[Math] Given a joint moment generating function, find $P(X

Question

Let random variables $X$ and $Y$ have joint MGF:
$$M(t_1,t_2) = 1/2e^{t_1+t_2} + 1/4e^{2t_1+t_2} + 1/12e^{t_2} + 1/6e^{4t_1+3t_2}$$ I now need to find $P(X<Y)$. I know how to find the moments as I had to find $V[X]$ for part of this problem, but I am not quite sure how to figure out probability from this.

Answer 1

Let $Z$ be a discrete random variable taking values $1\dots,n$ with respective probabilities $p_1,\dots,p_n$ (of course $p_1+\dots+p_n =1$).

Define $X$ and $Y$ as follows: let $x_i,y_i,~i=1,\dots,n$ be a sequence of numbers, and let $X=x_i,Y=y_i$ whenever $Z=i$.

Note then that $X<Y$ if and only if $Z$ is equal to an $i$ for which $x_i<y_i$.

The MGF of $(X,Y)$ is easily obtained by conditioning on $Z$:

$$E[e^{t_1 X+ t_2 Y} ] = \sum_{i=1}^n p_i e^{t_1 x_i + t_2 y_i}.$$

In our case $n=4$, $p_1=\frac 12=\frac{6}{12},p_2=\frac 14=\frac{3}{12},p_3=\frac 1{12},p_4=\frac {1}{6}=\frac{2}{12}$, and $x_1=1,y_1=1$, $x_2=2,y_2=1$, $x_3=0,y_3=1$ and $x_4=4,y_3=3$.

The only value of $i$ such that $x_i<y_i$ is $i=3$. Since $p_3=\frac{1}{12}$, it follows that $P(X<Y) = p_3= \frac{1}{12}$.

Can you see what is $P(X=Y)$ and $P(Y<X)$ ?