Please refer to an excerpt from a text titled Essentials of Stochastic Process by Durret.

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Here comes a Lemma, built on the above argument.

Lemma: $N\left ( s \right )$ has a poisson distribution with mean $\lambda s$


$N\left ( s \right ) = n$ IFF $T_{n}\leq s< T_{n+1}$ This follows from the fact that the number of arrival of customers into a bank is discrete.

Since $\tau_{n+1} = T_{n}-T_{n+1}$ and $T_{n} = t, T_{n+1}=s$ so, $\tau_{n+1} > s-t$

Here, and where I am having difficulties understanding, the author claims that

$P\left ( N(s) = n\right )=\int_{0}^{s}f_{T_{n}}\left ( t \right )P\left ( t_{n+1}>s-t \right ).dt$

First, I do not understand why the probability density is required in the integral and second why the probability that the time of arrival between the n and n+1 customer being greater than the times s-t is required.

Secondly, how does the above fit with the fact that the probability of the number of customer by time s is n?

I hope someone would help explain the integral above from a physical approach.


The reasoning here uses the law of total probability to obtain:

$$\begin{equation} \begin{aligned} \mathbb{P}(N(s) = n) &= \mathbb{P}(T_n \leqslant s < T_{n+1}) \\[6pt] &= \int \limits_0^s \mathbb{P}(T_{n+1} > s | T_n = t) p(T_n = t) dt \\[6pt] &= \int \limits_0^s \mathbb{P}(T_{n+1} - T_n > s - t | T_n = t) p(T_n = t) dt \\[6pt] &= \int \limits_0^s \mathbb{P}(\tau_{n+1} > s-t) f_{T_n}(t) dt. \end{aligned} \end{equation}$$

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  • $\begingroup$ Can you explain why the second line follows from the third? I have zero intuition about the "common senseness" of the probability condition that follows in the second line. I'm trying to make the jump using physical reasoning. $\endgroup$ – Physkid Mar 14 '18 at 8:50
  • $\begingroup$ Because you are conditioning on the fact that $T_n = t$ you can subtract $T_n$ from one side of the inequality and $t$ from the other. $\endgroup$ – Ben Mar 14 '18 at 9:45

This answer is adapted from a similar answer I wrote here.

In a Poisson process we have events occurring at some specified rate $\lambda > 0$ and we can analyse the process by looking either at the time between events, or the number of events in a given time. To do the former, let $\tau_1, \tau_2, \tau_3, ... \sim \text{IID Exp} (\lambda)$ be the time between events in the process, and define the partial sums $T_n \equiv \sum_{i=1}^n \tau_i$, which represent the time taken for the first $n$ events. Then we have $T_n \sim \text{Ga} (n, \lambda)$ so that:

$$\begin{equation} \begin{aligned} \mathbb{P}(T_n \leqslant t) = 1 - \mathbb{P}(T_n >t) &= 1 - \int\limits_t^{\infty} \text{Ga} (s|n, \lambda) ds \\ &= 1-\frac{\lambda^n}{\Gamma (n) } \int\limits_t^{\infty} s^{n-1} \exp (- \lambda s) ds \\ &= 1-\frac{1}{\Gamma (n)} \int\limits_t^{\infty} (\lambda s) ^{n-1} \exp (- \lambda s) \lambda ds \\ &= 1-\frac{1}{\Gamma (n)} \int\limits_{\lambda t}^{\infty} r^{n-1} \exp \left( - r \right) dr \\ &= 1-\frac{\Gamma(n, \lambda t)}{\Gamma (n)}. \\ \end{aligned} \end{equation}$$

Using integration by parts, the upper incomplete gamma function follows the recurrence:

$$\begin{matrix} \Gamma (n, x) = (n-1) \Gamma(n-1, x) + x^{n-1} \exp (-x) & & \Gamma (1, x) = \exp(-x). \end{matrix}$$

For integer $n$, repeated application of this recurrence yields:

$$\Gamma (n, x) = \Gamma (n) \exp (-x) \sum_{k=0}^{n-1} \frac{x^k}{k!}.$$

So, letting $N(t) \sim \text{Pois} (\lambda t)$ we have:

$$\begin{equation} \begin{aligned} \mathbb{P}(T_n \leqslant t) &= 1- \exp (-\lambda t) \sum_{k=0}^{n-1} \frac{(\lambda t)^k}{k!} \\ &= 1 - \sum_{k=0}^{n-1} \text{Pois} (k|\lambda t) \\ &= \sum_{k=n}^{\infty} \text{Pois} (k|\lambda t) \\ &= \mathbb{P} (N(t) \geqslant n). \end{aligned} \end{equation}$$

This gives us a basic intuitive result for the Poisson process. If the time taken for the first $n$ events is no greater than $t$ then the number of events that have occurred at time $t$ is at least $n$. If the time between events follows an exponential distribution then the number of events at a given time follows a Poisson distribution.

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