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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?

afsdf dfsaf
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2 Answers2

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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 $g^{-1}(y)=-\log y$.

JPi
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  • what is the " convolution results "? – afsdf dfsaf May 13 '14 at 03:55
  • Convolutions are standard results that give you the distribution of a sum of two or more iid random variables. – JPi May 13 '14 at 03:59
  • Can you be a little bit more explicit on step 3? – afsdf dfsaf May 13 '14 at 18:25
  • 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 like your step 4 result...Could you see what I did wrong?

     – afsdf dfsaf May 14 '14 at 14:20
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    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. – JPi May 15 '14 at 12:59
  • If you're unfamiliar with the change of variables, note that if $z=g(y)$ with $g$ differentiable with everywhere nonzero derivative, then $f_Z(z)= f_Y\bigl( g^{-1}(z) \bigr) / \bigl| g'\bigl( g^{-1}(z)\bigr) \bigr|$. – JPi May 15 '14 at 13:01
  • could you expand on what you said here? – afsdf dfsaf May 15 '14 at 15:26
  • Yes. Note that for increasing $g$ (decreasing $g$ is analogous), $$\Pr(Z\leq z)=\Pr(g(Y)\leq z)=\Pr(Y\leq g^{-1}(z)) = F(g^{-1}(z)).$$ Now differentiate with respect to $z$ to get $f_Z(z) = f_Y(g^{-1}(z)) g^{-1}'(z)= f_Y(g^{-1}(z))/g'(g^{-1}(z))$. In your case $g=\exp(-y)$ which is decreasing in $y$ and hence you need to replace the denominator with $|g'(g^{-1}(z))|$. – JPi May 15 '14 at 15:36
  • you could just revised your answer under step 4...so that I can make it my favor answer – afsdf dfsaf May 15 '14 at 15:38
  • I've done just that. – JPi May 15 '14 at 15:50
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    how do you make that to become $(-\log y)^{n-1}/(n-1)$? – afsdf dfsaf May 15 '14 at 15:55
  • fixed typo, grrr, thanks – JPi May 15 '14 at 16:13
  • just to confirm: $$\frac{f_Y (g^{-1}(z))}{g'(g^{-1}(z)} = (-\log y)^{n-1}/(n-1)!$$? – afsdf dfsaf May 15 '14 at 16:30
  • Yes, that is correct – JPi May 15 '14 at 17:12
  • could you show the intermediate steps? – afsdf dfsaf May 15 '14 at 18:05
  • Just figure it out, what is the range for this pdf in this case, is it Z >0 because $e^{-Y} > 0$? – afsdf dfsaf May 17 '14 at 16:42
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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).

Clement C.
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