# Thread: MGF of a multiplication of RV

1. ## MGF of a multiplication of RV

Hi, I am at a loss for a problem...

If Y1 and Y2 are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of U=Y1Y2. Use this moment-generating function to find E(U) and V(U). Check the result by evaluating E(U) and V(U) directly from the density functions for Y1 and Y2.

I try to do Mu(t) = E(e^tu) the normal way, but then I get: E(e^ty1y2) which doesn't help me in the least...
I know that E(U) = E(Y1)*E(Y2) = 0 and V(U) = 1 similarly, but I cannot find the MGF. Any pointers? Am I missing something obvious, or not so obvious?

2. Originally Posted by Volcanicrain
Hi, I am at a loss for a problem...

If Y1 and Y2 are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of U=Y1Y2. Use this moment-generating function to find E(U) and V(U). Check the result by evaluating E(U) and V(U) directly from the density functions for Y1 and Y2.

I try to do Mu(t) = E(e^tu) the normal way, but then I get: E(e^ty1y2) which doesn't help me in the least...
I know that E(U) = E(Y1)*E(Y2) = 0 and V(U) = 1 similarly, but I cannot find the MGF. Any pointers? Am I missing something obvious, or not so obvious?
You're meant to know that if $U = X+Y$ and X and Y are independent then $m_U(t) = m_X(t) \cdot m_Y(t)$.

3. Hello,
Originally Posted by mr fantastic
You're meant to know that if $U = XY$ and X and Y are independent then $m_U(t) = m_X(t) \cdot m_Y(t)$.
Isn't it for when U=X+Y, but not XY ?

But I admit that I don't understand the order of the questions since you're first asked to find the mgf and then the expectations and all the stuff. And finding that mgf is more difficult than finding the expectation if it's indeed a product of normal distributions

4. I hate to say, but katie is right.
http://comics.com/pearls_before_swine/2010-01-19/
However, my guess is that there should be a plus sign in there.
http://comics.com/pearls_before_swine/2010-01-29/

5. Originally Posted by Moo
Hello,

Isn't it for when U=X+Y, but not XY ?

But I admit that I don't understand the order of the questions since you're first asked to find the mgf and then the expectations and all the stuff. And finding that mgf is more difficult than finding the expectation if it's indeed a product of normal distributions
*sigh* I imagined a plus sign that isn't there (but probably is meant to be there).

6. I e-mailed my teacher and he said there was no mistake... :S that it had to do with conditional probability. Oye :S
I appreciate your insight though ^^

7. I would let W=Y1 and make a 2-2 transformation and then integrate out the W.

8. Maybe this will work.
I'll use X and Z as standard normal rvs.
Let U=XZ. You want ...

$M_U(t)=E( e^{XZt} )=E(E(e^{XZt}|X))=E(M_Z(Xt)|X))$

What I'm doing is fixing the X which makes the argument now Xt.
So plug in Xt into the MGF of Z and see what you get.

Using the MGF of Z and plugging in Xt for t, I get

$E( e^{.5X^2t^2} )$

and now using the density of X...

$={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{.5x^2t^2-.5x^2}dx$

$={1\over \sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-.5x^2(1-t^2)}dx$

which you can compute without integrating by observing the st deviation here as $(1-t^2)^{-.5}$

9. Wow, thank you very much.
I've been flirting around with the "given X" part of your answer, I just hadn't pinpointed that yet haha.

I'm just a tad confused at the end - how did you solve that final integral? The integral is equal to the MGF of U, right?
Thanks a million for your persisted help in this problem

10. Originally Posted by Volcanicrain
Wow, thank you very much.
I've been flirting around with the "given X" part of your answer, I just hadn't pinpointed that yet haha.

I'm just a tad confused at the end - how did you solve that final integral? The integral is equal to the MGF of U, right?
Thanks a million for your persisted help in this problem
Compare the final integral to the integral of a Gaussian (note that t is treated as a constant ....)

11. Thank you both so very much!
I've learned a lot in this thread - a ton actually! and now I can finally solve this problem I'd been thinking about since last Monday.

Thank you thank you thank you!

12. both?

13. Well to you and to mr fantastic for helping me with the integral