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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.

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1 Answer 1

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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

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  • $\begingroup$ what calculation? Do you mean plugging them into the Poisson distribution formulas or something else? The way you said it sounded so ominous! ;) $\endgroup$ Commented Dec 21, 2017 at 16:51
  • $\begingroup$ :) .. yes I am sorry for that. I mean the explicit calculation of the probability plugging in the poisson pmf. $\endgroup$
    – gioxc88
    Commented Dec 21, 2017 at 16:53
  • $\begingroup$ I was wondering, if I wanted to find $P(T>t)$ using the Poisson pmf, how would I do it? I know how to do that for a finite number of trials, but if I don't know how many trials there are, like in this problem: stats.stackexchange.com/questions/319918/… I am at a loss. Thank you. $\endgroup$ Commented Dec 21, 2017 at 17:02
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    $\begingroup$ $P(T>t)$ is simply $1 - P(T \le t)$ $\endgroup$
    – gioxc88
    Commented Dec 21, 2017 at 17:05
  • $\begingroup$ What is T? Where did it appear from all of a sudden? $\endgroup$
    – Zahava Kor
    Commented Dec 21, 2017 at 19:14

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