Difference of two discrete random variables Given two random variables $X$ and $Z$ on non negative integers. Supposing that 


*

*$Z$ is a Poisson distribution of Poisson constant $\lambda$.

*$X\le Z$ and that $\forall n\ge0,\  \forall k\le n,\ P(X=k|Z=n)=\displaystyle \binom nk \cdot p^k (1-p)^{n-k},\ \ 0<p<1.\ $


Show that $X$ and $Y=Z-X$ are independent Poisson variables.
I was able to show that $X \sim \mathcal P(\lambda p)$ 
but for $Y=Z-X$, I found some problems, I tried the following
$$P(Z-X=k)=\sum^{+\infty}_{i=0}P(X=i)\cdot P(Z=k+i)$$ I got an expression that I couldn't sum to infinity and then I remembered that I assumed independence in this approach, now I'm lost and I dunno what to do.
 A: A start: One can compute. Multiply the conditional probability $\Pr(X=k|Z=n)$ by $\Pr(z=n)$, which you know, to find the probability that $X=k$ and $Z=n$.
Let $Y=Z-X$.  We may want to find the distribution of $Y$. As you were doing, we find an expression for $\Pr(Y=y)$. This is the sum over all $n \ge y$ of the probability that $Z=n$ and $X=n-y$. My calculation gives
$$\sum_{n=y}^{\infty} e^{-\lambda} \frac{\lambda^n}{n!} \binom{n}{n-y}p^{n-y}(1-p)^y.$$
This is less fearsome than it looks.  Here are some steps to the simplification. 
(i) Take the $e^{-\lambda}$ outside the sum, also the $(1-p)^y$.
(ii) Write $\lambda^n$ as $\lambda^y \lambda^{n-y}$, and take the $\lambda^y$ part outside.
(iii) Look at the $\dfrac{1}{n!}\dbinom{n}{n-y}$ part. Express the binomial coefficient in terms of factorials. The $n!$ cancel, which is nice. We are left with $\dfrac{1}{(n-y)!y!}$. Take the $y!$ part outside.
Inside we are left with $\sum_{n=y}^\infty \frac{1}{(n-y)!} (\lambda p)^{n-y}$.
This sum is easy to recognize. So now we know the distribution of $Y$. That will help with the independence issue.
It would be more direct to find the joint distribution function of $X$ and $Y$. 
A: As the last sentence in André's answer suggests, it is easier to find the joint
distribution of $X$ and $Y$. 
Suppose that we are given that $Z = n$. Then, the conditional distribution
of $X$ is given to be $\tt{Binomial}$$(n,p)$, while the conditional distribution
of $Y = Z-X = n-X$ can be deduced to be  $\tt{Binomial}$$(n,1-p)$.  Think of
this as $X$ is the number of successes and $Y$ the number of failures
on $n$ independent trials, where the probability of success is $p$.  Notice
also that $X$ and $Y$ are not conditionally independent random variables
given that $Z = n$; they are very much dependent random variables since
$X+Y=n$. But we do have that for $k, \ell \geq 0$,
$$p_{X,Y\mid Z=n}(k,\ell\mid Z=n) = P\{X=k, Y=\ell\mid Z=n\} 
= \begin{cases} \binom{n}{k}p^k(1-p)^{n-k}, & 0\leq k\leq n, \ell = n-k,\\
\quad \\
0, & \text{otherwise.}\end{cases}$$
Consequently, the unconditional joint probability mass function can be found
via the law of total probability as
$$\begin{align*}
p_{X,Y\mid Z=n}(k,\ell) 
&= \sum_{n=0}^\infty p_{X,Y\mid Z=n}(k,\ell\mid Z=n)P\{Z=n\}\\
&= \binom{k+\ell}{k}p^k(1-p)^{\ell}\cdot \exp(-\lambda)\frac{\lambda^{k+\ell}}{(k+\ell)!}\\
&= \frac{(k+\ell)!}{k!\ell!}(\lambda p)^k(\lambda(1-p))^{\ell}\cdot(\exp(-\lambda p)\exp(-\lambda(1-p))\frac{1}{(k+\ell)!}\\
&= \exp(-\lambda p)\frac{(\lambda p)^k}{k!}
\cdot\exp(-\lambda(1-p))\frac{(\lambda(1-p))^{\ell}}{\ell!}
\end{align*}$$
where, of course, we have assumed that $k, \ell \geq 0$.
The result shows that $X$ and $Y$ are independent random 
variables; in fact,
independent Poisson random variables with parameters $\lambda p$ and
$\lambda(1-p)$ respectively.
