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This is a problem from A First Course in Probability, Sheldon Ross Ed. 7 Problem 7.75. I am really stumped on this one.

The MGF of X is given by $M_X(t)=exp(2e^t-2)$ and the MGF of Y is $M_Y(t)= (\dfrac{3}{4}e^t+\dfrac{1}{4})^{10}$ If X and Y are independent, what is P{X+Y=2}?

I know, from the MGFs that X~Poisson(2) and Y~Binomial(10,3/4). I know that the MGF of X+Y is the product of the two MGFs. I do not recognize that product as the MGF of a well known distribution. I do not know what the distribution of the product of a poisson and binomial is. Any thoughts are appreciated!

Max
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You want $\Pr(X+Y)=2$. This is $$\Pr(X=0)\Pr(Y=2)+\Pr(X=1)\Pr(Y=1)+\Pr(X=2)\Pr(Y=0).$$ You know the distributions of $X$ and of $Y$, so the required probabilities can be computed.

André Nicolas
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  • Haha so simple. I was reading into this way too hard. Thanks. Is there any way to derive a pdf/cdf from an MGF? – Max Nov 12 '13 at 07:08
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    There are ways, inverse Laplace transform. I doubt it would give anything useful in this case. – André Nicolas Nov 12 '13 at 07:14