Find E[XY] using the moment generating function of bivariate normal distribution

Question

From Introduction to Mathematical Statistics Page 198

Let (X,Y) follow a bivariate normal distribution with $X \sim N(\mu_{1}, \sigma_{1}^{2})$ and $Y \sim N(\mu_{2}, \sigma_{2}^{2})$ . Then the moment generating function is

$M_{X,Y}(t_{1},t_{2}) = exp \left \{t_{1}\mu_{1}+t_{2}\mu_{2}+\frac{1}{2}(t_{1}^{2}\sigma_{1}^{2}+2t_{1}^{}t_{2}^{}\rho \sigma_{1}^{}\sigma_{2}^{}+t_{2}^{2}\sigma_{2}^{2}) \right \}$

Then,

$E(XY)={\frac{\partial^2 M_{X,Y}(0,0) }{\partial t_{1} \partial t_{2}}}{} = \rho\sigma_{1}\sigma_{2}+\mu_{1}\mu_{2}$

Please help me with the partial derivative. It seems to me that the $t_{1}\mu_{1}+t_{2}\mu_{2}$ by the symmetry of second derivatives zero out as constants and I get $E(XY)=\rho\sigma_{1}\sigma_{2}$. Am not doing something right?

Thanks

Answer 1

Write $M(u,v)=\exp(\phi(u,v))$, where $$\left.\frac{\partial\phi}{\partial u}\right|_{(0,0)}=\mu_1\\ \left.\frac{\partial\phi}{\partial v}\right|_{(0,0)}=\mu_2\\ \left.\frac{\partial^2\phi}{\partial u\,\partial v}\right|_{(0,0)}=\rho\sigma_1\sigma_2. $$ Then, by the chain rule, $$\frac \partial{\partial u} M=\frac {\partial\phi}{\partial u}e^\phi=e^\phi \frac{\partial\phi}{\partial u}$$ and by the product rule $$\frac \partial{\partial v}\frac \partial{\partial u} M = \left(\frac \partial{\partial v}e^\phi\right)\left(\frac{\partial\phi}{\partial u}\right) + e^\phi\left( \frac{\partial}{\partial v}\frac{\partial\phi}{\partial u}\right)\\ \frac {\partial^2}{\partial u\,\partial v} M = e^\phi\left(\frac {\partial}{\partial u}\phi\right)\left(\frac{\partial\phi}{\partial v}\right) + e^\phi\left( \frac{\partial^2}{\partial u\,\partial v}\phi\right) $$ and so on. When I evaluate at $(u,v)=(0,0)$ I get the same expression you do. What is maybe confusing you is that $E[XY]$ is not the covariance $\rho\sigma_1\sigma_2$ but rather $\mu_1\mu_2$ plus that.

Answer 2

Let the $M_{X,Y}(t_1,t_2)= \exp \left \{t_{1}\mu_{1}+t_{2}\mu_{2}\color{red}-\frac{1}{2}(t_{1}^{2}\sigma_{1}^{2}+2t_{1}^{}t_{2}^{}\rho \sigma_{1}^{}\sigma_{2}^{}+t_{2}^{2}\sigma_{2}^{2}) \right \}=e^{f(t_1,t_2)}$

Then the first partial derivative w.r.t $t_1$ is

$$\frac{\partial M}{\partial t_1}=\underbrace{(\mu_1-t_1\sigma_{1}^{2}-t_{2}^{}\rho \sigma_{1}^{}\sigma_{2}^{})}_{u(t_2)}\cdot \underbrace{e^{f(t_1,t_2)}}_{v(t_2)}$$ Next we apply the product rule: $(u(t_2)\cdot v(t_2))^{'}=u(t_2)^{'}\cdot v(t_2)+u(t_2)\cdot v(t_2)^{'}$

$$\frac{\partial M}{\partial t_1 \partial t_2}=(-\rho \sigma_{1}^{}\sigma_{2}^{})\cdot e^{f(t_1,t_2)}+(\mu_1-t_1\sigma_{1}^{2}-t_{2}^{}\rho \sigma_{1}^{}\sigma_{2}^{})\cdot (\mu_2-t_2\sigma_{2}^{2}-t_{1}^{}\rho \sigma_{1}^{}\sigma_{2}^{})\cdot e^{f(t_1,t_2)}$$

Setting $t_1=t_2=0$

$$\frac{\partial M}{\partial t_1 \partial t_2}=-\rho \sigma_{1}^{}\sigma_{2}^{}\cdot 1+(\mu_1)\cdot (\mu_2)\cdot 1=\mu_1\cdot \mu_2-\rho \sigma_{1}^{}\sigma_{2}^{}$$