Distribution of the time of the first point of the process, conditional on the event that exactly one point occurs in the interval $[0,t]$

Question

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

1)What is $P(N_s = 1|N_t =1)$ for $0<s<t$?

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 $\lambda$?

Answer 1

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.

Answer 2

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.