Comprehensive Study Notes: Stochastic Processes and the Poisson Process

Chapter 4: The Poisson Process

The Poisson process is considered one of the most fundamental models in probability theory and stochastic processes. It provides a mathematical description for the occurrence of random events over time. Common examples include:

• The arrival of customers at a service desk.

• The emission of particles from a radioactive source.

• Phone calls incoming to a call centre.

Two central ingredients underlie the Poisson process:

• The number of arrivals in a given time interval follows a Poisson distribution.

• The waiting time between successive arrivals is exponentially distributed.

4.1 The Exponential Distribution

The exponential distribution is the canonical model for the waiting time until the first occurrence of a random event that happens at a constant average rate.

Definition 4.1.1

A random variable $X : \Omega \rightarrow \mathbb{R}$ is said to follow an exponential distribution with parameter \lambda > 0 if:

$P(X \in [a, b]) = \int_a^b \lambda e^{-\lambda\beta} \, d\beta$

for all $0 \leq a \leq b$. Equivalently, $X$ has the probability density function:

$f_X(x) = \lambda e^{-\lambda x}$

for all $x \geq 0$. The notation used is $X \sim \exp(\lambda)$.

Lemma 4.1.2: Properties of the Exponential Distribution

Let $X$ be exponentially distributed with parameter \lambda > 0. Then:

• (a) Cumulative Distribution Function (CDF):   $F_X(\beta) = \begin{cases} 1 - e^{-\lambda\beta} & \text{if } \beta \geq 0 \\ 0 & \text{else} \end{cases}$

• (b) Expectation and Variance:   $E[X] = \frac{1}{\lambda}$   $\text{Var}(X) = \frac{1}{\lambda^2}$

• (c) Moment Generating Function (MGF):   $G_X(\beta) := E[e^{\beta X}] = \frac{\lambda}{\lambda - \beta}$   for all \beta < \lambda.

Proof of Lemma 4.1.2:

• (a) For $\beta \geq 0$:   $F_X(\beta) = \int_0^\beta f_X(\alpha) \, d\alpha = \int_0^\beta \lambda e^{-\lambda\alpha} \, d\alpha = 1 - e^{-\lambda\beta}$.

• (b) Integrating by parts with $u = \beta$, $dv = \lambda e^{-\lambda\beta} d\beta$:   $E[X] = \int_0^\infty \beta \lambda e^{-\lambda\beta} \, d\beta = [-\beta e^{-\lambda\beta}]_0^\infty + \int_0^\infty e^{-\lambda\beta} \, d\beta = \frac{1}{\lambda}$.   Similarly, $E[X^2] = \int_0^\infty \beta^2 \lambda e^{-\lambda\beta} \, d\beta = \frac{2}{\lambda^2}$. Thus, $\text{Var}(X) = \frac{2}{\lambda^2} - (\frac{1}{\lambda})^2 = \frac{1}{\lambda^2}$.

• (c) For \beta < \lambda:   $G_X(\beta) = \int_0^\infty e^{\beta \alpha} \lambda e^{-\lambda \alpha} \, d\alpha = \lambda \int_0^\infty e^{-(\lambda-\beta)\alpha} \, d\alpha = \frac{\lambda}{\lambda - \beta}$.

Theorem 4.1.3: Lack of Memory Property

An exponential random variable has a unique feature known as the lack-of-memory (or memoryless) property. Informally: "If we have already waited $t$ units of time, the probability of having to wait at least $s$ further units is the same as if no time had elapsed."

If $X \sim \exp(\lambda)$, then for all $s, t \geq 0$:

P(X > t + s | X > t) = P(X > s)

Proof: By Lemma 4.1.2, P(X > t) = 1 - F_X(t) = e^{-\lambda t}. Thus:

P(X > t + s | X > t) = \frac{P(X > t + s)}{P(X > t)} = \frac{e^{-\lambda(t+s)}}{e^{-\lambda t}} = e^{-\lambda s} = P(X > s).

Among all continuous probability distributions with a density, the exponential is the only one possessing this property. In the discrete setting, the geometric distribution is the unique distribution with this property.

Examples of the Exponential Distribution

• Example 4.1.4 (Bank Session): Let $X$ denote the time (in minutes) a customer spends in a bank, assumed exponential with mean 10 minutes (implying $\lambda = \frac{1}{10}$).

  • Probability spending more than 15 minutes: P(X > 15) = e^{-\lambda(15)} = e^{-3/2} \approx 0.220.

  • Probability spending more than 15 minutes given the customer is still there after 10 minutes: P(X > 15 | X > 10) = P(X > 5) = e^{-\lambda(5)} = e^{-1/2} \approx 0.604.

