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Let $N_t$ be a Poisson process of rate $\lambda$.

1)What is $P(N_s = 1|N_t =1)$ for $0

2) Find the distribution of the time of the first point of the process, conditional on the event that exactly one point occurs in the interval [0,t]?

I know $P(N_t = 1)=e^{-\lambda t}(\lambda t)$. I was wondering based on the memoryless property for Poisson processes whether $P(N_s = 1|N_t =1)=e^{-\lambda t}(\lambda t)$ as well since the distribution of $(N_t,t>t_0)$ conditional on the process $(N_t,t\ge t_0)$ depends only on the value of $N_{t_0}$

Additionally, I am not sure about part 2), would this also be a Poisson process of rate $\lambda$?

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    Look at http://math.stackexchange.com/a/379366/413376 – 2017-02-28

2 Answers 2

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Conditionally on $\{N_t=1\}$, the first jump is uniformly distributed inside $[0,\,t]$. The probability that it belongs to $[0,s]$ is the ratio of lengths: $$P(N_s = 1|N_t =1) = \frac{s}{t}.$$

The answer to the second question is already contained here.

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    how is the second question contained in $s/t$? The above is the probability of the counting process, whereas the second question is asking about the distribution of the time of the first point of the process conditioned on one point occurring in the $[0,t]$ interval. I'm confused by this relation. Thank you for your help. – 2017-02-28
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    @z8910 The c.d.f. of the time $T_1$ of the first point of the process, conditional on the event that exactly one point occurs in the interval $[0,t]$ is $P(T_1\leq s | N_t=1)$. It is the same as the probability that exactly one point occurs in the interval $[0,s]$ given that exactly one point occurs in the interval $[0,t]$: $P(T_1\leq s | N_t=1)=P(N_s =1 | N_t=1)=s/t.$ It means that this conditional distribution is Uniform on $[0,t]$. – 2017-03-01
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I know $P(N_t = 1)=e^{-\lambda t}(\lambda t)$.

Yes.   Also written as $~\mathsf P(N_{(0;t]} = 1)~=~e^{-\lambda t}(\lambda t)~$, where $N_{(0;t]}$ is the count of point-events in the interval $(0;t]$.

I was wondering based on the memoryless property for Poisson processes whether $P(N_s = 1|N_t =1)=e^{-\lambda t}(\lambda t)$ as well since the distribution of $(N_t,t>t_0)$ conditional on the process $(N_t,t\ge t_0)$ depends only on the value of $N_{t_0}$

No.   Not in the slightest.

However, $~\mathsf P(N_{(0;s]}=1)~=~ e^{-\lambda s}\lambda s~$ and $~\mathsf P(N_{(s;t]}=0)~=~e^{-\lambda(t-s)}~$, because the counts of events in an interval depends only on the length of the interval (which is what you were thinking of).

Also the counts of Poisson point-events in disjoint intervals are independent.

Use that and the definition of conditional probability.

$$\mathsf P(N_{(0;s]}=1\mid N_{(0;t]}=1) ~=~\dfrac{\mathsf P(N_{(0;s]}=1, N_{(s;t]}=0)}{\mathsf P(N_{(0;t]}=1)}$$

This, as NCh spoils, simplifies into a nice result.