Showing that $Var(log(Y_i)) \approx \frac{1}{\alpha}$ if $Y_i = x_i^T\beta\epsilon_i$ with $\epsilon_i \sim N\left(1, \frac{1}{\alpha}\right)$?

Question

Suppose that $Y_i$ follows a multiplicative error model:

$$ Y_i = \mu_i \epsilon_i $$

where

$$ \mu_i = x_i^T\beta \ \ \text{and} \ \ \epsilon_i \sim N\left(1, \frac{1}{\alpha}\right) $$

I would like to show why:

$$ Var(log(Y_i)) \approx \frac{1}{\alpha} $$

How can I use the Delta method to achieve this?

Answer 1

By the delta method $\log(Y_i)-\log(\mu_i)$ is approximately distributed as $N(0,\sigma^2\cdot h)$ where $$\sigma^2=Var(\mu_i\varepsilon_i)=\frac{\mu_i^2}{\alpha},$$ and $$h=[(\log{\mu_i})']^2=\frac{1}{\mu_i^2}.$$ In particular, $$Var(log(Y_i))=Var(log(Y_i)-\log(\mu_i))\approx\sigma^2h=\frac{1}{\alpha}.$$

Answer 2

Let $g(Y) = \ln Y$, you can approximate $g(Y)$ at $E\mu \epsilon = \mu$ with first order Taylor series $$ g(Y) \approx g(\mu)+ g'(\mu)(Y-\mu), $$ rearranging the equation and taking square you get $$ (g(Y) - g(\mu))^2\approx (g'(\mu))^2(Y-\mu)^2, $$ by taking an expectation of each side you get $$ Var(g(Y))\approx (g'(\mu))^2Var(Y) = \frac{1}{\mu^2}\frac{\mu^2}{\alpha} = \frac{1}{\alpha}. $$