A question involving inter-arrival times of a Poisson process I can't demonstrate that the inter-arrival times of a Poissom process are i.i.d. How can I demonstrate it? Or, where can I find the demonstration?
Thank you!

The distribution of the number of arrivals in any time period of length $t$ is $\mathrm{Poisson}(\lambda t)$ and the numbers of arrivals in non-overlapping time intervals are independent.
Let $X_t$ be the number of arrivals between now and now plus $t$.
Let $T_x$ be the time of the $x$th arrival after now.
Then
$$
\underbrace{\Pr(T_1>t) = \Pr(X_t=0)}_{\begin{array}{c}\text{Since the events $[T_1>t]$ and} \\ \text{$[X_t=0]$ are the same event.}\end{array}} = \frac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t}.
$$
Thus $T_1$ is exponentially distributed with rate $\lambda$ and expected value $1/\lambda$.  Now look at
$$
\Pr(T_{x+1}-T_x>t) = \operatorname{E}(\Pr(T_{x+1}-T_x>t \mid T_x)) = \operatorname{E}(\Pr(N_{T_x+t} - N_{T_x} =0 \mid T_x)).
$$
To find $\Pr(N_{T_x+t} - N_{T_x} =0 \mid T_x)$ as a function of $T_x$, consider $\Pr(N_{T_x+t} - N_{T_x} =0 \mid T_x= s)$.  This is the same as $\Pr(N_{s+t}-N_s=0)$, and that is $\dfrac{(\lambda t)^0 e^{-\lambda t}}{0!} = e^{-\lambda t}$.  Hence $T_{x+1}-T_x$ is also exponentially distributed with rate $\lambda$.  Furthermore, notice that if we seek $\Pr(T_{x+1}-T_x>t\mid T_1,\ldots,T_x)$, then the same reasoning above leads to the same distrtibution, and we see that the distribution does not depend on $T_1,\ldots,T_x$.  Thus the $x$th interarrival time is independent of the history.
