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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 $Y$?

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    $\begingroup$ $Y=e^{X_1+\cdots+X_n}$ 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. $\endgroup$ – drhab May 21 '18 at 15:28
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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 $\sigma = \sqrt{n}$

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  • $\begingroup$ (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 $\sqrt{n}$. $\endgroup$ – Mike Earnest May 21 '18 at 15:49
  • $\begingroup$ @MikeEarnest You're right, fixed it $\endgroup$ – caverac May 21 '18 at 15:51

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