Proof of the Lucas' Cost of Business Cycles

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

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.

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!

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