Superposition and Thinning of Poisson Processes Given two arrival processes $N_t = X_t + Y_t$, $X_t$ and $Y_t$ independent, with $Y_t$,$X_t$ each occuring with $P = 0.5$, what is $E[N_t|X_t = n]$? 
To answer this, I'm trying to find  find $P(N_t|N_s=n)$, given $t,s \geq 0$, $N\sim\text{Poisson}(\lambda)$.
$$\frac{P(N_t = M)P(N_t-N_s = n - m)}{P(N_s=n)}$$
$$\frac{e^{-t}(t)^n}{m!} \frac{e^{-\lambda (s-t)}\lambda (s-t)^{m-n}}{(n-m)!}$$
which reduces to
$$ \binom{n}{m}(\frac{t}{s})^m (1-\frac{t}{s})^{n-m}$$
which is clearly a binomial. 
However, I think I might have approached this problem from the wrong direction (I fail to see how the above is useful for solving this problem), and I'm not sure what to do now.
 A: Just apply the Linearity of Expectation.
$$\begin{align}\mathsf E(N_t\mid X_t) & = \mathsf E(X_t+Y_t\mid X_t) \\[1ex] & = \mathsf E(X_t\mid X_t)+\mathsf E(Y_t\mid X_t) \\[1ex] & = \ldots\end{align}$$
Which, if $X_t$ and $Y_t$ are independent, then simplifies into an easy solution.   Are they?
A: 
Yeah, I was thinking that too, which is why I'm confused. I'm not even sure if the two are even remotely related. The expectation I was looking for in that part of my answer isn't related to $X_t$ - it's something along the lines of $E[N_2 =1\mid N_4 =3]$. I know that $N_t$ is a superposition of $X_t$ and $Y_t$, and that each separate process occurs with probabilityrate $0.5$ . 

If $X_t \perp Y_t,\textsf{ and } X_t, Y_t \mathop{\sim}^\textsf{iid} \mathcal P(0.5t)$ , then:
$$\begin{align}\mathsf P(N_t=k) & = \sum_{j=0}^k \frac{(0.5t)^{j} e^{-0.5t}}{j!}\frac{(0.5t)^{k-j} e^{-0.5t}}{(k-j)!} \\[1ex] & = (0.5t)^k e^{-t}\sum_{j=0}^{k}\frac{1}{j!(k-j)!} \\[1ex] & = (0.5t)^k e^{-t} 2^k \frac 1 {k!} \\ & =\dfrac{t^k e^{-t}}{k!}\end{align}$$
Which is to say, $N_t\sim \mathcal{Pois}(t)$ (as expected: the sum of the counts of two i.i.d. poison processes with rate $\lambda$ is the count of a poison process with rate $2\lambda$.)

So are you looking for $$\mathsf E(\{N_2=1\} \mid \{N_4=3\}) = \dfrac{ \mathsf P(N_2=1)\cdot\mathsf P(N_4-N_2=2) }{ \mathsf P(N_4=3) }$$
Or $$\mathsf E(N_2\mid N_4=4) = \sum_{n=0}^{3} \dfrac{ n\;\mathsf P(N_2=n)\cdot\mathsf P(N_4-N_2=3-n) }{ \mathsf P(N_4=3) }$$
