Conditional expectation of poisson procces problem

Question

Let $N_{t}^{i}$ - be three independent Poisson processes of intensity $1$. $\tau$ = $\inf\{t: \,N_{t}^{3} = 1\}$, $X^{i}$ = $N_{\tau}^{i}$ (means that $X^{i}$ - the values of the first two processes at the moment the third level reaches $1$). Find the conditional expectation $E(X^{1}|X^{2})$.

Answer 1

First of all, since $N^3$ is a Poisson process with intensity $1$, $\tau$ is exponentially distributed with parameter $1$ and is independent of $N^1$ and $N^2$. Since $X^2$ takes values in $\mathbb{N}$, we have $$\mathbb{E}[X^1|X^2] = \sum_{n=0}^\infty \frac{\mathbb{E}[X^1 \mathbf{1}_{X^2=n}]}{\mathbb{P}(X^2=n)} \mathbf{1}_{X^2 = n}. $$ Start by computing $\mathbb{P}(X^2 = n)$: $$\begin{align}\mathbb{P}(N^2_\tau = n) &= \int_0^\infty \mathbb{P}(N^2_t = n|\tau = t)\mathbb{P}(\tau \in dt) \\ &= \int_0^\infty \mathbb{P}(N^2_t = n) e^{-t} \, dt \\ &= \int_0^\infty \frac{t^n}{n!}e^{-2t}\, dt = \frac{1}{2^{n+1}}. \end{align}$$ Here I used the independence of $N^2$ and $\tau$ in the second equality. Similarly, $$\begin{align} \mathbb{E}[N^1_\tau \mathbf{1}_{N^2_\tau = n}] &= \int_0^\infty \mathbb{E}[N_t^1 \mathbf{1}_{N_t^2 = n} | \tau = t] e^{-t} \, dt \\ &= \int_0^\infty \mathbb{E}[N_t^1]\mathbb{P}(N_t^2 = n) e^{-t} \, dt\\ &= \int_0^\infty \frac{t^{n+1}}{n!} e^{-2t} \, dt = \frac{n+1}{2^{n+2}}. \end{align}$$ Finally, we get that $$\mathbb{E}[X^1|X^2] = \sum_{n=0}^\infty \frac{n+1}{2} \mathbf{1}_{X^2 = n} = \frac{X^2+1}{2}.$$