Derive mgf bivariate normal distribution

[Lewis7879]

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

 

[mathman]

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

 

[Lewis7879]

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

I'm actually suppose to do it like

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

great idea
it is possible to do it this way ?
its a very long step.
M(s,t)= e^(xs+yt) ∫∫ f_XY (x,y) dxdy
= e^(xs+yt) ∫∫ (1/(2π√(1-ρ²)σ_xσ_y)) exp [ [-1/2(1-ρ²)] [(x-μ_x/σ_x)² + (y-μ_y/σ_y)² - 2ρ(x-μ_x/σ_x)(y-μ_y/σ_y)}

 

[mathman]

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) ∫∫ f_XY (x,y) dxdy
= e^(xs+yt) ∫∫ (1/(2π√(1-ρ²)σ_xσ_y)) exp [ [-1/2(1-ρ²)] [(x-μ_x/σ_x)² + (y-μ_y/σ_y)² - 2ρ(x-μ_x/σ_x)(y-μ_y/σ_y)}

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

 

[Lewis7879]

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

I'm actually suppose to do it like
M(s,t)=

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

yes I know that but will I get the solution of mgf of bivariate normal distribution ? I'm stuck current trying to derive it.

 

[mathman]

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.

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