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The moment generating function of $X$ is given by $M_X(t)=e^{2e^t-2}$ and that of $Y$ by $M_Y(t)=\left(\frac34e^t+\frac14\right)^{10}$. If $X$ and $Y$ are independent, what are $P(XY=0)$?

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For every nonnegative random variable $Z$, $\mathbb P(Z=0)=\lim\limits_{t\to-\infty}M_Z(t)$, called $M_Z(-\infty)$.

Using this identity for $X$ and $Y$, the independence of $X$ and $Y$ leads to $$ \mathbb P(XY\ne0)=\mathbb P(X\ne0)\mathbb P(Y\ne0)=(1-M_X(-\infty))(1-M_Y(-\infty)). $$ In the present case, $$ \mathbb P(XY=0)=1-(1-\mathrm e^{-2})(1-4^{-10})=\mathrm e^{-2}+4^{-10}-\mathrm e^{-2}4^{-10}. $$

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haha I just figured it out but thank you! –  Nerlbao Nerl Nov 19 '12 at 7:51
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Hint: $X$ has a Poisson distribution. You probably have a typo in your $M_Y(t)$, but it has a discrete distribution with a finite number of possible values, and you can find the probability of $Y=0$.

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You were right, I had a typo. –  Nerlbao Nerl Nov 19 '12 at 7:04
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$X$ is Poisson with $\lambda = 2$

$Y$ is Binomial with $p = \frac{3}{4}$ and $n=10$ given these I think you can solve for $P(XY=0)$

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