How to calculate conditional expectation of two independent poison r.v. ($E(N_1|N_1+N_2)$)

Question

Problem:

$N_1, N_2$ are 2 independent Poisson random variables, where $N_i \sim \mathcal{P}(\lambda_i)$. Please calculate $E(N_1\mid N_1+N_2)$ and $E(N_1+N_2\mid N_1)$.

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I stuck at $$E(N_1\mid N_1+N_2) = \sum_{i=0}^niP(N_1=i\mid N_1+N_2=n)=\sum_{i=0}^ni\binom{n}{i}\frac{\lambda_1^n\lambda_2^{n-i}}{{(\lambda_1+\lambda_2)}^n},$$ since I cannot simplify this formula.

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Please help me (A totally different method is also fine.), thx.

Answer 1

It can be derived that $N_1|N_1 + N_2$ ~ $Bin(n,\frac{\lambda_1}{\lambda_1 + \lambda_2 })$ where $T = N_1 + N_2$ and $n$ is the observed value of $T$. i.e. $$P(N_1 = i|N_1 +N_2 = n) = \binom{n}{i}\frac{\lambda_1^i}{{(\lambda_1 + \lambda_2)}^i}*\frac{\lambda_2^{n-i}}{(\lambda_1 + \lambda_2)^{n-i}}$$

Therefore $$ E(N_1|N_1 +N_2) = n \frac{\lambda_1}{\lambda_1 + \lambda_2} $$

Since $N_1$ and $N_2$ are independent, $$E(N_1 + N_2|N_1) = E(N_1|N_1) + E(N_2|N_1)$$ $$\implies E(N_1 + N_2|N_1) = N_1 + E(N_2) = N_1 + \lambda_2$$

Answer 2

Hint: $ \sum\limits_{i=0}^{n} i\binom n i t^{i}=\sum\limits_{i=1}^{n} \frac {n!} {(i-1)!(n-i)!} t^{i}=nt \sum\limits_{j=0}^{n-1}\binom {n-1}j t^{j}=nt(1+t)^{n-1}$. Take $t=\frac {\lambda_1} {\lambda_2}$.

$E(N_1+N_2|N_1)=E(N_1|N_1)+E(N_2|N_1)=N_1+EN_2=N_1+\lambda_2$.