How do I find the pdf of $\prod_{i=1}^n X_i$, where $X_is$ are independent uniform [0,1] random variables.

I know X~U[0,1], -ln(x) is exponential(1). I also know the sum of two or more independent exponential random variable is gamma.

For $Y = \sum_{i=1}^n -ln(x_i)$, which is a gamma(n, 1), I found the pdf for Y is $$\int_0^\infty \frac{1}{\Gamma (n)} y^{n-1} e^{-y} dy$$.

Let $Z = e^Y$

I am trying to the pdf for Z, what I found is $$\int_1^\infty \frac{1}{\Gamma (n)} ln(z)^{n-1} \frac{1}{z^2} dz$$, which does not look right to me. Could someone check it?

  • $\begingroup$ See this. $\endgroup$ – Clement C. May 13 '14 at 2:15
  • $\begingroup$ Thank you for the resource! The material is too math intensive $\endgroup$ – afsdf dfsaf May 13 '14 at 2:16
  • $\begingroup$ Can you rederive the result when $n=2$? (using this directly, for instance?) And for $n=3,4,...$ using the trick of taking the logarithm of the product, and computing the convolution of the pdfs of the logarithms of uniform r.v.'s? $\endgroup$ – Clement C. May 13 '14 at 2:22
  • $\begingroup$ Can you show me some steps? $\endgroup$ – afsdf dfsaf May 13 '14 at 2:23
  • $\begingroup$ Related math.stackexchange.com/questions/738797/… $\endgroup$ – leonbloy May 15 '14 at 18:33

Step 1: Note that $-\log X_j$ has an exponential(1) distribution.

Step 2: Note that $\sum_{j=1}^n (-\log X_j)$ is a sum of iid random variables, so convolution results apply.

Step 3: Figure out the convolution for the exponential distribution, by looking it up, by using characteristic functions, or moment-generating functions, whichever you're familiar with. You'll find $f_X(x)=x^{n-1} e^{-x} / (n-1)!$ as the density of $-\log Y$, which is a $\Gamma$ distribution.

Step 4: Transform this back to get something like $(-\log y)^{n-1}/(n-1)!$.

Note that for increasing $g$ (decreasing $g$ is analogous), $$\Pr(Y\leq y)=\Pr(g(X)\leq y)=\Pr(X\leq g^{−1}(y))=F_X(g^{−1}(y)).$$

Now differentiate with respect to $z$ to get

$$ f_Y(y) = f_X(g^{-1}(y)) g^{-1}\;'(y)= \frac{f_X(g^{-1}(y))}{g'(g^{-1}(y))}. $$

In your case $g(x)=\exp(−x)$ which is decreasing in $x$ and hence you need to replace the denominator with $|g′(g^{−1}(y))|$.

So $g'(x)=-\exp(-x)$ and $g^{-1}(y)=-\log y$.

  • $\begingroup$ what is the " convolution results "? $\endgroup$ – afsdf dfsaf May 13 '14 at 3:55
  • $\begingroup$ Convolutions are standard results that give you the distribution of a sum of two or more iid random variables. $\endgroup$ – JPi May 13 '14 at 3:59
  • $\begingroup$ Can you be a little bit more explicit on step 3? $\endgroup$ – afsdf dfsaf May 13 '14 at 18:25
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    $\begingroup$ Don't integrate. The density function is the integrand, not the integral. Aside from that, I think it should be $Z=e^{-Y}$. By the standard change of variable formula for density functions I get for $y= -\log z$ that $$y^{n-1} e^{-y} / (\Gamma(n) e^{-y}) = (-\log z)^{n-1}/(n-1)!$$ Hope this helps. $\endgroup$ – JPi May 15 '14 at 12:59
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    $\begingroup$ how do you make that to become $(-\log y)^{n-1}/(n-1)$? $\endgroup$ – afsdf dfsaf May 15 '14 at 15:55

An attempt:

For the case $n=2$, using directly the fact that (for $Y=X_1X_2$, where $X_1,X_2\sim\operatorname{Unif}([0,1])$ are independent) $$ \forall z\in[0,1],\quad f_Y(z)=\int_0^1 f_{X_1}(x)f_{X_2}\left(\frac{z}{x}\right)\frac{dx}{x} $$ you get that for all $z\in(0,1]$ $$ f_Y(z)=\int_0^1 1\cdot \mathbb{1}_{\{\frac{z}{x}\in[0,1]\}}\frac{dx}{x} = \int_z^1 \frac{dx}{x} = \ln\frac{1}{z} $$

For the general case, you can apply the trick of taking the logarithm of $Y=\prod_{k=1}^n X_k$, i.e. consider $Z\stackrel{\rm def}{=} \ln Y = \sum_{k=1}^n \ln X_k = \sum_{k=1}^n Z_k$ for $Z_k\stackrel{\rm def}{=} \ln X_k$, so that for all $z\in(-\infty,0]$ $$ f_Z(z)=(f_{X_1}\ast\cdot\ast f_{X_k})(z) = (f_{X_1}\ast\cdot\ast f_{X_1})(z) $$ where $\ast$ is the convolution operator and $$ \begin{align*} f_{X_1}(x) &= \frac{d}{dx}\int_{-\infty}^x \mathbb{P}\left\{\ln X_1 \in dt\right\} = \frac{d}{dx} \mathbb{P}\left\{\ln X_1 \leq x\right\}= \frac{d}{dx} \mathbb{P}\left\{X_1 \leq e^x\right\}\\ &= \frac{d}{dx}\mathbb{P}\left\{X_1 \leq e^x\right\}= \frac{d}{dx} e^x = e^x \end{align*} $$ (caveat: I haven't carefully checked my calculations) Once (after a lot of fun) you figure $f_Z$, you can go back to $f_Y$ by a similar process (using the cdf of $f_Z$, etc).

  • $\begingroup$ I just realized If X~U[0,1], -ln(x) is exponential(1)...Does this help reduce the work? $\endgroup$ – afsdf dfsaf May 13 '14 at 2:44
  • $\begingroup$ This is essentially what the last computation above shows (the one ending with $e^x$). $\endgroup$ – Clement C. May 13 '14 at 2:46
  • $\begingroup$ In the second line, how did you change the lower bound of the integral from $0$ to $z$? $\endgroup$ – NauticalMile Jan 23 '15 at 17:56
  • $\begingroup$ The indicator $\mathbb{1}_{\frac{z}{x}\in[0,1]}$ is zero if $0 < x < z$. $\endgroup$ – Clement C. Jan 23 '15 at 18:14
  • $\begingroup$ That was helpful, thanks. I think I found my mistake: $\frac{z}{x}$ is $z$ divided by $x$, not $z|x$, correct? My prof sometimes writes $\frac{X}{Y}$ to show $X$ given $Y$ for random variables, but in this case the simple arithmetic division caught me off guard. $\endgroup$ – NauticalMile Jan 26 '15 at 3:18

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