Let $X$ and $Y$ be independent random variable with respective moment generating function
$M_x(t) = \frac{(8+e^t)^2}{81} $ and $M_y(t) = \frac{(1+3e^t)^3}{64} , -\infty<t<\infty $
Then $ P(X+Y = 1) $equals
I know that using moment generating function we can find probability
$M_x(t) = P(X=0)e^{t*0} + P(X=1)e^{t*1}…..P(X=n)e^{t*n}$
Comparing this mgf we can get the particular probability.
But how do we do this question?
Best Answer
Hint: $X$ and $Y$ are non-negative integer valued random variables. Hence $$P(X+Y=1)=P(X=1,Y=0)+P(X=0,Y=1)$$ $$=P(X=1)P(Y=0)+P(X=0)P(Y=1).$$ Now note that $M_X(t)=\frac {64+16e^{t}+e^{2t}} {81}$. Since $Ee^{tX}=\sum e^{nt}P(X=n)$ we see that $P(X=0)$ and $P(X=1)$ are the coefficients of $e^{0t}$ and $e^{t}$. Can you finish?