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Let $X$ and $Y$ be independent Random Variables, where $X$ is Poisson distributed and $Y$ is geometrically distributed. Determine $P(X=Y)$.

Since the Poisson distribution is defined on $N_{0}$ and and the geometric distribution on $N$ or on $N_{0}$ depending on the power. Hence I choose the way where the geometric distribution is defined on $N_{0}$.

Thus,

$\begin{align}\mathsf P(X=Y) &= \sum_{k=0}^\infty \mathsf P(X=k,Y=k) \\[1ex] &= \sum_{k=0}^\infty \mathsf P(X=k)~\mathsf P(Y=k)\\[1ex] &= \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k}}{k!}\cdotp(1-\theta)\theta^k \\[1ex] & = e^{-\lambda}(1-\theta) \sum_{k=0}^\infty \frac{(\lambda \theta)^{k}}{k!} \\[2ex] &= e^{-\lambda}(1-\theta)~e^{\lambda \theta} \\[2ex] &= e^{\lambda(\theta -1)}(1-\theta)\end{align}$

Is this correct? I would appreciate any hint!

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  • 5
    $\begingroup$ Yes. $ $ $ $ $ $ $\endgroup$ – Did Dec 4 '16 at 10:50
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Yes, as long as they are independent, then you can use this formula to do calculations.

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