Conditional distribution of event times in Poisson Process

Question

I want to derive the conditional distribution of the event times $T_1, T_2, ..., T_n$ of a Poisson process $(N(t):t\ge0)$ with intensity function $\rho(t)$, given that $N(\tau)=n$, using the properties of Poisson processes.

Is the below approach correct? How is this solved properly?

Let me break down the steps to find the conditional distribution:

1. Definition of Intensity Function $\rho(t)$: The intensity function $\rho(t)$ of a Poisson process represents the expected rate of events occurring at time $\rho(t)$. It is given that the intensity function is $\rho(t)$.

2. Distribution of Event Times: The time between consecutive events in a Poisson process follows an exponential distribution with parameter $\rho(t)$. In other words, if $W_i$ represents the time between the $i-1$-th and $i$-th event, then $W_i$ is exponentially distributed with parameter $\rho(t)$.

3. Conditional Distribution: Given that $N(\tau)=n$, it means there are $n$ events in the interval $[0, \tau]$. Therefore, $T_1$ will be the time of the first event, $T_2$ will be the time of the second event, and so on, up to $T_n$ which will be the time of the $n$-th event. These event times are the order statistics of $n$ exponential random variables with parameter $\rho(t)$.

The probability density function (PDF) of the first event time $T_1$ can be calculated as the PDF of the minimum of $n$ exponential random variables:

$f_{T_1|N(\tau) = n}(t) = n \cdot \rho(t) \cdot\exp\left(-n \int_0^t \rho(u) du\right)$

For $T_2$, given that $T_1 = t_1$, the time of the first event, the remaining $n-1$ events must occur in the interval $[t_1, \tau]$, and their order statistics follow the same reasoning as above. The conditional PDF of $T_2$ is:

$f_{T_2|N(\tau) = n, T_1 = t_1}(t) = (n-1) \cdot \rho(t) \cdot \exp\left(-(n-1) \int_{t_1}^t \rho(u) du\right)$

Similarly, for $T_3$ given $T_1 = t_1$ and $T_2 = t_2$, and so on, the conditional PDFs follow the same pattern:

$f_{T_k|N(\tau) = n, T_1 = t_1, \ldots, T_{k-1} = t_{k-1}}(t) = (n-k+1) \cdot \rho(t) \cdot \exp\left(-(n-k+1) \int_{t_{k-1}}^t \rho(u) du\right)$

This process continues up to $T_n$.

So, in summary, the conditional distribution of the event times $T_1, T_2, ..., T_n$ of a Poisson process $(N(t):t\ge0)$ with intensity function $\rho(t)$, given that $N(\tau) = n$, follows the above derived conditional PDFs for each event time.