Bivariate moment generating function

Question

For two random variables $(X,Y)$, the MGF can be defined as $M_{XY}(s,t) = E[e^{sX+tY}]$.

Find $M_{XY}(s,t)$ when $X$ and $Y$ are two jointly normal random variables with $E[X] = μ_X,E[Y] = μ_Y ,Var(X) = σ_X^2 ,Var(Y) = σ_Y^2 ,ρ(X,Y) = ρ.$

Answer 1

Hint: Let $Z=Y-cX$. Choose $c$ such that $EZX-EZEX=0$. We now have independence of $X$ and $Z$ since jointly normal variables are independent iff their covariance is $0$. Now $Ee^{sX+tY}=Ee^{(s+ct)X+tZ}=Ee^{(s+ct)X} Ee^{tZ}$.

Answer 2

If $X, Y$ are normal random variables, then

$$ \left[ \begin{array}{c} X\\ Y \end{array} \right] \sim \mathcal{N} (\left[ \begin{array}{c} \mu_1\\ \mu_2 \end{array} \right], \left[ \begin{array}{cc} \sigma^2_X & \rho\sigma_Y\sigma_X\\ \rho\sigma_Y\sigma_X & \sigma^2_Y \end{array} \right]) $$

If $\mathbf{\Sigma}$ is the covariance matrix and $\mathbf{X}$ the column vector of the random variables, the probability density function will be:

$$f(\mathbf{X}) =\frac{e^{-\frac{1}{2}(\mathbf{X} - \mathbf{\mu})^T\mathbf{\Sigma^{-1}}(\mathbf{X} - \mathbf{\mu})}}{2\pi|\mathbf{\Sigma}|^\frac{1}{2}}$$

We also know that

$$E[e^{sX + tY}] = \iint f_{X, Y}(s, t)e^{sx + ty}dxdy$$

From then on, it's just a matter of calculations.