<p>I'll first present the assumptions Lucas made. First assume</p>
<p><span class="math-container">$$
c_t = Ae^{\mu t}e^{-(1/2)\sigma^2}\varepsilon_t,
$$</span>
where <span class="math-container">$\log \varepsilon_t \sim N(0, \sigma^2)$</span>. Under these assumptions, we have <span class="math-container">$\mathrm{E}e^{-(1/2)\sigma^2}\varepsilon_t = 1$</span> and <span class="math-container">$\mathrm{E}c_t = Ae^{\mu t}$</span>. The costs of business cycles are then defined as the <span class="math-container">$\lambda$</span> that solves (I'm sure Lucas gives enough intuition for why one defines it as such)</p>
<p><span class="math-container">$$
\mathrm{E}\left\{\sum_{t=0}^\infty \beta^t \frac{[(1 + \lambda)c_t]^{1 - \gamma}}{1 - \gamma}\right\} = \sum_{t=0}^\infty \beta^t \frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}, 
$$</span></p>
<p>which holds if for every <span class="math-container">$t$</span></p>
<p><span class="math-container">$$
\mathrm{E}\left\{\beta^t\frac{[(1 + \lambda)c_t]^{1 - \gamma}}{1 - \gamma}\right\} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$</span></p>
<p>Before we get to the cancelling step, plug in the assumed process for <span class="math-container">$c_t$</span>:</p>
<p><span class="math-container">$$
\mathrm{E}\left\{\beta^t\frac{[(1 + \lambda) Ae^{\mu t}e^{-(1/2)\sigma^2}\varepsilon_t]^{1 - \gamma}}{1 - \gamma}\right\} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$</span></p>
<p>Since the only random variable is <span class="math-container">$\varepsilon_t$</span>, you can pull everything that does not depend on <span class="math-container">$\varepsilon_t$</span> out of the expectation to get</p>
<p><span class="math-container">$$
\left\{\beta^t\frac{[(1 + \lambda) Ae^{\mu t}e^{-(1/2)\sigma^2}]^{1 - \gamma}}{1 - \gamma}\right\}\mathrm{E}\varepsilon_t^{1 - \gamma} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$</span></p>
<p>Comparing the left and right hand sides, we can cancel out a bunch of terms to get</p>
<p><span class="math-container">$$
(1 + \lambda)^{1 - \gamma}e^{-(1 - \gamma)(1/2)\sigma^2}\mathrm{E}\varepsilon_t^{1 - \gamma} = 1.
$$</span></p>
<p>Now write (recall that <span class="math-container">$\log\varepsilon_t$</span> is normally distributed, and <span class="math-container">$\mathrm{E}e^{tX} = e^{\tilde{\mu} t + \tilde{\sigma}^2 t^2/2}$</span> for any normally distributed random variable <span class="math-container">$X$</span> with mean <span class="math-container">$\tilde{\mu}$</span> and variance <span class="math-container">$\tilde{\sigma}^2$</span>)</p>
<p><span class="math-container">\begin{align}
\mathrm{E}\varepsilon_t^{1 - \gamma} &amp;= \mathrm{E}e^{(1-\gamma)\log \varepsilon_t} \\
&amp;= e^{(1- \gamma)^2 \sigma^2 /2}
\end{align}</span></p>
<p>Plugging that in and rearranging, we have</p>
<p><span class="math-container">$$
(1 + \lambda)^{1- \gamma} = e^{\gamma (1 - \gamma)(1/2)\sigma^2}.
$$</span></p>
<p>Now take logs of both sides</p>
<p><span class="math-container">$$
(1 - \gamma) \log(1 + \lambda) = \gamma (1 - \gamma) (1/2) \sigma^2,
$$</span></p>
<p>Cancelling terms and noting that <span class="math-container">$\log(1 + \lambda) \approx \lambda$</span> for small <span class="math-container">$\lambda$</span>, we get</p>
<p><span class="math-container">$$
\lambda \approx \frac{1}{2}\gamma \sigma^2.
$$</span></p>
<p>Hope this helped!</p>

CC BY-SA 4.0
2023-06-08T19:21:40.667Z
I was getting it wrong when considering the expectation of epsilon.
Thank you very much!
Diogo Ferreira
CC BY-SA 4.0
2023-06-08T19:29:56.920Z
@DiogoFerreira No problem!
Wittgenstein's Poker
CC BY-SA 4.0
2023-06-08T18:11:11.580Z
2023-06-08T19:29:35.767Z
2023-06-08T19:29:35.767Z
Wittgenstein's Poker
<p>I am trying to derive the parameter used by Lucas to measure the cost of business cycles, namely:</p>
<p><a href="https://i.sstatic.net/5mGfM.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/5mGfM.png" alt="![Text]()" /></a></p>
<p>derived in the paper &quot;Macroeconomic Priorities&quot;.
I already searched in several papers but I am unable to find the entire derivation that leads to this expression.</p>
<p>Can anyone please help me or point to me where I can find it? I now that in the paper it is said that it comes from &quot;canceling, taking logs,and collecting terms&quot;, but I am really unable to put it together.</p>
<p>Thank you in advance.</p>

