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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?

  • 1
    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](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) 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.2012-12-12
  • 0
    anyone with any idea, suggestion, hint...2012-12-21
  • 0
    Unanswered for over 5 years ...2018-03-09
  • 0
    @wolfies Indeed...2018-03-10

0 Answers 0