Probability – Characteristic Function of Z=XY for Independent Non-Standard Normal Variables

characteristic-functionsnormal distributionprobabilityprobability distributionsprobability theory

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.

Best Answer

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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