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Let $X\sim N(\mu,\sigma^2)$ be independent of $Y \sim \text{Chisquared}(k)$.

I seek the pdf of the product $Z = X Y$


$X$ has pdf:

$$f_X(x)={1\over\sigma\sqrt{2\pi}}e^{-{1\over2}({x-\mu\over\sigma})^2}$$

$Y$ has pdf over the positive real line: $$f_Y(y)={y^{(k/2)-1}e^{-y/2}\over{2^{k/2}\Gamma({k\over2})}}$$

One way to find the solution is to use Rohatgi's well known result (1976,p.141) if $f_{XY}(x,y)$ be the joint pdf of continuous RV's $X$ and $Y$, the pdf of $Z$ is $$f_Z(z) = \int_{-\infty}^{\infty}{{1\over|y|}f_{XY}({z\over y},y)dy} $$

Since, $X$ and $Y$ are independent $f_{XY}(x,y)=f_X(x)f_Y(y)$ $$f_Z(z) = \int_{-\infty}^{\infty}{{1\over|y|}f_X({z\over y})f_{Y}(y)dy} $$ $$f_Z(z) = {1\over\sigma_x\sqrt{2\pi}}{1\over{2^{k/2}\Gamma({k\over2})}}\int_{0}^{\infty}{{1\over|y|}e^{-{1\over2}({{z\over y}-\mu_x\over\sigma_x})^2} {y^{(k/2)-1}e^{-y/2}}dy} $$ ... where we face the problem of solving the integral $\int_{0}^{\infty}{e^{-{1\over2}({{z\over y}-\mu_x\over\sigma_x})^2} {y^{(k/2)-2}e^{-y/2}}dy}$.

Can anyone help me with this problem?

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    $\begingroup$ I have no idea what you are trying to write there, so unfortunately I cannot help you format the mathematics which is clearly "off". Please take a look at this FAQ and try to edit your question using our MathJax capabilities. As it stands it is quite difficult to tell what exactly it is that you are asking about. $\endgroup$ – Willie Wong Dec 12 '12 at 15:04
  • $\begingroup$ anyone with any idea, suggestion, hint... $\endgroup$ – robin Dec 21 '12 at 15:05
  • $\begingroup$ Unanswered for over 5 years ... $\endgroup$ – wolfies Mar 9 '18 at 16:20
  • $\begingroup$ @wolfies Indeed... $\endgroup$ – user85798 Mar 10 '18 at 4:02

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