A convolution involving binomials Given $$f(i)\gt0,\:g(i)>0,\:i =0,1,2,3,...\:$$and$$\sum_{i=0}^{\infty}f(i) = 1,\sum_{i=0}^{\infty}g(i) = 1$$Prove that, if$$\frac{g(l-k)f(k)}{\sum_{i=0}^{l}f(i)g(l-i)}=\binom{l}{k}p^k(1-p)^{l-k}\: when\:0\le k\le l,\:where\: 0\lt p\lt1$$then$$f(i) = e^{-ru}\frac{(ru)^i}{i!},\: g(i) = e^{-u}\frac{(u)^i}{i!},\:i =0,1,2,3,...\:where\:u\gt 0$$ 
Progress: Let $F(s)$ and $G(s)$ be the generating functions of $f(i)$ and $g(i)$ then the problem becomes $$\frac{F^{(k)}(0)G^{(l-k)}(0)}{\frac{1}{l!}\sum_{i=0}^{l}\binom{l}{i}F^{(i)}(0)G^{(l-i)}(0)} = p^{k}(1-p)^{(l-k)}$$ So I eliminated one binomial but introduced another!
 A: Let $F,G$ be independent random variables distributed according to $f,g$, and let $H \sim F+G$. You are given that for all $k,\ell$,
$$
\Pr[F=k|F+G=\ell] = \binom{\ell}{k} p^k (1-p)^{\ell-k}.
$$
That gives us
$$
\Pr[F=k] = \sum_{\ell=0}^\infty \binom{\ell}{k} p^k (1-p)^{\ell-k} \Pr[F+G=\ell] = \frac{p^k}{(1-p)^k} E[\binom{H}{k} (1-p)^H].
$$
Let $\rho = p/(1-p)$.
So we must have
$$
1 = \sum_{k=0}^\infty \Pr[F=k] = E[\sum_{k=0}^\infty \rho^k \binom{H}{k} (1-p)^H] = E[(1+\rho)^H(1-p)^H],
$$
which holds since $1+\rho = 1/(1-p)$. In the same way we can compute $E[F]$:
$$
E[F] = E[\sum_{k=0}^\infty H \rho^k \binom{H-1}{k-1} (1-p)^H] = \rho E[H (1+\rho)^{H-1} (1-p)^H] = \frac{\rho}{1+\rho} E[H] = pE[H].
$$
More generally, the same computation gives
$$
\begin{align*}
E[F(F-1)\cdots(F-t+1)] &= p^t E[H(H-1)\cdots(H-t+1)], \\
E[G(G-1)\cdots(G-t+1)] &= (1-p)^t E[H(H-1)\cdots(H-t+1)].
\end{align*}
$$
We also know that $H = F+G$. For $t = 2$, we get
$$
E[H(H-1)] = E[(F+G)(F+G-1)] = E[F(F-1)] + E[G(G-1)] + 2E[F]E[G] = (p^2+(1-p)^2)E[H(H-1)] + 2p(1-p)E[H]^2.
$$
Rearranging, we get $2p(1-p) E[H(H-1)] = 2p(1-p)E[H]^2$, and so
$$
E[H(H-1)] = E[H]^2.
$$
In this way, we can get all moments of $H$ as functions of $E[H]$; once we do that, we derive all the moments of $F$ and $G$. If you do that then you should discover that $F,G,H$ are all Poisson random variables, with $E[F] = pE[H]$ and $E[G] = (1-p)E[H]$.
