Computing the following $\sum\limits_{k=0}^\infty k\binom{n}{k}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n}$

Question

Let $X,Y$ be two independet Poisson variables with parameters $\mu,\lambda>0$. Let $N:=Y+X$ what is $\mathbb{E}(X\vert N=n)$?

I already computed $P(X=k\vert N=n)$ for $k,n\in \mathbb{Z}_{+}$ which is $$P(X=k\vert N=n)=\binom{n}{k}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n}$$ if $n>k$ else $0$.

I know that $n=k+j$. But now I get stucked $$\mathbb{E}(X\vert N=n)=\sum\limits_{k=0}^\infty k\binom{n}{k}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n}=\sum\limits_{k=0}^\infty k\frac{n!}{k!(n-k)!}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n}$$ $$\sum\limits_{k=0}^\infty \frac{(k+j)!}{(k-1)!(j)!}\frac{\mu^{j}\lambda^k}{(\mu+\lambda)^{k+j}}$$

How can I compute the expected value?

Answer 1

To deal with sums over $k$ whose summands contain powers in $k$, you should remember the identity:

$$k^m\lambda^k = \left(\lambda\frac{\partial}{\partial \lambda}\right)^m\left[\lambda^k\right]$$

Together with binomial theorem, you can compute your sum as: $$\begin{align} \sum_{k=0}^\infty k\binom{n}{k} \frac{\mu^{n-k} \lambda^k}{(\mu+\lambda)^n} &= \frac{1}{(\mu+\lambda)^n}\sum_{k=0}^n k\binom{n}{k} \mu^{n-k}\lambda^k\\ &= \frac{1}{(\mu+\lambda)^n}\sum_{k=0}^n \lambda\frac{\partial}{\partial \lambda}\left[\binom{n}{k} \mu^{n-k}\lambda^k\right] = \frac{1}{(\mu+\lambda)^n}\lambda\frac{\partial}{\partial \lambda}\left[\sum_{k=0}^n \binom{n}{k} \mu^{n-k}\lambda^k\right]\\ &= \frac{1}{(\mu+\lambda)^n}\lambda\frac{\partial}{\partial \lambda}(\mu+\lambda)^n = \frac{n\lambda}{\mu+\lambda} \end{align} $$

Answer 2

You may use ${\frac {1}{(1-z)^{\alpha +1}}}=\sum _{n=0}^{\infty }{n+\alpha \choose n}z^{n}$ as follows:

\begin{align} \mathbb{E}(X\vert N=n)&=\sum\limits_{k=0}^\infty k\binom{n}{k}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n}=\sum\limits_{k=0}^\infty k\frac{n!}{k!(n-k)!}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n} = \sum\limits_{k=1}^\infty k\frac{n!}{k!(n-k)!}\frac{\mu^{n-k}\lambda^k}{(\mu+\lambda)^n} \\ &=\sum\limits_{k=1}^\infty \frac{(k+j)!}{(k-1)!(j)!}\frac{\mu^{j}\lambda^k}{(\mu+\lambda)^{k+j}} =(j+1)\sum\limits_{k=1}^\infty \frac{(k+j)!}{(k-1)!(j+1)!}\frac{\mu^{j}\lambda^k}{(\mu+\lambda)^{k+j}} \\ &=(j+1)\sum\limits_{k=1}^\infty \binom{k+j}{k-1} \frac{\mu^{j}\lambda^k}{(\mu+\lambda)^{k+j}} =(j+1)\sum\limits_{k=1}^\infty \binom{k+j}{k-1} (\frac{\mu}{\mu+\lambda})^j (\frac{\lambda}{\mu+\lambda})^k \\ & =(j+1)(\frac{\mu}{\mu+\lambda})^j \sum\limits_{k=1}^\infty \binom{k+j}{k-1} (\frac{\lambda}{\mu+\lambda})^k = (j+1)(\frac{\mu}{\mu+\lambda})^j \sum\limits_{n=0}^\infty \binom{n+j+1}{n} (\frac{\lambda}{\mu+\lambda})^{n+1} \\ &= (j+1)(\frac{\mu}{\mu+\lambda})^j (\frac{\lambda}{\mu+\lambda})\sum\limits_{n=0}^\infty \binom{n+j+1}{n} (\frac{\lambda}{\mu+\lambda})^{n} \\ &= (j+1)(\frac{\mu}{\mu+\lambda})^j (\frac{\lambda}{\mu+\lambda})\dfrac{1}{(1-\frac{\lambda}{\lambda +\mu})^{j+2}} = (j+1)(\frac{\mu}{\mu+\lambda})^j (\frac{\lambda}{\mu+\lambda})(\dfrac{\lambda+\mu}{\mu})^{j+2} \\ &=(j+1)(\frac{\mu}{\mu+\lambda})^j (\frac{\lambda}{\mu+\lambda})(\dfrac{\lambda+\mu}{\mu})^{j+2} = (j+1)\frac{\lambda(\lambda +\mu)}{\mu^2} \end{align}