Superposition and Thinning of Poisson Processes

Question

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.

Answer 1

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?

Answer 2

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 probability^rate $0.5$ .

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 $$\mathsf E(N_2\mid N_4=4) = \sum_{n=0}^{3} \dfrac{ n\;\mathsf P(N_2=n)\cdot\mathsf P(N_4-N_2=3-n) }{ \mathsf P(N_4=3) }$$