Q. 7.52 - A First Course in Probability

Question

Show how to compute $Cov(X,Y)$ from the joint moment generating function of$X$ and $Y$. 

Step-by-Step Solution

Answer

The Compute $Cov(X,Y)$from the joint moment generating function value are $Cov(X,Y)=\left(\frac{{\partial }^2}{\partial t_1\partial t_2}{M}_{X,Y}-\frac{\partial }{\partial t_1}{M}_{X,Y}\frac{\partial }{\partial t_2}{M}_{X,Y}\right)(0,0)$.

Step 1: Given Information

The joint moment generating function of $X$ and $Y$.

Step 1: Given Information

We have that the joint moment generating function of $X$ and $Y$ is,

$M_{X,Y}\left(t_1,t_2\right)=E\left(e^{t_{1}X+t_{2}Y}\right)$

Observe that,

$\frac{\partial }{\partial t_1}{M}_{X,Y}\left(t_1,t_2\right)=E\left(X{e}^{t_{1}X+t_{2}Y}\right)$

Which implies

$E\left(X\right)=\frac{\partial }{\partial t_1}{M}_{X,Y}(0,0)$ and 

$E\left(Y\right)=\frac{\partial }{\partial t_2}{M}_{X,Y}(0,0)$.

Step 3: Explanation

Now, consider what happens if we partially differentiate$M_{X,Y}\left(t_1,t_2\right)$ respective to $t_1$ and then to $t_2$. We end up with

$\frac{{\partial }^2}{\partial t_1\partial t_2}{M}_{X,Y}\left(t_1,t_2\right)=E\left(XY{e}^{t_{1}X+t_{2}Y}\right)$

Which implies,

$\frac{{\partial }^2}{\partial t_1\partial t_2}{M}_{X,Y}(0,0)=E\left(XY\right)$

Finally we have that, $Cov(X,Y)=E\left(XY\right)-E\left(X\right)E\left(Y\right)$

$=\left(\frac{{\partial }^2}{\partial t_1\partial t_2}{M}_{X,Y}-\frac{\partial }{\partial t_1}{M}_{X,Y}\frac{\partial }{\partial t_2}{M}_{X,Y}\right)(0,0)$.

Step 4: Final answer

The$Cov(X,Y)$ joint moment generating function value are $Cov(X,Y)=\left(\frac{{\partial }^2}{\partial t_1\partial t_2}{M}_{X,Y}-\frac{\partial }{\partial t_1}{M}_{X,Y}\frac{\partial }{\partial t_2}{M}_{X,Y}\right)(0,0)$.