I need to calculate $P[N_{s}=k||N_{u},\,u \geq t]\,(s \leq t)$ for the Poisson process. However, I have been instructed to use the following example in order to do so:
Example: The Poisson process $[N_{t}:t \geq 0]$ has independent increments. Suppose that $0 \leq t_{1} \leq \cdots \leq t_{k} \leq u$. The random vector $(N_{t_{1}},N_{t_{2}}-N_{t_{1}}, \cdots , N_{t_{k}} – N_{t_{k-1}})$ is independent of $N_{u}-N_{t_{k}}$, and so $(N_{t_{1}},N_{t_{2}}, \cdots , N_{t_{k}})$ is independent of $N_{u}-N_{t_{k}}$. If $J$ is the set of points $(x_{1},\cdots , x_{k}, y)$ in $\mathbb{R}^{k+1}$ such that $x_{k}+y \in H$, where $H \in \mathcal{R}^{1}$ $[Note: (\Omega,\mathcal{F}) = (\mathbb{R}^{1}, \mathcal{R}^{1}]$.
And, if $\nu$ is the distribution of $N_{u}-N_{t_{k}}$, then $P[(x_{1},\cdots , x_{k}, N_{u}-N_{t_{k}}) \in J] = P[x_{k}+N_{u}-N_{t_{k}} \in H] = \nu(H-x_{k})$.
This also holds if $k =1$, and hence $P[N_{u} \in H|| N_{t_{1}}, \cdots , N_{t_{k}}] = P[N_{u} \in H||N_{t_{k}}]$.
The Poisson process therefore also has the Markov Property (which is a consequence solely of the independence of the increments): $$\text{for}\,\, k\geq 1, 0 \leq t_{1} \leq \cdots \leq t_{k} \leq u,\, \text{and}\, H \in \mathcal{R}^{1},\\ P[X_{u} \in H || X_{t_{1}}, \cdots , X_{t_{k}}] = P[X_{u} \in H || X_{t_{k}}]$$
The extended Markov Property also follows: $$P[X_{u} \in H || X_{s},\, s \leq t] = P[X_{u} \in H || X_{t}], \,\, t \leq u. $$
However, I'm really not sure how this will help.
I'm starting with $P[N_{s}=k||N_{u}]$, and I know that this means that I am trying to find the probability of $k$ occurrences before $s$ given the number of occurrences before $u$? But what would that be? I know that if I had something like $N_{S} = 0$ and $N_{t} =i$, $i = 0,1,2, \cdots$, due to independent and stationary increments, I would obtain $P(N_{s}=0 || N_{t}=i) = \frac{P(N_{s}=0,\, N_{t}=i)}{P(N_{t}=i)} = \left( 1 – \frac{s}{t}\right)^{t}$, but in this case, I don't know what to do.
Could somebody please help me finish this problem? We glazed over Poisson processes and conditional expectations/conditional probability in general very quickly at the end of the semester, and now I'm responsible to know how to do them. So, the more detailed your answer, the better.
I thank you ahead of time for your time and patience.
Best Answer
I would make the following reasoning:
$ P(N_s = k \mid N_u) = \dfrac{P(N_s = k, N_u = h)}{P(N_u = h)}$
Of course if $h<k$ then $ P(N_s = k \mid N_u) = 0$
We can write $N_u = N_s + N_u - N_s$ so that
the event $N_s = k \cap N_u = h$
is equivalent to $N_s = k \cap N_u - N_s = h -k$ of course if $h \ge k$
Clearly $N_s$ and $N_u - N_s$ are independent hence we can write:
$P(N_s = k , N_u - N_s = h -k) = P(N_s = k)P(N_u - N_s = h -k)$
and finally:
$ P(N_s = k \mid N_u) =\dfrac{P(N_s = k)P(N_u - N_s = h -k)}{P(N_u = h)}$
Of course you know that
$N_s \sim Pois(\lambda s)$
$N_u \sim Pois(\lambda u)$
$N_u - N_s \sim Pois(\lambda(u-s))$
And I will leave to you the calculation.
Greetings