Let (X,Y) be a random vector having the joint moment generating function
M(t_1,t_2) = \left( \frac{1}{2}e^{-t_1} + \frac{1}{2} e^{t_1} \right)^2\left( \frac{1}{2} + \frac{1}{2} e^{t_2} \right)^2, \qquad t_1,t_2 \in \mathbb{R}^2 Then P( |X+Y|=2 ) equals _______________________ (round off to 2 decimal places)
Since,
M(t_1,t_2) = M_1(t_1)*M_2(t_2) \\
;\textit{where, } M_1(t_1)= \left( \frac{1}{2}e^{-t_1} + \frac{1}{2} e^{t_1} \right)^2 , \\
M_1(t_2)= \left( \frac{1}{2} + \frac{1}{2} e^{t_2} \right)^2 The two random variables X and Y are independent and have moment generating functions M1(t) and M2(t) respectively. Simultaneously their pgf’s are given by:
P_X(s)= \left( \frac{1+s^2}{2s} \right)^2 = \frac{1}{4s} + \frac{s}{2} +\frac{ s^3}{4} \\
\implies Pr[X=-1]=\frac{1}{4} , P[X=1]=\frac{1}{2},Pr[X=3]=\frac{1}{4} \\
P_Y(s)= \left( \frac{1+s}{2} \right)^2 = \frac{1}{4} + \frac{s}{2} +\frac{ s^2}{4} \\
\implies Pr[Y=0]=\frac{1}{4} , P[Y=1]=\frac{1}{2},Pr[Y=2]=\frac{1}{4} \\Thus,
P \left[ |X+Y|=2 \right] = P(X=1 ,Y=1) = \frac{1}{2} \frac{1}{2} =0.25