# Multivariate Moment Generating Function

Let $X$ and $Y$ be two independent random variables both have Laplace distribution. What is the moment generating function of $U=X+Y$ and $V=X-Y$?

Initially, I want to work out the $f_{U,V}(u,v)$, and then work out the $M_{U,V}(s,t)=E_{f_{U,V}}(e^{sU+tV})$. But they are hard to compute.

So I try another way: $$M_{U,V}(s,t)=E_{f_{U,V}}(e^{sU+tV})=E_{f_{U,V}}(e^{s(X+Y)+t(X-Y)})=E_{f_{U,V}}(e^{(s+t)X+(s-t)Y)})$$

But at this point I am not so sure about whether I can make it become $E_{f_{X}}(e^{(s+t)X})E_{f_{Y}}(e^{(s-t)Y})$.

Can I do this? Why? Thanks in advance.

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Guide: Consider what to make $g,h,T,$ and $S$ to make the following points applicable:

1. If $X,Y$ are independent variables, then $g(X),h(Y)$ are independent variables too.

2. If $T,S$ are independent variables, then $f_{T,S}(x,y)=f_T(x)f_S(y)$ and so we have $$\mathbb{E}(TS)=\iint f_T(x)f_S(y)xydxdy=\int f_T(x)xdx\int f_S(y)ydy=\mathbb{E}(T)\mathbb{E}(S).$$

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What I confuse about is how I deal with the subscript: at beginning I have $E_{f_U,f_V}[g(U,V)]$ then it becomes $E_{f_U,f_V}[h(X,Y)]$. Can I just change it to $E_{f_X,f_Y}$? –  John Mar 31 '12 at 1:36
@John: I'm not familiar with your putting functions in subscripts. What does that signify? From what I understand, we put random variables as subscripts to signify which variables we let vary, as opposed to which we fix intermittently. Fixing neither $U$ nor $V$ allows $X$ and $Y$ to vary over their whole range of values. Also, it doesn't look like you understood my guide, seeing where you put $g$ and $h$ and didn't bother with separation. Note that $t,s$ are fixed for our purpose, and $e^{(s+t)X}$ is a function of $X$ while $e^{(s-t)Y}$ a function of $Y$. These are intended to be $T$ and $S$. –  anon Mar 31 '12 at 1:49
In summary: $$\mathbb{E}_{U,V}(\alpha(U,V))=\mathbb{E}_{X,Y}(\alpha(X+Y,X-Y))$$ because $$f_{U,V}(u,v)=f_{X,Y}(x+y,x-y).$$ –  anon Mar 31 '12 at 2:09

you need to multiply the Jacobian to do the transforamation from U,V to X,Y

$$E_{U,V}[F(u,v)]=\int{F(u,v) du dv}=\int{F(x,y) J(x,y) dx dv}=E_{X,Y}[FJ]$$

where $$F(x,y)$$ is obtained just by putting u and v's vaue in terms of x and y. in your case $$F`(x,y)=F(u/2+v/2,u/2-v/2)$$ and J is jacobain. In your case J is independent of x and y, so you can divide the product term in two factors.

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