2
$\begingroup$

Given a Poisson distribution with parameter $\lambda$ (basically a Poisson process), how can I prove that this Poisson process is independent and stationary increment? Or the memoryless property: $$P(N(t+s)|N(t))=P(N(s))$$ From what I learned, the basic definition of a Poission process includes self increment and then from that you can get the 'Poisson distribution' form of the process. I am just wondering how the reverse proof goes?

$\endgroup$
  • 1
    $\begingroup$ A proof is a sequence of statements each of which is deduced logically from its predecessor. To start a proof you therefore must have a set of assumptions and definitions. What, then, are you assuming as the start of your proof? What definition of a Poisson process do you propose to begin with? $\endgroup$ – whuber Sep 8 '13 at 17:04
  • $\begingroup$ @whuber what I have is actually a poisson distribution(that is, only know the pmf and the mean in this case becomes λ*t ). Then I want to treat is as a stochastic process and deduce reversely the independent and stationary increment property from the pmf. $\endgroup$ – Junting Zhu Sep 8 '13 at 17:18
  • 2
    $\begingroup$ You do not have a process, just a random variable for a given value of time $t$. You have not specified how the potential random variables for different values of $t$ relate to each other. $\endgroup$ – Aniko Sep 8 '13 at 22:09
  • $\begingroup$ Are you defining it to have exponentially distributed interarrival times? If so memoryless is basically assumed. $\endgroup$ – dsaxton Oct 19 '16 at 0:35
2
$\begingroup$

This is modified from my answer to a related question. The only thing this answer assumes you already know about the Poisson process is that $\mathbb{P}(N_t=n)$ is a Poisson-distributed random variable (for a proof, see page 4 here). It uses the construction of the Poisson process using exponential inter-arrival times.

It turns out that this construction has independent increments (as I show below) and other properties, and that these properties actually uniquely characterize the process. I do not prove or address or use this unique characterization of the process at all in my answer below.

Proof of independent increments (following Durrett's Probability Theory and Examples, pp. 155-156, 4th edition) $\newcommand{\Prob}{\mathbb{P}}$

$N_t = \max \{ n \ge 0: T_n \le t \}$ where $T_0 = 0$ and $T_n = S_1+ \dots S_n$, $S_n \sim \exp(\lambda)$ i.i.d. RV's.

Let $T_1' = T_{N_t +1} - s$ be the amount of time that elapses after time $s$ until the next arrival. The following computation shows that $T_1'$ is independent of $N_s$: $$\Prob(T_1' \ge t|N_s = n) = \Prob(T_{N_s+1} \ge (s+t) | N_s = n) \\ = \Prob(T_{n+1} \ge (s+t), T_n \le s )/ \Prob(N_s = n)\,.$$ The numerator equals: $$ \Prob(T_{n+1} \ge (s+t), T_n \le s ) = \int_0^s f_{T_n}(u) \Prob(S_{n+1} \ge (s+t) - u)du \\= \int_0^s \frac{\lambda^n u^{n-1}}{(n-1)!}e^{-\lambda u} e^{-\lambda( (s+t) -u)}du = e^{-\lambda (s+t)}\frac{\lambda t^n}{n!}. $$ Using the fact that $\mathbb{P}(N_t=n)$ is Poisson-distributed, the denominator is $$\Prob(N_t=n) = e^{-\lambda s}(\lambda s)^n/n!\,,$$ so: $$\Prob(T_{n+1} \ge (s+t)| N_s = n) = \frac{e^{- \lambda(s+t)}}{e^{-\lambda s}} = e^{-\lambda t} =\Prob(T_1' \ge t) \,. $$

For $k \ge 2$, define $T_k' = T_{N_t +k} - T_{N_t +k-1}$, the amount of time elapsing between the $(k-1)$th arrival after time $s$ and the $k$th arrival after time $s$. Observing that: $$\Prob(T_n \le s, T_{n+1} \ge u, T_{n+k} - T_{n+k-1} \ge v_k, k=2,\dots,K) = \Prob(T_{n} \le s, T_{n+1} \ge u) \prod_{k=2}^{K} \Prob(S_{n+k} \ge v_k) \,, $$ it follows that the $T_1', T_2', \dots$ are i.i.d. and independent of $N_t$. In other words, arrivals after time $s$ are independent of $N_s$ and have the same distribution as the original sequence.

From this follows the desired conclusion of independent increments, namely that if $0:= s_0 < s_1 < \dots < s_m$ then $N_{s_i} - N_{s_i - 1}$, $i = 1, \dots, n$ are independent.

This is because the vector $( N_{s_2} - N_{s_1}, \dots, N_{s_m} - N_{s_m -1} )$ is measurable with respect to $\sigma(T_k', k \ge 1)$ and thus independent of $N_{s_1}$. Then it follows by induction: $$\Prob(N_{s_i} - N_{s_{i- 1}} = k_1, i = 1, \dots, n) = \prod_{i=1}^m \exp (-\lambda (s_i - s_{i-1})) \frac{\lambda (s_i - s_{i-1} )^{k_i}}{k_i!}\\ = \prod_{i=1}^{m} \Prob(N_{s_i} - N_{s_{i-1}} = k_i) \,. \square $$

The fact that $T_1'$ is independent of $N_s$ amounts essentially to the memoryless property of the exponential distribution. Note: I imagine this is similar to the argument that you used to show that $N_{s+t}-N_s$ is independent of $N_s$. The only difference here is that now we are renewing the process at a random time, as opposed to a deterministic time.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.