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Let $(N_t)_t$ be a Poisson process with parameter λ = 2. By $τ_k$ denote the time of the k-th arrival (k = 1, 2, . . .), and by $ρ_k = τ_k −τ_{k−1}$ - the interarrival time between the (k−1)th and kth arrival (k = 1, 2, . . .), with $τ_0 = 0$ (as in the construction of Poisson process).

Find the following:

(a) $E(N_3N_4)$
(b) $E(ρ_3ρ_4)$
(c) $E(τ_3τ_4)$

I can understand (a), and (b). But, couldn't understand (c).

I collected the following from a friend:
(c) $ E(τ_3τ_4) \\ = E[τ_3(τ_3+ρ_4)] \\ = E(τ^2_3 + ρ_4) \\ = E(τ^2_3) + E(ρ_4) \\ = (k/\lambda)^2 + 1/ \lambda $

First of all, is this a correct solution? If not, what is the correct one?

The processing of $τ_3$ doesn't look okay to me.

It seems to me that $ρ_4$ is introduced here because $τ_3$ and $τ_4$ are not independent. Why are not they independent?

Suppose, the 3rd item arrived at 9:05 AM and the 4th one arrived at 9:10 AM. If the process started at 9:00 AM, the 3rd one took 5 minutes and the 4th one took 10 minutes to arrive. How are they not independent?

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$\tau_3$ and $\tau_4$ are not independent because these are arrival times and they proceed additively, i.e. $\tau_4=\tau_3+\rho$. For example, if you know that $\tau_3=20$, it means $\tau_4\geq20$.

$\tau_3$ is Gamma distributed with $(3,\lambda)$ because it is sum of $\rho_1,\rho_2,\rho_3$, which are iid exponentially distributed RVs. In general sum of $\alpha$ iid exponentials is Gamma with $(\alpha,\lambda)$. For a Gamma RV, we have $E[X^2]=\frac{k+k^2}{\lambda^2}$, so here $E[\tau_3^2]=\frac{12}{\lambda^2}$.

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  • $\begingroup$ What are $N_3$ and $N_4$ in this context? I mean what are their relationships with $\tau_3$ and $\tau_4$? $\endgroup$ – user366312 Apr 14 at 19:21
  • $\begingroup$ What is $t$'s relationship with $\tau$? $\endgroup$ – user366312 Apr 14 at 19:27
  • $\begingroup$ Speaking generally, $N_t$ refers to number of arrivals which occur by time t. t is just a variable, just like k. And, it's not a random variables. You shouldn't seek for t's relation to $\tau$. $\endgroup$ – gunes Apr 14 at 19:29
  • $\begingroup$ So, $t$ is just an index, like what we use in programming arrays? $\endgroup$ – user366312 Apr 14 at 19:37
  • $\begingroup$ similar, except that it is continuous. One can say $N_{3.512}$ for example, which will mean the number of events by $t=3.512$. $\endgroup$ – gunes Apr 14 at 19:38

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