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.