Finding $E(N_{2}\mid N_{1} + N_{3})$ where $N_{t}$ is a Poisson process $N_{t}$ is a Poisson process.
So: We can try to find it by definition:
$$
\phi (z) = E(N_{2}\mid N_{1} + N_{3} = z)
$$
And use total probability rule, but it'll just leave us with some series, which might diverge:
\begin{align}
\phi(z) &= \sum_{k = 0}^{\infty}kP(N_{2}=k\mid N_{1}+N_{3} = z) 
\\&=  \sum_{k=0}^{\infty} k\sum_{j=0}^{\infty}\frac{P(N_{1}=j, N_{2} - N_{1} = k - j, N_{3} - N_{2} = z - k - j)}{P(N_{1}=j, N_{3}-N_{1} = z - 2j)} 
\\&= \cdots =\sum_{k=0}^{\infty} k\sum_{j=0}^{\infty}C_{z-2j}^{k-j}2^{2j - z}
\end{align}
And I was wondering if there was some easier way to compute this conditional expectation.
 A: Consider the three increments
$$X = N_1, \qquad Y = N_2 - N_1, \qquad Z = N_3 - N_2 .$$
These are independent Poisson random variables each with parameter $\lambda$. Instead of conditioning on $N_1 + N_3$, suppose we condition on both $N_1$ and $N_3 - N_1$. This amounts to conditioning on both $X$ and on $Y+Z$:
$$ \begin{align*} E(N_2 | N_1, N_3-N_1) &= E(X+Y | X,Y+Z) \\ &= X + E(Y | X, Y+Z) \\ &= X + \frac 1 2 (Y+Z) . \end{align*}$$
The final line comes from the fact that if $U$ and $V$ are independent Poisson variables with paremeters $\lambda$ and $\mu$, respectively, then the distribution of $U$ conditional on $U+V$ is binomial with probability $\frac{\lambda}{\lambda + \mu}$ and with $U+V$ trials. See the second bullet point here for this fact. This has now worked out very conveniently for us, because
$$ X + \frac 1 2 (Y+Z) = \frac 1 2 (N_1 + N_3) $$
so that, by repeated conditioning,
$$ \begin{align*} E(N_2 | N_1 + N_3) &= E\Big( E(N_2 | N_1, N_3-N_1) \, \Big| \, N_1 + N_3 \Big) \\ &= E \Big( \frac 1 2 (N_1 + N_3) \, \Big| \, N_1 + N_3 \Big) \\ &= \frac 1 2 (N_1 + N_3) . \end{align*}$$
