Let $(N(t),\ t\geq 0)$ be a Poisson counting process with rate $\lambda=1$. Determine \begin{equation} \mathbb{P}\left(N(1,4]=6,\ N(3,7]=13\right)\label{eq:one}\tag{1} \end{equation}
Attempt:
The intervals $(1,4]$ and $(3,7]$ obviously overlap, so I cannot use the independent increment property directly. Trying to get some intuition for the problem, I sought to instead consider the (disjoint) intervals $(1, 3]$, $(3, 4]$, and $(4, 7]$ and then enumerate the particular outcomes in each interval, for which summing their probabilities will give $\eqref{eq:one}$, that is:
\begin{array}{c | c | c} N(1,3] & N(3,4] & N(4,7] \\ \hline 6 & 0 & 13 \\ 5 & 1 & 12 \\ 4 & 2 & 11 \\ 3 & 3 & 10 \\ 2 & 4 & 9 \\ 1 & 5 & 8 \\ 0 & 6 & 7 \end{array}
So that: \begin{equation} \{N(1,4]=6,\ N(3,7]=13\} = \bigsqcup_{i=0}^{6}{\{N(1,3]=6-i,\ N(3,4]=i,\ N(4,7]=13-i\}} \end{equation}
And so: \begin{align} \mathbb{P}\left(N(1,4]=6,\ N(3,7]=13\right)&=\sum_{i=0}^6{\mathbb{P}\left(N(1,3]=6-i,\ N(3,4]=i,\ N(4,7]=13-i\right)} \\ &= \sum_{i=0}^6{\mathbb{P}\left(N(1,3]=6-i\right)\mathbb{P}\left(N(3,4]=i\right)\mathbb{P}\left(N(4,7]=13-i\right)} \end{align}
Since the intervals are disjoint. From this point I believe it is a matter of replacing each probability with the Poisson PMF with the appropriate parameter for each interval and evaluating the sum. My questions are:
- Is my intuition valid here?
- If so, this seems to be a very brute force approach. Is there a deeper - or perhaps more concise - probabilistic interpretation of this problem?