Assume we have a probability space $(\Omega,\mathcal{F},\mathbb{P})$ where $\mathcal{F} =(\mathcal{F}_t)_{0 \leq t \leq T}$ is a Filtration, with $T < \infty$.
On that prbability space we want to define a Poisson Process $N$ with rate $\lambda$.
We want to derive probabilites for specific stopping times. First we define last and first hitting times. Those are my own definitions and can be confirmed or be corrected: \begin{align*} \tau^{last} := \tau^{l} := \sup \{ t \in [0,T] \mid \Delta N(t) = 1 \} \end{align*} with the convention that $\sup \{ \emptyset \} := 0$. And the first hitting time: \begin{align*} \tau^{first} := \tau^{f} := \inf \{ t \in [0,T] \mid \Delta N(t) = 1 \} \end{align*} with the convention that $\inf \{ \emptyset \} := + \infty$.
We know that the process $(N_{T}-N_{T-t})_{0\leq t \leq T}$ has the same distribution as $(N_{t})_{0\leq t \leq T}$. (correct me if I am wrong please). This is sometimes called the time reversal law.
Following wants to be proven: Show that $T-\tau^{l}$ has the same distribution as $\min\{\tau^{f},T\}$.
As below you can find my approach, please correct or confirm:
From the time reversal law we get that $(N_{T}-N_{T-\tau^{l}})$ has the same distribution as $(N_{\tau^{l}})$. And since $\tau^{l}$ is a first hitting time, we know that $(N_{\tau^{l}}-N_{0}) = (N_{\tau^{l}}) \sim \text{Poi}(\lambda)$.
When we have event $\{ \tau^{l} > 0 \}$ then it follows that $\{ \tau^{f} > 0 \}$. On that event we have that $(N_{T-\tau^{l}})$ has the same distribution as $(N_{\tau^{f}})$. On the event $\{ \tau^{l} = 0 \}$ we have that $\{ \tau^{f} = +\infty \}$ through the convention in the definition. Thats why $(N_{T-\tau^{l}})$ has the same distribution as $(N_{T})$. On that event we also have $\min\{T,\tau^{f}\} = T$.
So we get that $(N_{T-\tau^{l}})$ has the same distribution as $(N_{\tau^{l}})$ which again as the same distribution as $ (N_{\min \{ T,\tau^{f} \}})$.
Thanks for your help. Any suggestions are greatly appreciated FrakChris
I think that you need to define $(T-\tau^l)^+$ (maximum with zero) because there may be no jumps and then the second random variable is not well defined.
You are generally on the right track with the time-reversal property:
As $N(t)$ is a Poisson process the time between jumps is exponentially distributed, and in particular the time until the first jump, $\tau^f$, is exponential with rate $\lambda$. Therefore $\min\{\tau^f,T\}$ is distributed as an exponential random variable truncated by the constant $T$.
As you say, by the time reversibility of the Poisson process you have that $N(T-t)$ is also a Poisson process with the same rate. This means that the time until the first jump in the time-reversed process, $T-\tau^l$ is also exponential with rate $\lambda$. Therefore, the random variable $(T-\tau^l)^+$ is also distributed as an exponential random variable truncated by the constant $T$.