Probability Generating functions (I know it's a bit long...):

Question

anonymous · Answer

If X and Y are 2 random variables, the joint probability generating function (pgf), $G$, of X and Y is defined to be:$$G_{X,Y}(z_1,z_2)=E\left(z_1^Xz_2^Y\right), \text{where } E \text{ is the expected value}$$ If X and Y are independent, then $G_{X,Y}(z_1,z_2)=G_X(z_1)G_Y(z_2)  ~~~(**)$

In the univariate case, $G_X(z)=E(z^X)$

Suppose X and Y are independent random variables, and suppose you have $W=X+Y$ and $U=XY$. 
I feel though by using $(**)$ it looks as if the pgf's of $T$ and of $U$ look the same. This can't be, however, since I know the distribution of $T$ and $U$ are different and that two different distributions can't have the same pgf. 

My problem is finding the pgf of $U$.  Let's assume continuous random variables. Let $W = X+Y$. Showing the pgf of W is standard and found in any textbook dealing with pgfs:  $$\begin{align}G_W(z)&=E\left(z^W \right) \&=E\left(z^{X+Y} \right)\&=E\left(z^Xz^Y \right)\&=E\left(z^X \right)E\left(z^Y \right)\text{by independence}\&=\int\limits_{x\in R_X}z^xf_X(x)\, dx\int\limits_{y \in R_Y}z^yf_Y(y)\,dy,~~ R_X, R_Y \text{ are the support of }X \text{ and }Y\&=G_X(z)G_Y(z)\end{align} $$ I guess here the pgf of W matches $(**)$ as $z_1=z_2=z$?
Now, I tried doing the same for U:
Let $U=XY$. Then $$\begin{align}G_U(z)&=E\left(z^U \right) 
\&=E\left(z^{XY} \right)
\&=G_{XY}(z)
\&=\iint\limits_{(x,y)\in R_{XY}}z^{xy}f_{XY}(x,y)\,dx \,dy, ~~R_{XY}\text{ is the joint support of } X\text{ and } Y
\&=\iint\limits_{(x,y)\in R_X \times R_Y}z^{xy}f_X(x)f_Y(y)\,dx\,dy, ~~\text{by independence}
\&=\int\limits_{x\in R_X}~\int\limits_{y \in R_Y}z^{xy}f_X(x)f_Y(y)\,dy\,dx
\&=\int\limits_{x\in R_X }f_X(x)\int\limits_{y \in R_Y}z^{xy}f_Y(y)\,dy \,dx\end{align} $$
How can I match the pgf of U to match $(**)$

amistre64 · Answer

ow,  i think i just burned a retina .....

anonymous · Answer

@amistre64 lol

anonymous · Answer

@SithsAndGiggles @kirbykirby @aum

anonymous · Answer

Aren't PGFs only defined for discrete variables?

xapproachesinfinity · Answer

I  don't even know what that is xD