# If $X\thicksim \text{Exp}(\lambda)$ and $Y\thicksim\text{Geom}(p)$ are independent, find $\mathbb P(\lfloor X\rfloor=Y)$

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 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.

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 $$p-pe^{-\lambda}$$.