# The Simple Math of Aggregation on Idiosyncratic Shocks

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.

• The mathematics of idiosyncratic shocks is not that simple if you care about measure-theoretic details. Nov 4, 2021 at 21:44
• @Alalalalaki. I honestly think this post could do with a title change. Nov 5, 2021 at 16:17

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

• 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? Nov 4, 2021 at 16:53
• The integration should be over $\ln(\varepsilon)$. I corrected it.
– tdm
Nov 4, 2021 at 17:02