# Moment generating function of X*Y, normal distributions

• Jun 29th 2013, 10:25 AM
MagisterMan
Moment generating function of X*Y, normal distributions
Let $\displaystyle$X$$and \displaystyle Y$$ be jointly normally distributed random variables with correlation $\displaystyle$0 \le \rho \le 1$$and variance \displaystyle \sigma^2 = 1$$. How do I find the moment generating function for $\displaystyle$X \cdot Y$$? I first thought of doing a transformation of variables so that \displaystyle U=X \cdot Y$$ and $\displaystyle$V=Y \imples$$, finding the joint density \displaystyle f_{U,V}(u,v) and integrating over \displaystyle V$$, but unfortunately this seems to lead to very messy calculations. Is there some cleaner way of dealing with this problem?
• Jun 29th 2013, 05:04 PM
chiro
Re: Moment generating function of X*Y, normal distributions
Hey MagisterMan.

Did you try and calculate E[e^(XYt)]?

If you are having problems you should also search the literature/google for the product distribution PDF for a normal*normal and calculate E[e^(tW)] where W = XY.

I would first try finding E[e^(XYt)] using the joint distribution.
• Jun 29th 2013, 06:06 PM
MagisterMan
Re: Moment generating function of X*Y, normal distributions
Yes, $\displaystyle$E(e^{XYt})$$is what I'm trying to find. I've done some googling and I've found out how to solve this in the independent case. I'd do the following then (using Law of total expectation): Let \displaystyle W=XY$$ and $\displaystyle$\psi$$be the moment generating function. \displaystyle \psi_W(t)=E(e^{XYt}) = E(E(e^{XYt}|Y))=E(\psi_X(tY))$$

$\displaystyle$ E(\psi_X(tY)) = E(e^{t^2 Y^2 /2})$$using the mgf for a normal distribution and in the end after solving the integral for expectation one ends up with: \displaystyle  \psi_W(t)=(1-t^2)^{-\frac{1}{2}}$$. Not sure really sure if this method with conditional expectation can be applied to the general case (correlation not equal to 0).
• Jun 29th 2013, 06:45 PM
chiro
Re: Moment generating function of X*Y, normal distributions
You should be able to apply the law of total expectation even in the correlated case.

If something was independent then you would have E[X|Y] = E[X] regardless of what Y is if the two are independent.
• Jun 29th 2013, 07:26 PM
MagisterMan
Re: Moment generating function of X*Y, normal distributions
To clarify a bit on the calculations:

$\displaystyle$E(e^{t^2 Y^2 /2}) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{t^2 y^2 /2} e^{-y^2} dy = \int_{-\infty}^{\infty} e^{-y^2/2 (1- t^2)} dy = \frac{1}{\sqrt{1-t^2}} $$I know that the correlated case is not the same as the case of independence, so basically what's confusing me is that I don't see where I use the assumption that X and Y are independent, yet I don't arrive at the answer for the general case (which depends on \displaystyle \rho$$).
• Jun 29th 2013, 08:07 PM
chiro
Re: Moment generating function of X*Y, normal distributions
Try getting the marginal distribution and integrate out Y to get E[e^(XYt)|Y] which is a function of X and then integrate this out with respect to X to get your final expectation.

To do this you need to integrate the expectation over all values of y and this is known as the marginal distribution.

You should be able to do this since you are only dealing with one integration dimension (instead of two).

You should do this instead of using the characteristic function to convince yourself that the answer is what it is since it is a step by step approach that you can justify (and find errors if they crop up).
• Jun 30th 2013, 10:32 AM
MagisterMan
Re: Moment generating function of X*Y, normal distributions
Thanks, I'll try that.