Finding $P(X \leq a | Y = k)$ where $X\sim\operatorname{Exp}(\lambda)$ and $Y = [X]$ where $[\cdot]$ rounds up to the nearest integer. $Y = [X]$ where $[\cdot]$ rounds up to the nearest integer. It's given that $X \sim \operatorname{Exp}(\lambda)$. The first part asks me to show that $Y \sim \operatorname{Geo}(1-e^{-\lambda})$. The question asks me to find $P(X \leq a | Y = k) \forall x,y \in \Bbb R$ and any $a>0$. I know the formula for this in terms of the joint distribution and the marginal pmf of $Y$. I tried splitting it into cases where $a<k-1$, $k-1 \leq a < k$, $a \geq k$. For which I think $P(k-1 \leq X \leq a)$ is relevant. Am I on the right track here or not?
Thanks!
 A: From what you claim is required to show. I think $[.]$ refers to the ceiling function.
Then $P(Y=k)=P(\lceil X\rceil = k) = P(k-1< X\leq k) = \int_{k-1}^{k}\lambda e^{-x\lambda}dx = e^{-(k-1)\lambda}-e^{-k\lambda} = e^{-(k-1)\lambda}(1-e^{-\lambda})$.
So the pmf of $Y$ follows pmf of  $\text{Geo}(1-e^{-\lambda})$ distribution.
Next $Y=k$ implies $k-1< X \leq k$.
So $P(X\leq a | Y=k)=\int_{k-1}^{a}\lambda e^{-x\lambda}dx = e^{-(k-1)\lambda}-e^{a\lambda}\,\,$ if $k-1<a< k$ and $P(X\leq a | Y=k)=1$ if $a\geq k$ and is $0$ for all other  values of $a$.
So to sum up you have
$\begin{align}P(X\leq a | Y=k)=
 & {} e^{-(k-1)\lambda}-e^{-a\lambda}\,\,,k-1<a<k\\
& {} =1\,\,, a\geq k\\
&{}=0\,\,,a \leq k-1
\end{align}$
To find the conditional expectation you see that you have already evaluated the conditional cdf.
So the conditional pdf is found by just differentiating wrt a.
Or $f_{X|Y}(x)=\lambda e^{-x\lambda}\,\,,k-1<x\leq k$. And $0$ elsewhere.
So you just find the expectation by simply integrating.
$\int_{k-1}^{k}x\lambda e^{-x\lambda}dx$.
Further more you will get an expression in terms of $k$ and $\lambda$. Then if you sum that expression over all $k$.(Make some rearrangements to make it telescope) . You will get $\frac{1}{\lambda}$. Which is what is "expected" as the expectation of conditional expectation is just expectation of X.
$E[X]=E[E[X|Y]]$.
Verify this.
