As discretization of Bernoulli Process

\(\quad\)Prob. of \(k\) arrivals in interval duration of \(\tau\). Note \(N_\tau \approx \) Binomial(\(n, p\)).

\[ \begin{aligned} \mathbb{P}(N_\tau = k) &= \mathbb{P}(k,\tau) & \text{where } \displaystyle\sum_{k=0}^\infty\mathbb{P}(k,\tau) = 1 \\ &= \frac{(np)^k}{k!} e^{-\displaystyle np} \\ &= \frac{(\displaystyle\color{blue}{\frac{\tau}{\delta}}\cdot\color{red}{\lambda\delta})^k}{k!} e^{-\displaystyle\color{blue}{\frac{\tau}{\delta}}(\color{red}{\lambda\delta)}} & \text{where }n = \displaystyle\color{blue}{\frac{\tau}{\delta}}\text{ and }p = \color{red}{\lambda \delta} + O(\delta^2) \\ \mathbb{P}(k, \tau) &= \frac{(\lambda\tau)^k}{k!} e^{-\lambda\tau} & \text{which is the Poisson PMF with parameter } (\lambda, \tau)\\ \end{aligned} \]

\(\quad\)\(\mathbb{E}[N_\tau] \approx np \approx \lambda \tau \qquad var(N_\tau) \approx np(1-p) \approx \lambda\tau \qquad \)Note \(\mathbb{E}[N_\tau] = \displaystyle\sum_{k=0}^\infty k \frac{(\lambda\tau)^k e^{-\lambda\tau}}{k!}\)

Small interval probabilities

\[ \begin{aligned} \mathbb{P}(k,\delta) \approx \begin{cases}1 - \lambda\delta & k=0 \\ \color{red}{\lambda\delta} & k = 1 \\ 0 & k > 1 \end{cases} \\ \end{aligned} \]

by using very small \(\delta\) so that the probability of having more than one arrival becomes negligible. \(\lambda\), the arrival rate, can be perceived as the probability of arrival per unit time.

Time \(T_1\) until the \(1^{st}\) arrival

\[ \begin{aligned} \mathbb{P}(T_1 \le t) &= 1 - \mathbb{P}(T_1 > t) = 1 - \mathbb{P}(0, t) = 1 - e^{-\lambda t} \\ f_{T_1}(t) &= \frac{d}{dt} \mathbb{P}(T_1 \le t) = \lambda e^{-\lambda t} = \text{Exponential}(\lambda) \end{aligned} \]

Memorylessness: \(f_{T_1}(t' - t\,|\, t' > t) =\) Exponential(\(\lambda\)).

Time \(Y_k\) of the \(k^{th}\) arrival

\[ \begin{aligned} \mathbb{P}(Y_k \le y) &= \sum_{n=k}^\infty \mathbb{P}(n, y) & \text{CDF argument} \\ f_{Y_k}(y) \delta &\approx \mathbb{P}(y \le Y_k \le y + \delta) = \mathbb{P}(k-1, y) \mathbb{P}(1, \delta) = \mathbb{P}(k-1, y) \cdot \color{red}{\lambda\delta} & \text{more intuitive argument}\\ \therefore f_{Y_k}(y) &\approx \mathbb{P}(k-1, y) \cdot \lambda = \frac{\lambda^k y^{k-1} e^{-\lambda y}}{(k-1)!} \qquad \text{for }y \ge 0 & \color{blue}{\text{Erlang distribution }} \text{of order } k \\ \end{aligned} \]

\[ \begin{aligned} Y_k &= T_1 + \cdots + T_k & \text{sum of i.i.d. exponentials} \\ \mathbb{E}[Y_k] &= \frac{k}{\lambda} \qquad var(Y_k) = \frac{k}{\lambda^2} \\ \end{aligned} \]

Sum of independent Poisson r.v's with means/parameters \(\mu\) and \(\nu\)

\(\quad\)Poisson(\(\mu\)) + Poisson(\(\nu\)) = Poisson(\(\mu + \nu) \qquad\) with mean/parameter \(\mu + \nu\)

In general, the sum of two independent Poisson random variables is also a Poisson random variable as if the two given r.v.'s are numbers of arrivals in disjoint time intervals. This is a special property similar to that of Normal r.v.'s, and most other distributions do not have.

Splitting a Poisson process

Splitting a Bernoulli process into two would result in two Bernoulli processes that are dependent, since the arrival in a given time slot of one process would imply the non-arrival of the other process in the same time slot. Similarly, splitting a Poisson process into two would also result in two Poisson processes. Surprisingly, however, they are independent, as time is continuous.

Source: MITx 6.041x, Lecture 22, 23.