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# Lecture 15 ## The Poisson Process ### 1. Bernoulli Process-Poisson Process #### 1.1 Bernoulli Process * Consider the number of events occurring in a time interval $(0, t]$. * Divide the interval into $n$ subintervals of length $\Delta t$, so that $n = \frac{t}{\Delta t}$. * Suppose that at m...

# Lecture 15 ## The Poisson Process ### 1. Bernoulli Process-Poisson Process #### 1.1 Bernoulli Process * Consider the number of events occurring in a time interval $(0, t]$. * Divide the interval into $n$ subintervals of length $\Delta t$, so that $n = \frac{t}{\Delta t}$. * Suppose that at most one event can occur in each subinterval. * Assume that the probability that an event occurs in each subinterval is $p$, and the occurrences in different subintervals are independent. * Let $X$ be the number of events in $(0, t]$. Then, $X \sim Bin(n, p)$. The probability mass function (PMF) of $X$ is given by: $P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$, where $k = 0, 1, 2,..., n$. #### 1.2 Poisson Process * The Poisson process is obtained from the Bernoulli process by taking the following limits: $n \rightarrow \infty$ (or equivalently $\Delta t \rightarrow 0$), and $p \rightarrow 0$ in such a way that $np = \lambda$, where $\lambda$ is a constant. * In this case, for $k = 0, 1, 2,...$ $P(X = k) = \lim_{n \rightarrow \infty} \binom{n}{k} p^k (1 - p)^{n - k}$ $= \lim_{n \rightarrow \infty} \frac{n(n - 1)...(n - k + 1)}{k!} (\frac{\lambda}{n})^k (1 - \frac{\lambda}{n})^{n - k}$ $= \lim_{n \rightarrow \infty} \frac{n^k + lower \space order \space terms}{k!} \frac{\lambda^k}{n^k} (1 - \frac{\lambda}{n})^n (1 - \frac{\lambda}{n})^{-k}$ $= \frac{\lambda^k}{k!} e^{-\lambda} (1)^{-k}$ $= \frac{\lambda^k e^{-\lambda}}{k!}$ * Then, $X$ is a Poisson random variable with parameter $\lambda$, i.e. $X \sim Poi(\lambda)$. ### 2. Definition * A Poisson process with rate $\lambda > 0$ is a counting process $\{N(t), t \geq 0\}$ satisfying: 1. $N(0) = 0$. 2. The process has independent increments. 3. The number of events in any interval of length $t$ is Poisson distributed with mean $\lambda t$. That is, for all $s, t \geq 0$, $P(N(t + s) - N(s) = n) = \frac{e^{-\lambda t} (\lambda t)^n}{n!}$, $n = 0, 1, 2,...$ * Here, $N(t)$ represents the number of events that occur by time $t$. * Independent increments means that the number of events which occur in disjoint time intervals are independent. For example, $N(b) - N(a)$ is independent of $N(d) - N(c)$ if $(a, b]$ and $(c, d]$ are disjoint. * Note: For the Poisson process, the mean and variance are equal: $E[N(t)] = Var[N(t)] = \lambda t$ ### 3. Interarrival Times and Memoryless Property * Let $T_1$ be the time of the first event. * Let $T_i$ be the time between the $(i-1)$th and the $i$th event, $i \geq 2$. * $T_i$'s are called interarrival times. * We want to find the distribution of $T_1$. * $P(T_1 > t) = P(N(t) = 0) = \frac{e^{-\lambda t} (\lambda t)^0}{0!} = e^{-\lambda t}$. * Then, $F_{T_1}(t) = P(T_1 \leq t) = 1 - e^{-\lambda t}$. * By taking the derivative with respect to t, we have $f_{T_1}(t) = \lambda e^{-\lambda t}$, $t \geq 0$ * So, $T_1$ has an exponential distribution with parameter $\lambda$. * Since the Poisson process has independent increments, it possesses the memoryless property. * $P(T_2 > t \mid T_1 = s) = P(N(t + s) - N(s) = 0) = e^{-\lambda t}$ * In general, $T_i$'s are independent and identically distributed (i.i.d.) exponential random variables with parameter $\lambda$.