<s>[INST] Finding Variance from a joint moment generating function 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.
 [/INST] 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 $$\operatorname{Var}[X] = \left(\frac{1^2}{2} + \frac{2^2}{4} + \frac{4^2}{6} \right) - \left(\frac{1}{2} + \frac{2}{4} + \frac{4}{6}\right)^2 = \frac{25}{18}.$$
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