents to one, $\int\_{I} d \mu^{i}=1$. Agents are subject to idiosyncratic productivity shocks to skills. Let $\varepsilon\_{t}^{i}$ be the shock of agent $i$, and suppose shocks are iid, with mean 1, and defined over the set $E$.

Using Negishi approach, we can found a fictitious RA with utility $$\frac{C\_{t}^{1-\gamma}-1}{1-\gamma}+\phi \Phi \frac{\left(1-N\_{t}\right)^{1-\sigma}-1}{1-\sigma}$$ where $\Phi=\frac{\left[\int\_{I}\left(\alpha^{i}\right)^{\frac{1}{\sigma}}\left(\varepsilon\_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}\right]^{\sigma}}{\left[\int\_{I}\left(\alpha^{i}\right)^{\frac{1}{\gamma}} d \mu^{i}\right]^{\gamma}}$.

In the case of no initial wealth heterogeneity $\alpha^{i}=1$ for all $i$, $\Phi=\left[\int\_{I}\left(\varepsilon\_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}\right]^{\sigma}$. Suppose that $\log \varepsilon \sim N\left(-v\_{\varepsilon} / 2, v\_{\varepsilon}\right)$, then $\Phi=\exp \left(\sigma\left(\frac{\sigma-1}{\sigma}\right)\left(\frac{\sigma-1}{\sigma}-1\right) \frac{v\_{\varepsilon}}{2}\right)=\exp \left(\frac{1-\sigma}{\sigma} \frac{v\_{\varepsilon}}{2}\right)$, which shows how the variance of the shocks affects the taste for leisure of the fictitious representative agent.
"""

I guess the assumption of the log normal distribution is to make the result simple but I don't understand how to derive the last equation.

**Tags:** macroeconomics
> **Comments:**
> **username_1**: The mathematics of idiosyncratic shocks is not that simple if you care about measure-theoretic details.  
> **username_2**: @Alalalalaki. I honestly think this post could do with a title change.  

---

## 1 answers

### Answer by username_3 on 2021-11-04. Score: 5 | Accepted answer

The normal distribution with mean $\mu$ and standard deviation $s$ has density:
$$
f(\delta) = \frac{1}{\sqrt{2\pi}s} \exp(-\frac{1}{2}\left(\frac{\delta - \mu}{s}\right)^2)
$$
Now, let $\delta = \ln(\varepsilon)$ so:
$$
\varepsilon^{1-1/\sigma} = \exp\left(\delta(1-1/\sigma)\right)
$$
Then we can write:
$$
\begin{align\*}
&\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon),\\
&= \frac{1}{\sqrt{2\pi} \sqrt{v}}\int \exp(\delta(1 - 1/\sigma)) \exp\left(-\frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2\right) d \delta,\\
&= \frac{1}{\sqrt{2\pi} \sqrt{v}}\int \exp\left(\delta(1 - 1/\sigma) -\frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2\right) d \delta,\\
\end{align\*}
$$
Now we manipulate the term in the exponent:
$$
\begin{align\*}
&\delta(1 - 1/\sigma) - \frac{1}{2}\left(\frac{\delta + v/2}{\sqrt{v}}\right)^2,\\
&= \frac{1}{2v}\left[2 \delta v(1 - 1/\sigma) - \delta^2 - \delta v- \frac{v^2}{4}\right]\\
&= \frac{1}{2v}\left[-\delta^2 + 2 \delta[v - v/\sigma - v/2] - \frac{v^2}{4}\right]\\
&= \frac{1}{2v}\left[-\delta^2 + 2 \delta[v/2 - v/\sigma] \underbrace{-[v/2 - v/\sigma]^2 + [v/2 - v/\sigma]^2}\_{=0}- \frac{v^2}{4}\right],\\
&= -\frac{1}{2v}\left(\delta-[v/2 - v/\sigma]\right)^2 +\frac{1}{2v}[v/2 - v/\sigma]^2 - \frac{v^2}{8v}
\end{align\*}
$$
The last two terms give:
$$
\begin{align\*}
&\frac{v}{2}[1/2 - 1/\sigma]^2 - \frac{v}{8},\\
&=\frac{v}{2}[1/4 - 1/\sigma + 1/\sigma^2 - 1/4],\\
&= \frac{v}{2}\frac{1 - \sigma}{\sigma^2}
\end{align\*}
$$
So:
$$
\begin{align\*}
&\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon),\\
&= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right) \frac{1}{\sqrt{2\pi}\sqrt{v}} \int \exp\left(-\frac{1}{2}\left(\frac{\delta - [v/2 - v/\sigma]}{\sqrt{v}}\right)^2\right) d \delta,\\
&= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right)\int g(\delta) d \delta,\\
&= \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma^2}\right)
\end{align\*}
$$
Here $g(\delta)$ is the distribution of a random normal variable with variance $v$ and mean $v/2 - v/\sigma$, so it integrates to 1.

Then:
$$
\Phi = \left(\int \varepsilon^{1-1/\sigma} f(\ln(\varepsilon)) d (\ln \varepsilon) \right)^\sigma = \exp\left(\frac{v}{2}\frac{1 - \sigma}{\sigma}\right).
$$

> **Comments:**
> **username_0**: Sorry I cannot get around why $\int_{I}\left(\varepsilon_{t}^{i}\right)^{1-1 / \sigma} d \mu^{i}$ is equivalent  to $\int \varepsilon^{1-1 / \sigma} f(\ln (\varepsilon)) d \varepsilon$? Whatever the measure $\mu$ would be, it will always be the probability measure of the random variable?  So even if there are heterogeneous agents, any aggregate term of their shocks would be irrelevant to the distribution?  
> **username_3**: The integration should be over $\ln(\varepsilon)$. I corrected it.  

---

