Product of N independent uniform random variables

Question

We have N independent random variables of $X_1 , X_2 , ... X_N$ with uniform distribution $U(0,1)$. the distribution of $Y$ as $$ Y = \frac{1}{X_1X_2X_3 ... X_N} $$ is desired.
To solve this problem, I write: $$ T = \ln(Y) = - \sum_{i=1}^N \ln(X_i) = \sum_{i=1}^N Z_i $$ where $ Z_i =-\ln(X_i) $. Then the distribution of $Z_i$ random variables could be written as $$ f_{Z_i}(Z_i) = \frac{f_{X_i}(e^{-Z_i})}{|e^{Z_i}|} = e^{-Z_i}f_{X_i}(e^{-Z_i}) $$ But I do not know how to solve it any further. Would be helpful to utilize its Characteristic function in order to rewrite the sum of natural-log sum of random variables? Or Should I solve it in a different way?
Thanks for your help in advance.

Answer 1

As you noted,

$$logY=\Sigma_i (-logX_i)$$

$$W_i=-logX_i\sim Exp(1)$$

Thus $log(Y)=\Sigma_i W_i\sim Gamma(n;1)$

Now, knowing the distribution of $logY$ it is easy to find the requested density

$$f_Y(y)=\frac{log^{n-1}y}{\Gamma(n)y^2}\cdot \mathbb{1}_{(1;+\infty)}(y)$$

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P.S.: I used the notation $log$ but Natural logarithm is meant

Answer 2

I've found another answer for this question, for whom get here by search.
If we have $$ \phi_T(s) = E\{e^{sT}\} = E\{e^{s\sum_{i=0}^N Z_i}\} = \prod_{i=0}^NE\{e^{sZ_i}\} $$ $$ = \prod_{i=0}^N\phi_{Z_i}(s) = \left( \phi_{Z_i}(s)\right)^N = (\frac{1}{1-s})^N $$ in Laplace transform, there is this duality $$ f_T(t) \leftrightarrow L^{-1}\{\phi_T(-s)\} \rightarrow f_T(t) = \frac{1}{(N-1)!}t^{N-1}e^{-t}u(t) $$ So finally we could write this $$ T = \ln(Y) \rightarrow Y = e^T $$ $$ f_Y(y) = \frac{f_T(\ln(y))}{|e^{\ln(y)}|} = \frac{1}{(N-1)!}\frac{(\ln(y))^{N-1}}{y^2}u(y-1) $$