Poisson process : law of $Z_t=t- T_{N_t}, W_t= T_{N_{t+1}}-t$

Question

• $N_t$ is a Poisson process with intensity $\lambda$, $(T_n)_n$ are its jumps
• $Z_t=t- T_{N_t}, W_t= T_{N_{t+1}}-t$ with $T_0=0$

1. Show that for $w\leq 0$ and $z \in[0,t[$, $P(Z_t \leq z, W_t \leq w) = \exp( - \lambda (z+w))$, deduce their law, and that they are independants
2. Compute $ \lim\limits_{t \to \infty} \mathbb{E}(W_t+Z_t)$, deduce that $T_{N_{t+1}} -T_{N_t}$ has a different law that $T_{n+1}- T_n$
3. Show that $ \{ T_{N_t+1} -t, (T_{N_t+k+1} - T_{N_t+k})_{k \geq 1} \}$ are iid with same law than $T_{n+1}- T_n$

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• $P(Z_t=t)= P( T_1 >t)= e^{ - \lambda t}$ $Z_t$ has an atom.
• $E \sim \varepsilon( \lambda)$
• $ \gamma_n$ is the density of a $\gamma(n ,\lambda)$

$ \begin{align*} \mathbb{P}(T_{N_t} \leq t-z, T_{N_t+1} \geq t+w, N_t=n) &= \mathbb{P}(T_n \leq t-z, T_n+E \geq t+w) \\ &= \int_{\mathbb{R}_+^2} 1_{x \leq t-z} 1_{x+y \geq t+w} \gamma_n(x) \gamma_1(y) \, dx \, dy \\ &= \int_{\mathbb{R}_+} 1_{x \leq t-z} \gamma_n(x) \int_{\mathbb{R}_+} 1_{y \geq t+w-x} \gamma_1(y) \, dy \, dx \\ &= \int_{\mathbb{R}_+} 1_{x \leq t-z} \gamma_n(x) e^{-\lambda(t+w-x)} \, dx \\ &= e^{-\lambda(t+w)} \int_0^{t-z} \frac{\lambda^n x^{n-1}}{(n-1)!} \, dx \\ &= e^{-\lambda(t+w)} \frac{\lambda^n (t-z)^n}{n!} \end{align*} $

$ \begin{align*} \mathbb{P}(T_{N_t} \leq t-z, T_{N_t+1} \geq t+w) &= \sum_{n=0}^{\infty} \mathbb{P}(T_{N_t} \leq t-z, T_{N_t+1} \geq t+w, N_t=n) \\ &= \sum_{n=0}^{\infty} e^{-\lambda(t+w)} \frac{\lambda^n (t-z)^n}{n!} \\ &= e^{-\lambda(t+w)} e^{\lambda (t-z)} = e^{-\lambda(w+z)} \\ \end{align*} $

Or, we can say that $\{ T_{N_t} \leq t-z, T_{N_t+1} \geq t+w \} = \{ N_{(t+w)-}-N_{(t-z)+}=0\}$, the ponctual process has no pint in $I=]t-z, t+w[$ and the points falling in $I$ follow a Poisson distribution of parametre $\lambda l(I)$,

• so $P(N_{(t+w)-}-N_{(t-z)+}=0) =e^{ - \lambda( z+w) }$

• $P( W_t \geq w) =P(Z_t \geq 0, W_t \geq w)= e^{- \lambda w}$ so $W_t$ follows an exponential distribution of parameter $\lambda$

• $P_{Z_t}(t) = 1- e^{ - \lambda z } \mathbb{1}_{0 \leq z<t} + e^{ - \lambda t} \mathbb{1}_{w=z}$

• $(Z_t, W_t)$ are independent as their joint repartition functions writes as a product of functions of each variable.

2. $E(Z_t+W_t) \xrightarrow{ t \to \infty } \dfrac{2}{\lambda}$, so $T_{N_t+1} -T_{N_t}$ has not the same law than $T_{n+1}-T_n$

Answer 1

We consider : $\{ T_{N_t+1} -t, (T_{N_t+k+1} - T_{N_t+k})_{k \geq 1} \}$

$W_t=T_{N_t+1} -t \sim \varepsilon(\lambda)$ due to previous question

$(T_{N_t+k+1} - T_{N_t+k})_{k \geq 1}$ iid and same law $\varepsilon(\lambda)$ due the Markov property with stopping time $T_{N_t}$

They may be all the successive jumps of $M_u=N_{N_t+u} - N_{N_t}$