• Example 4.1.5 (Lightbulb Lifetime): Lifetime $X$ (in hours) is exponential with mean 10 ($X \sim \exp(1/10)$). You find the bulb working. The probability it lasts another 5 hours is P(X > t + 5 | X > t) = P(X > 5) = e^{-5/10}. If $X$ were not exponential, the probability would depend on $t$: $\frac{1 - F_X(t + 5)}{1 - F_X(t)}$.

Mathematical Properties of Exponential Variables

• Proposition 4.1.6 (Sum of Exponentials): Let $X_1, \dots, X_n$ be independent $\exp(\lambda)$ random variables. Then $S_n = X_1 + \dots + X_n$ follows a Gamma distribution with shape parameter $n$ and rate parameter $\lambda$. The density is:   $f_{S_n}(\beta) = \frac{\lambda^n}{(n-1)!} \beta^{n-1} e^{-\lambda \beta}$

• Proposition 4.1.7 (Minimum of Exponentials): Let $X_1, \dots, X_n$ be independent $\exp(\lambda_k)$ variables. Then $M = \min(X_1, \dots, X_n)$ is exponentially distributed with parameter $\sum_{k=1}^n \lambda_k$. Interpretation: If $n$ independent exponential clocks start with rates $\lambda_k$, the first clock to ring occurs after an exponential time with the sum of the rates.

• Proposition 4.1.8 (Competition of Exponentials): Let $X \sim \exp(\lambda)$ and $Y \sim \exp(\mu)$ be independent. Then:   P(X < Y) = \frac{\lambda}{\lambda + \mu}   Proof: Using joint density $f_{X,Y}(\alpha, \beta) = \lambda e^{-\lambda \alpha} \mu e^{-\mu \beta}$, we compute:   P(X < Y) = \int_0^\infty \int_\alpha^\infty \lambda e^{-\lambda \alpha} \mu e^{-\mu \beta} \, d\beta \, d\alpha = \int_0^\infty \lambda e^{-\lambda \alpha} e^{-\mu \alpha} \, d\alpha = \frac{\lambda}{\lambda + \mu}.

Application Examples

• Example 4.1.9 (Post Office): Two clerks are busy, service times are $\exp(\lambda_1)$ and $\exp(\lambda_2)$. You are served when one becomes available. Total time $T = \min(R_1, R_2) + S$, where $R_i$ is remaining service time and $S$ is your own service time. Due to memorylessness, $R_i \sim \exp(\lambda_i)$. Expected time:   $E[T] = E[\min(R_1, R_2)] + E[S] = \frac{1}{\lambda_1 + \lambda_2} + \frac{2}{\lambda_1 + \lambda_2} = \frac{3}{\lambda_1 + \lambda_2}$.

• Example 4.1.10 (Competing Tasks): Check-up (mean 20 mins, $\lambda_1 = 1/20$) and Consultation (mean 30 mins, $\lambda_2 = 1/30$) start at the same time.

  • Probability check-up finishes first: P(R_1 < R_2) = \frac{1/20}{1/20 + 1/30} = \frac{3}{5}.

  • Expected time until both finish ($T = \max(R_1, R_2)$): E[T] = E[\min(R_1, R_2)] + P(R_1 < R_2)E[R_2] + P(R_2 < R_1)E[R_1].   $E[T] = 12 + \frac{3}{5}(30) + \frac{2}{5}(20) = 38 \text{ minutes}$.

4.2 The Poisson Distribution

The Poisson distribution models the number of events in a fixed interval of time or space occurring independently at a constant average rate. It is often described as the "law of rare events."

Definition 4.2.1

A random variable $X$ is Poisson distributed, written $X \sim \text{Poi}(\lambda)$, if:

$P(X = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

for all $k \in \mathbb{Z}_{\geq 0}$. The parameter $\lambda$ represents the expected number of occurrences.

Proposition 4.2.2: Moments of Poisson

For $X \sim \text{Poi}(\lambda)$, \lambda > 0:

• (a) Factorial Moments: $E[X(X - 1) \dots (X - k + 1)] = \lambda^k$. Specifically, $E[X] = \lambda$.

• (b) Variance: $\text{Var}(X) = \lambda$.

Proof of (b): Since $E[X(X-1)] = \lambda^2$, we have: $\text{Var}(X) = E[X(X-1)] + E[X] - (E[X])^2 = \lambda^2 + \lambda - \lambda^2 = \lambda$.

