If $X_i$ is a log-normal r.v., show $X=\prod_{i=1}^{n} X_i$ is also a log-normal r.v.

Question

A random variable $X$ has log-normal distribution if $Y=ln(X)$ has a normal distribution with density function
$$f_Y(y)=\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(y=\mu)^2}{2\sigma^2}} \ \ \ \ y\in(-\infty,\infty)$$

I have calculated that the density function of $X$ is
$$f_X(x)=\frac{1}{x\sqrt{2\pi\sigma^2}}e^{-\frac{(lnx-\mu)^2}{2\sigma^2}} \ \ \ \ \ \ \ \ \ x>0$$

If the $X_i$ are independent log-normal random variables, show that the product $X=\prod_{i=1}^{n} X_i$ is also a log-normal random variable.

I thought that if I could find the density function, $f_{X_i}(x_i)$, I could deduce the answer.

I took $$X=X^n_i$$
So
$$f_{X_i}(x_i)=f_X(x_i)\Big|\frac{dx}{dx_i}\Big|=\frac{x}{\sqrt{2\pi\sigma^2}}e^{-\frac{(lnx_i-\mu)^2}{2\sigma^2}}x^{n-2}_i \ \ \ \ \ \
\ \ \ x_i>0$$
Which I looks similar to what I was expecting (a density function of a log-normal random variable).

How should I approach this problem?

Answer 1

Hint:

$\ln(\prod_{i=1}^nX_i)=\sum_{i=1}^n\ln(X_i)$

Note that $\ln(X_i)$ have normal distribution and the $\ln(X_i)$ are independent.

Then what can be said of the sum?