Considering a Poisson process with parameter $\lambda$, let $N(t_2)$ denote the number of events in $(0,t_2]$ and $N(t_1, t_3)$ denote the number of events in $(t_1,t_3]$, under the assumption that $0<t_1<t_2<t_3$. I want to compute the conditional density $E(Y\mid X)$.
\begin{align} E(Y\mid X) = E(Y\mid X=x) &= \sum_y yP_{Y\mid X}(y\mid x) \\ &= \sum_y y P(Y=y\mid X=x)\\ &= \sum_y y \frac{P(Y=y,X=x)}{P(X=x)} \\ &= \sum_y y \frac{P(N(t_1, t_3)=y,N(t_2)=x)}{P(N(t_2)=x)} \\ \end{align} At this point, it is immediate how the denominator is to be computed (since $N(t_2)$ is a discrete Poisson r.v.). The problematic point is the joint distribution in the numerator $P(N(t_1, t_3)=y,N(t_2)=x)$ since the intervals are overlapping and hence the corresponding r.v.s are not independent.
One way I thought of tackling this issue was by splitting the intervals into disjoint pieces,i.e. $(0, t_1]$, $(t_1, t_2]$ and $(t_2, t_3]$ and consider the corresponding random variables $N(t_1), N(t_1, t_2)$ and $N(t_2, t_3)$. Then I want to transform the joint distribution to:
\begin{align}P(N(t_1, t_3)=y,N(t_2)=x) &= P(N(t_1, t_2) + N(t_2, t_3)=y,N(t_1) + N(t_1, t_2)=x) \\ &= P(N(t_1, t_2)=y - N(t_2, t_3), N(t_1, t_2)=x-N(t_1)) \\ &= P(y - N(t_2, t_3) = x-N(t_1)) \\ &= P(N(t_1) + N(t_2, t_3) = x+y) \end{align}
First of all, is there a mistake I am not spotting? And second, is this approach worth it (assuming it's correct)? Please feel free to suggest alternative approaches that might make this faster and more elegant/computationally less painful. Many thanks in advance.
Yes as the comment pointed out:
Note that $N(t_1, t_3) = N(t_1, t_2) + N(t_2, t_3)$. $N(t_2, t_3)$ is independent of $N(0, t_2)$ and has a Poisson distribution (independent increment): $$ N(t_2, t_3) \sim \text{Poisson} (\lambda(t_3 - t_2))$$ Conditional on $N(0, t_2)$, $N(t_1, t_2)$ has a Binomial distribution (a well known result which can be easily proved): $$ N(t_1, t_2)|N(0, t_2) = x \sim \text{Binomial}\left(x, \frac {t_2 - t_1} {t_2}\right) $$ Therefore,
$$ \begin{align*} \mathbb{E}[N(t_1, t_3)|N(0, t_2) = x] & = \mathbb{E}[N(t_1, t_2) + N(t_2, t_3)|N(0, t_2) = x] \\ & = \mathbb{E}[N(t_1, t_2)|N(0, t_2) = x] + \mathbb{E}[N(t_2, t_3)] \\ & = x \frac {t_2 - t_1} {t_2} + \lambda(t_3 - t_2) \end{align*} $$ Since this holds for every $x$ in the support of $X$, you can replace the $x$ by the random variable $X$ for calculating $\mathbb{E}[Y|X]$.