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Suppose that $X$ and $Y$ are independent, $X\thicksim \text{Exp}(\lambda)$ and $Y\thicksim\text{Geom}(p).$ Let $\lfloor x\rfloor$ be the floor function (largest interger which is at most $x$). Find $\mathbb P(\lfloor X\rfloor=Y).$

$\textbf{My Attempt:}$

First I observe that \begin{align*}\mathbb P(\lfloor X\rfloor=n) &=\mathbb P(n\leq X\leq n+1)\\ &=\mathbb P(n< X\leq n+1)\\ &=\mathbb P(X\leq n+1)-\mathbb P(X<n)\\ &=e^{-n\lambda}-e^{-(n+1)\lambda}, \end{align*} for all non-negative integers $n$.

Then using the independence of $X$ and $Y$, I compute the following \begin{align*} \mathbb P(\lfloor X\rfloor=Y)&=p-e^{\lambda}+\sum_{n=1}^{\infty}\mathbb P(\lfloor X\rfloor=n)\mathsf(Y=n)\\ &=p-e^{-\lambda}+\sum^{\infty}_{n=1}p(1-p)^{n}(e^{-n\lambda}-e^{-(n+1)\lambda})\\ &=p+e^{-\lambda}+p\left[\sum^{\infty}_{n=1}(e^{-\lambda}-e^{-\lambda}p)^{n}-e^{-\lambda}\sum^{\infty}_{n=1}(e^{-\lambda}-e^{-\lambda}p)^{n}\right]\\ &=p-e^{-\lambda}+p\left[\frac{1}{1-e^{-\lambda}+e^{-\lambda}p}-\frac{e^{-\lambda}}{1-e^{-\lambda}+e^{-\lambda}p}\right]. \end{align*}


Is my work above correct? I am doubtful about my summation.

Any feedback is much appreciated. Thank you for your time.

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There is a mistake is in the formula for a geometric sum. The correct formula is $\sum\limits_{k=1}^{\infty} r^{n}=\frac r {1-r}$ and not $\frac 1 {1-r}$.

Also the first part should be $p-pe^{-\lambda}$.

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