<macroeconomics><utility><expected-utility><business-cycles><proof>
Proof of the Lucas' Cost of Business Cycles
Diogo Ferreira
# Proof of the Lucas' Cost of Business Cycles

(Asked by: username_0 on 2023-06-08)

I am trying to derive the parameter used by Lucas to measure the cost of business cycles, namely:

[![![Text]()](https://i.sstatic.net/5mGfM.png)](https://i.sstatic.net/5mGfM.png)

derived in the paper "Macroeconomic Priorities".
I already searched in several papers but I am unable to find the entire derivation that leads to this expression.

Can anyone please help me or point to me where I can find it? I now that in the paper it is said that it comes from "canceling, taking logs,and collecting terms", but I am really unable to put it together.

Thank you in advance.

**Tags:** macroeconomics, utility, expected-utility, business-cycles, proof


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## 1 answers

### Answer by username_1 on 2023-06-08. Score: 1 | Accepted answer

I'll first present the assumptions Lucas made. First assume

$$
c\_t = Ae^{\mu t}e^{-(1/2)\sigma^2}\varepsilon\_t,
$$
where $\log \varepsilon\_t \sim N(0, \sigma^2)$. Under these assumptions, we have $\mathrm{E}e^{-(1/2)\sigma^2}\varepsilon\_t = 1$ and $\mathrm{E}c\_t = Ae^{\mu t}$. The costs of business cycles are then defined as the $\lambda$ that solves (I'm sure Lucas gives enough intuition for why one defines it as such)

$$
\mathrm{E}\left\{\sum\_{t=0}^\infty \beta^t \frac{[(1 + \lambda)c\_t]^{1 - \gamma}}{1 - \gamma}\right\} = \sum\_{t=0}^\infty \beta^t \frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma},
$$

which holds if for every $t$

$$
\mathrm{E}\left\{\beta^t\frac{[(1 + \lambda)c\_t]^{1 - \gamma}}{1 - \gamma}\right\} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$

Before we get to the cancelling step, plug in the assumed process for $c\_t$:

$$
\mathrm{E}\left\{\beta^t\frac{[(1 + \lambda) Ae^{\mu t}e^{-(1/2)\sigma^2}\varepsilon\_t]^{1 - \gamma}}{1 - \gamma}\right\} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$

Since the only random variable is $\varepsilon\_t$, you can pull everything that does not depend on $\varepsilon\_t$ out of the expectation to get

$$
\left\{\beta^t\frac{[(1 + \lambda) Ae^{\mu t}e^{-(1/2)\sigma^2}]^{1 - \gamma}}{1 - \gamma}\right\}\mathrm{E}\varepsilon\_t^{1 - \gamma} = \beta^t\frac{(Ae^{\mu t})^{1 - \gamma}}{1 - \gamma}.
$$

Comparing the left and right hand sides, we can cancel out a bunch of terms to get

$$
(1 + \lambda)^{1 - \gamma}e^{-(1 - \gamma)(1/2)\sigma^2}\mathrm{E}\varepsilon\_t^{1 - \gamma} = 1.
$$

Now write (recall that $\log\varepsilon\_t$ is normally distributed, and $\mathrm{E}e^{tX} = e^{\tilde{\mu} t + \tilde{\sigma}^2 t^2/2}$ for any normally distributed random variable $X$ with mean $\tilde{\mu}$ and variance $\tilde{\sigma}^2$)

\begin{align}
\mathrm{E}\varepsilon\_t^{1 - \gamma} &= \mathrm{E}e^{(1-\gamma)\log \varepsilon\_t} \\
&= e^{(1- \gamma)^2 \sigma^2 /2}
\end{align}

Plugging that in and rearranging, we have

$$
(1 + \lambda)^{1- \gamma} = e^{\gamma (1 - \gamma)(1/2)\sigma^2}.
$$

Now take logs of both sides

$$
(1 - \gamma) \log(1 + \lambda) = \gamma (1 - \gamma) (1/2) \sigma^2,
$$

Cancelling terms and noting that $\log(1 + \lambda) \approx \lambda$ for small $\lambda$, we get

$$
\lambda \approx \frac{1}{2}\gamma \sigma^2.
$$

Hope this helped!

> **Comments:**
> **username_0**: I was getting it wrong when considering the expectation of epsilon. Thank you very much!  
> **username_1**: @DiogoFerreira No problem!  

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