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

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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.

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1 Answer 1

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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!

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  • $\begingroup$ I was getting it wrong when considering the expectation of epsilon. Thank you very much! $\endgroup$ Jun 8, 2023 at 19:21
  • $\begingroup$ @DiogoFerreira No problem! $\endgroup$ Jun 8, 2023 at 19:29

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