Finding Variance from a joint moment generating function

Question

The random vars X and Y have, for all real values of $T_1, T_2$, the joint mgf $M(T_1 , T_2) = \frac{1}{2} e^{T_1 +T_2} + \frac{1}{4} e^{2T_1 +T2} + \frac{1}{12}e^{T_2} + \frac{1}{6} e^{4T_1 +3T_2}$

I'm looking for V[X], but I believe that I am overlooking a way to simplify this problem. I know that variance is the second moment or the second derivative of the mgf,but I am not sure how to generate these derivatives and what to do after, given that it is a bivariate equation. Any pointers would be appreciated.

Answer 1

$$M(t_1, t_2) = E(e^{t_1 X + t_2 Y})$$

$$M(t_1, 0) = E(e^{t_1X}) = \frac{1}{2}e^{t_1} + \frac{1}{4}e^{2t_1} + \frac{1}{12} + \frac{1}{6}e^{4t_1} = M^{'}_{X}(t_1)$$

$$E(X^n) = \frac{\partial^n M^{'}_{X}(t_1)}{\partial t_1^n} \bigg\vert_{t_1=0}$$

$$\frac{\partial M^{'}_{X}(t_1)}{\partial t_1} = \frac{1}{2}e^{t_1} + \frac{2}{4}e^{2t_1} + \frac{4}{6}e^{4t_1}$$

$$\frac{\partial^2 M^{'}_{X}(t_1)}{\partial t_1^2} = \frac{1}{2}e^{t_1} + \frac{4}{4}e^{2t_1} + \frac{16}{6}e^{4t_1}$$

$$Var(X) = E(X^2)-E(X)^2 = (1/2 + 4/4 + 16/6) - (1/2 + 2/4 + 4/6)^2 = 25/18$$

Answer 2

Even more simply, you know from the form of the MGF that $X$ and $Y$ are joint discrete random variables: it is easy to see that $$\begin{align*}\Pr[(X,Y) = (1,1)] &= \frac{1}{2}, \\ \Pr[(X,Y) = (2,1)] &= \frac{1}{4}, \\ \Pr[(X,Y) = (0,1)] &= \frac{1}{12}, \\ \Pr[(X,Y) = (4,3)] &= \frac{1}{6}, \\ \end{align*}$$ and $0$ for any other values. Thus, the variance of $X$ is simply $$\operatorname{Var}[X] = \left(\frac{1^2}{2} + \frac{2^2}{4} + \frac{4^2}{6} \right) - \left(\frac{1}{2} + \frac{2}{4} + \frac{4}{6}\right)^2 = \frac{25}{18}.$$