Probabilities , Central limit theorem

Question

So I am working on a problem where I have to find the approximate law of $\frac{1}{\bar{X}}$ where $X_{j}$ follow the Normal distribution law. Everything is fine up until the point I have to check the region where $\mu$ is equal to zero. Can you help me. Thus far I derived the formula

$g(\overline{X}_n)\approx N(g(\mu),\sigma^2g(\mu)'^2/n)$

Thus I ll get in the general case where $\mu\ne0$, that $\frac{1}{\bar{X}}\approx N(\frac{1}{\mu},\frac{\sigma^2}{n\mu^2})$

Thank you in advance, for your contribution

Answer 1

I have to find the approximate law of $1/\overline{X}$

Applying Delta method, as you did, you get that

$$\frac{1}{\overline{X}_n}\dot{\sim}N\left(\frac{1}{\mu};\frac{\sigma^2}{n\mu^4} \right)$$

this because

$$[g'(\mu)]^2=\frac{1}{\mu^4}$$

Answer 2

If $X_1,\ldots, X_n$ are i.i.d. $N(\mu,\sigma^2)$, $$ \bar{X}_n\sim N(\mu, \sigma^2/n). $$ Thus, the exact distribution of $\bar{X}_n^{-1}$ is known. Specifically, the pdf of $\bar{X}_n^{-1}$ is given by: $$ f(z)=\frac{\sqrt{n}}{\sqrt{2\pi}\sigma z^2}\exp\!\left(-\frac{n}{2\sigma^2}\left(\frac{1}{z}-\mu\right)^2\right). $$