Poisson Process Definition and Properties

• The Poisson process models events occurring randomly in time.
• It's defined by the rate parameter $\lambda$, representing the average event count per unit time.
• Time can be divided in very short intervals of length $\Delta t$.
• The probability of an event in such interval is about $\lambda \Delta t$.
• The probability of more than a single event in interval is negligible.

Definition of the Poisson Process

• A Poisson process with rate $\lambda > 0$ is a continuous-time stochastic process ${N(t), t \geq 0}$ with certain properties.
• $N(0) = 0$ : meaning that the process starts from zero at time zero.
• The counts of events in non-overlapping intervals are independent from each other.
• the number of events $N(t + s) - N(s)$ in the interval $(s, t + s]$ has a Poisson distribution with mean $\lambda t$: $$P(N(t + s) - N(s) = n) = e^{-\lambda t} \frac{(\lambda t)^n}{n!}, \quad n = 0, 1, 2, \dots$$

Core Properties

• The number of events in an interval of length $t$ follows a Poisson distribution with mean $\lambda t$.
• $P(N(t) = n) = e^{-\lambda t} \frac{(\lambda t)^n}{n!}, \quad n = 0, 1, 2, \dots$
• Interarrival times (times between consecutive events) are independent and exponentially distributed with mean $1/\lambda$.
• If $T_i$ is the time between the $(i-1)$-th and $i$-th event, then $T_i \sim \text{Exp}(\lambda)$ and $E[T_i] = 1/\lambda$.
• Time $S_n$ of the $n$-th event, where $S_n = T_1 + \dots + T_n$, follows a Gamma distribution with parameters $n$ and $\lambda$.

Practical Example

• Customers arrive at a store according to a Poisson process with rate $\lambda = 10$ customers per hour.
• The probability that exactly 5 customers arrive between 8:00 AM and 8:30 AM is approximately 0.1755.
• Calculated using $P(N(0.5) = 5) = e^{-5} \frac{5^5}{5!}$ where $N(t)$ is the number of customers by time $t$ (in hours), and $N(0.5) \sim \text{Poisson}(10 \cdot 0.5)$.
• The probability that at least two customers arrive between 8:00 AM and 8:15 AM is about 0.7127.
• Computed as $P(N(0.25) \geq 2) = 1 - P(N(0.25) = 0) - P(N(0.25) = 1) = 1 - e^{-2.5} - 2.5 e^{-2.5}$, with $N(0.25) \sim \text{Poisson}(10 \cdot 0.25)$.