# Distribution of $Y=\Pi_{u=1}^n e^{X_i}$, where $(X_i)_i$ are i. i. d. with MGF $M_X(t)=e^{\frac{t^2}{2}}$

My working so far: $X_1=X_2=X_3=...=X_n$ are standard normal.

$Y=e^{X_1}e^{X_2}...e^{X_n}=e^{nX}$

I have tried to find the distribution of $Y$ using this method:$$F_Y(y)=Pr(e^{nX}\le y)$$ $$=Pr(X\le \frac{ln(y)}{n})$$ $$=\int_{-\infty}^{\frac{ln(y)}{n}}f_X(x)dx$$ $$\rightarrow f_y(y)=\frac{1}{n y}f(\frac{ln(y)}{n})$$ $$=\frac{1}{ny\sqrt{2\pi}}e^-\frac{ln^2(y)}{2n}$$

Is this correct? I feel that I'm missing something, namely I don't like that $n$ being in there but I don't know how to work this without that $n$ being there.

Additionally, how would I find the expected value and variance of $Y$?

• $Y=e^{X_1+\cdots+X_n}$ where $X_1+\cdots+X_n$ has normal distribution with mean $0$ and variance $n$. So the exponent has distribution $\sqrt{n}X$ where $X$ has standard normal distribution. – drhab May 21 '18 at 15:28

The moment generating function of a Normal distribution with mean $\mu$ and variance $\sigma^2$ is
$$M_X(t) = e^{t\mu + \sigma^2 t^2/2}$$
So comparing with your problem, you see that all variables $X_i$ are standard independent normal variables, moreover
$$Y = \prod_{i=1}^ne^{X_i}= e^{\sum_{i=1}^n X_i} = e^{Z}$$
Where $Z\sim \mathcal{N}(0,n)$. So $Y$ is distributed lognormal with parameters $\mu = 0$ and $\sigma = \sqrt{n}$
• (i) $\sum_i X_i\neq nZ$ It is distributed like $\sqrt{n} Z$. (ii) The parameters of the lognormal distribution are therefore $0$ and $\sqrt{n}$. – Mike Earnest May 21 '18 at 15:49