# Characteristic function of random variable $Z=XY$ where X and Y are independent non-standard normal random variables

I would like to find Characteristic function of random variable $Z=XY$ where X and Y are independent normal random variables, but they are not standard, i.e. $$X\sim N(\mu _x,\sigma_x)$$ $$Y\sim N(\mu _y,\sigma_y)$$ where $\mu _x\neq 0, \sigma _x\neq 1, \mu _y\neq 0,\sigma _y\neq 1$

For the standard case you can find the solution here , but I could not find any solution for the non-standard one. I would appreciate any help. Thank you.

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According to Maple:

$${{\rm e}^{{\dfrac {t \left( -t{\mu_{{x}}}^{2}{\sigma_{{y}}}^{2}-t{ \mu_{{y}}}^{2}{\sigma_{{x}}}^{2}+2\,i\mu_{{x}}\mu_{{y}} \right) }{2({t}^ {2}{\sigma_{{x}}}^{2}{\sigma_{{y}}}^{2}+1)}}}}{\frac {1}{\sqrt {{t}^{2} {\sigma_{{x}}}^{2}{\sigma_{{y}}}^{2}+1}}}$$

EDIT: As requested, here is Maple code for the characteristic function and the PDF:

with(Statistics):
X:= RandomVariable(Normal(mu[x],sigma[x]));
Y:= RandomVariable(Normal(mu[y],sigma[y]));
C:= simplify(CharacteristicFunction(X*Y,t)) assuming real;
f:= simplify(PDF(X*Y,z)) assuming real;


It doesn't return a closed form for the PDF. I don't know if there is one.

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Thank you a lot. – May Sep 17 '13 at 15:15
Actually my main problem is how to find PDF of XY. I tried different methods but I could not find it. Any help would be appreciated. Thank you. – May Sep 17 '13 at 15:36
@May: pdfs and cfs are related through inversion: en.wikipedia.org/wiki/… – Alex Sep 17 '13 at 17:18
@Alex Thanks, but I don't know how to find the following integral: $$f(u)=1/2\pi \int_{-\infty }^{\infty} {{\rm e}^{{\dfrac {t \left( -t{\mu_{{x}}}^{2}{\sigma_{{y}}}^{2}-t{ \mu_{{y}}}^{2}{\sigma_{{x}}}^{2}+2\,i\mu_{{x}}\mu_{{y}} \right) }{2({t}^ {2}{\sigma_{{x}}}^{2}{\sigma_{{y}}}^{2}+1)}}}}{\frac {e^{-itu}}{\sqrt {{t}^{2} {\sigma_{{x}}}^{2}{\sigma_{{y}}}^{2}+1}}} \;dt$$ – May Sep 17 '13 at 17:48
@Robert Israel: I am new to MAPLE, would you please give me your MAPLE code? Thanks. – May Sep 17 '13 at 18:35