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


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 $Y$?

  • 1
    $\begingroup$ $Y=e^{X_1+\cdots+X_n}$ 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

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 $\sigma = \sqrt{n}$

  • $\begingroup$ (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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.