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.