# Thread: Product of Random variables - Density

1. ## Product of Random variables - Density

A random variable $\displaystyle \xi$ has Gaussian distribution with mean zero and variance one, while a random variable $\displaystyle \eta$ has the distribution with the density:
$\displaystyle p_{\eta}(t)=$$\displaystyle \{ \begin{array}{cc}t e^{-t^2/2} & if\quad t\geq 0 \\ 0 & otherwise \end{array} What is the distribution of \displaystyle \zeta = \xi.\eta assuming that \displaystyle \xi and \displaystyle \eta are independent? One approach is to use the formula of change of variables (e.g. using Jacobian) for a product of random variables \displaystyle X,Y (see for example Grimmett and Stirzaker, p. 109) through the map: \displaystyle u = xy, v=x which gives: \displaystyle f_{U,V}(u,v)=f_{X,Y}(v,u/v)|v|^{-1} , use independence and integrate over \displaystyle v to obtain the result. But the formula is somewhat cumbersome as it gives 2 different expressions whether you assign \displaystyle \xi or \displaystyle \eta to \displaystyle X (no sign problem with \displaystyle \eta). I assume you get to the same result although the calculation doesn't seem tractable. Another (simpler?) approach is to use the formula of total probability and use independence. But again it doesn't seem you can go beyond an integral of the form: \displaystyle f_{\xi\eta}(a)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty } e^{-\frac{1}{2}((\frac{a}{x})^2+x^2)}dx Is this correct? Is it possible to do better? Thanks for any help. 2. Originally Posted by akbar A random variable \displaystyle \xi has Gaussian distribution with mean zero and variance one, while a random variable \displaystyle \eta has the distribution with the density: \displaystyle p_{\eta}(t)=$$\displaystyle \{ \begin{array}{cc}t\exp^{-t^2/2} & if\quad t\geq 0 \\ 0 & otherwise \end{array}$
What is the distribution of $\displaystyle \zeta = \xi.\eta$ assuming that $\displaystyle \xi$ and $\displaystyle \eta$ are independent?

One approach is to use the formula of change of variables (e.g. using Jacobian) for a product of random variables $\displaystyle X,Y$ (see for example Grimmett and Stirzaker, p. 109) through the map: $\displaystyle u = xy, v=x$ which gives:
$\displaystyle f_{U,V}(u,v)=f_{X,Y}(v,u/v)|v|^{-1}$ , use independence and integrate over $\displaystyle v$ to obtain the result.
But the formula is somewhat cumbersome as it gives 2 different expressions whether you assign $\displaystyle \xi$ or $\displaystyle \eta$ to $\displaystyle X$ (no sign problem with $\displaystyle \eta$). I assume you get to the same result although the calculation doesn't seem tractable.

Another (simpler?) approach is to use the formula of total probability and use independence. But again it doesn't seem you can go beyond an integral of the form:

$\displaystyle f_{\xi\eta}(a)=\frac{1}{\sqrt{2\pi}}\int_0^{\infty }\exp^{-\frac{1}{2}((\frac{a}{x})^2+x^2)}dx$

Is this correct? Is it possible to do better?
Thanks for any help.
Use characteristic functions: cf. here. In this case, if my quick computation is correct, you should get a Laplace distribution: $\displaystyle e^{-|x|}dx$.

3. Although the problem wasn't related (in the book) to the use of characteristic functions, I agree the result is almost immediate that way. I still wonder how you can get it without the transform...

One question though: you get $\displaystyle \Phi_{XY}(\lambda)=\frac{1}{1+\lambda^2}$ when you first use the Gaussian random variable, then evaluate the final expectation using the other distribution. Have you tried the other way round? e.g. first calculate the CF of the distribution (let me know your result), then evaluate the final expectation with the Gaussian distribution.

In short: do you get back on your feet?

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