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{$N_t$} with $t\in \mathbb{R}$ is a Poisson process with intensity $\lambda \in \mathbb{R^+}$, so that

1) $N(0)=0$,

2) {N(t) is with indipendent increments and omogeneous increments and

3) $\mathbb{P}(N(s,t) = i) ={ \frac{\lambda(t-s)^i}{i!}}e^{-\lambda(t-s)}$

{$T_n$} are the arrival times defined as follow

$T_0 = 0$ $\quad$ and $\quad$ $T_n$ = $inf${$ t : N(t) = n$}

Let $X_n = T_n - T_{n-1}$ so we can see $X_n$ as the interarrival time. I have to prove that the $X_1,X_2,X_3,...$ are all indipendent and has the same distribution which is an exponential with parameter $\lambda$.

So i have understood the part for the exponential distribution of the $X_i$ but now im not sure that the part about the indipendence is correct so if anyone can give any advice...

$\mathbb{P}(X_2>t | X_1 = t_1) = \mathbb{P}(N(t+t_1)=N(t_1)) = \mathbb{P}(N(t+t_1)-N(t_1) = 0) = \mathbb{P}(N(t)=0) = e^{-\lambda t}$

so $X_2$ is indipendent of $X_1$ andnow we go on by induction on $n+1$

$\mathbb{P}(X_{n+1}>t | X_1 = t_1,...,X_n = t_n) = \mathbb{P}(N(t+T)=N(T)) = \mathbb{P}(N(t+T)-N(T) = 0) = \mathbb{P}(N(t)=0) = e^{-\lambda t}$

where $T = t_1 + ...+ t_n$.

So i think i have proved that $X_{n+1}$ is indipendent of $X_1,...,X_n$.

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    $\begingroup$ Possible duplicate of Relationship between poisson and exponential distribution $\endgroup$ – jbowman Jun 16 '18 at 21:33
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    $\begingroup$ What is the definition of Poisson process that you are using? The answer will depend on that. $\endgroup$ – Dilip Sarwate Jun 16 '18 at 22:21
  • $\begingroup$ @DilipSarwate i have edited the question with the definition that im using $\endgroup$ – paolopazzo Jun 17 '18 at 7:39
  • $\begingroup$ @jbowman thank you man i read it and was very helpfull but im still missing the proof that all the $X_i$ are indipendent. $\endgroup$ – paolopazzo Jun 17 '18 at 7:40
  • $\begingroup$ I believe I fully address these issues in an answer at stats.stackexchange.com/questions/214421, even though the question is slightly different. In fact, I start with axiom (2) and derive (3) from it. $\endgroup$ – whuber Jun 21 '18 at 1:15

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