Distribution of the time for the first ocurrence in a Poisson Process given the number of events

Question

I am trying to solve this problem:

Let N(t) a Poisson Process of rate $\lambda$ where ocurrences are type I with probability $p$. Given that $N(t_0) = n$, what is the distribution of the waiting time for the first arrival of type I in the interval.

What I tried to do is to separate it in two processes, $N_1(t)$ of rate $\lambda p$ and $N_2(t)$ of rate $\lambda (1-p)$, so I did:

$$P(T_1 \le t | N_1(t_0) + N_2(t_0) = n) = \sum_{i=1}^n P(T_1 \le t | N_1(t_0) = n - i) P(N_2(t_0) = i)$$

Now I know that given $N_1(t_0) = n - i$, the distribution of the arrival times for $N_1$ are the same as the order statistics of $n-i$ uniform random variables. So in this case the conditional distribution of $T_1$ would be a Beta distribution or something like that.

However the beta distribution is not mentioned in my book before this question, so I think I cannot use that.

Answer 1

Time of the first arrival in Poisson process with rate $\lambda$ is exponential random variable $\sim Exp(\lambda)$. It is easy to prove $$P(T_1>t)=P(\text{no arrivals in }(0,t])=e^{-\lambda t}$$ so the CDF is $$ \begin{align} F_{T_1}(t) = \left\{ \begin{array}{l l} 1-e^{-\lambda t} & \quad t>0 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align}$$ In your case, you have Poisson process with rate $p\lambda$ (obtained after splitting the original Poisson process).

Answer 2

The distribution of the waiting time for the first arrival will be Exponential. It follows from the principal of count-time duality.

$N(t)$ is a Poisson Process with rate $\lambda$ where $N(t)$ is the number of occurrences at or before time $t$. Let the waiting time for the first occurrence ($T_1$) is greater than $t$. This means that there are no occurrences from time $0$ to $t$, making $N(t)=0$. This makes the event $T_1 > t$ and $N(t)=0$ same [called as count-time duality and connects a continuous ($T_1$) and discrete ($N(t)$) RV].

So, we have $P(T_1 > t) = P(N(t) = 0) = \frac{e^{-\lambda t (\lambda t)^0}}{0!} = e^{-\lambda t}$ [as $N(t)$ is a Poisson Process]. $P(T_1 \leq t) = 1 - e^{-\lambda t}$, making $T_1 \sim Expo(\lambda)$.

Hence, the time until the first arrival in a Poisson Process of rate $\lambda$ has an Exponential distribution with parameter $\lambda$.