Derive mgf bivariate normal distribution

1. Apr 25, 2015

Lewis7879

Does anyone know the proof of joint moment generating functions for bivariate normal distributions?
M_x,y (s,t)= E(e^(xs+yt))

2. Apr 25, 2015

mathman

Expand the exponential in a power series. E(each term) is that moment.

3. Apr 25, 2015

Lewis7879

I'm actually suppose to do it like
great idea
it is possible to do it this way ?
its a very long step.
M(s,t)= e(xs+yt) ∫∫ fXY (x,y) dxdy
= e(xs+yt) ∫∫ (1/(2π√(1-ρ2xσy)) exp [ [-1/2(1-ρ2)] [(x-μxx)2 + (y-μyy)2 - 2ρ(x-μxx)(y-μyy)}

4. Apr 26, 2015

mathman

No. $e^{(xs+yt)}$ must be under the integral sign.

5. Apr 26, 2015

Lewis7879

I'm actually suppose to do it like
M(s,t)=
yes I know that but will I get the solution of mgf of bivariate normal distribution ? I'm stuck current trying to derive it.

6. Apr 27, 2015

mathman

That will depend on what $f_{XY}(x,y)$ is. If it is bivariate normal, then you will get its moment generating function.