Proposition 4.2.3: Sum of Independent Poisson

If $X \sim \text{Poi}(\lambda)$ and $Y \sim \text{Poi}(\rho)$ are independent, then $X + Y \sim \text{Poi}(\lambda + \rho)$.

Proposition 4.2.4: Poisson as a Limit of Binomial

Let $X_n$ be Binomially distributed with parameter $n$ and $p_n = \lambda/n$. As $n \rightarrow \infty$:

$\lim_{n \rightarrow \infty} P(X_n = k) = \frac{e^{-\lambda} \lambda^k}{k!}$

This reveals that the Poisson distribution is appropriate for modeling many independent trials with small success probabilities ($p$) but a fixed expected number of successes ($np = \lambda$).

Example 4.2.5 (Approximation): For $n = 1000$, $p_n = 0.002$ ($\lambda = 2$). Comparing $X_n \sim \text{Bin}(1000, 0.002)$ and $Y \sim \text{Poi}(2)$, the probabilities are nearly identical (absolute differences are $\approx 10^{-4}$ to $0$).

4.3 Poisson Process

A counting process $(N(t) : t \geq 0)$ models the arrival of events over time.

Definition 4.3.2: Counting Process

A stochastic process is a counting process if:

• (a) $N(0) = 0$

• (b) $N(t) \geq 0$ for all $t \geq 0$

• (c) $N(s) \leq N(t)$ whenever $s \leq t$

Increment Properties

• Independent Increments: Numbers of events in disjoint time intervals are independent random variables. (e.g., goals in the first half vs. second half).

• Stationary Increments: The distribution of the number of events in an interval depends only on the length of the interval, not its position. (e.g., arrivals between 12-1 PM must be same as 3-4 PM).

Definition 4.3.4: Poisson Process

A counting process $N$ is a Poisson process with intensity \lambda > 0 if:

• (a) $N(0) = 0$

• (b) $N$ has independent and stationary increments.

• (c) For every $t \geq 0$, $N(t) \sim \text{Poi}(\lambda t)$, specifically: $P(N(t) = k) = \frac{e^{-\lambda t} (\lambda t)^k}{k!}$.

The parameter $\lambda$ represents the expected number of arrivals per unit time ($E[N(t)] = \lambda t$).

Arrival and Waiting Times

• Definition 4.3.5:

  • $T_k = \inf\{t \geq 0 : N(t) = k\}$ is the $k$-th arrival time.

  • $\tau_k = T_k - T_{k-1}$ is the $k$-th waiting time ($T_0 := 0$).

• Theorem 4.3.6:

  • (a) Waiting times $\tau_1, \tau_2, \dots$ are i.i.d. $\exp(\lambda)$.

  • (b) Arrival time $T_k$ follows a Gamma distribution with parameters $(k, \lambda)$, with density: $f_{T_k}(\beta) = \frac{\lambda^k}{(k-1)!} \beta^{k-1} e^{-\lambda \beta}$.

Theorem 4.3.9: Equivalent Characterizations

A counting process $N$ is a Poisson process with intensity $\lambda$ if and only if:

1. It has independent and stationary increments, and for small $h$, $P(N(h) = 1) = \lambda h + o(h)$ and $P(N(h) \geq 2) = o(h)$.

2. It results from independent exponentially distributed waiting times $\tau_k$.

Note on Little-o Notation (Definition 4.3.7): $f = o(g)$ as $\beta \rightarrow 0$ if $\lim_{\beta \rightarrow 0} \frac{f(\beta)}{g(\beta)} = 0$.

• Example: $e^h = 1 + h + o(h)$.

• This allows defining the Poisson process by its infinitesimal behavior over very short intervals.

4.4 Compound Poisson Processes

A compound Poisson process attaches a random quantity (jump size) to each arrival of a Poisson process.

Definition 4.4.2

Let $N(t)$ be a Poisson process with intensity $\lambda$ and $(Y_k)_{k=1}^\infty$ be an i.i.d. sequence of random variables independent of $N$. The compound Poisson process $S(t)$ is:

$S(t) = \sum_{k=1}^{N(t)} Y_k$

(If $N(t)=0$, then $S(t)=0$).

Theorem 4.4.4: Wald's Identity

If E[Y_1^2] < \infty, then for all $t \geq 0$:

• $E[S(t)] = \lambda t E[Y_1]$

• $\text{Var}(S(t)) = \lambda t E[Y_1^2]$

Example 4.4.5 (Marine Ecologist): Crabs emerge at scale $\lambda = 3$ per hour. Weights $Y_k$ have mean 4 lbs and standard deviation 2 lbs. Over 2 hours:

• $E[S(2)] = (3)(2)(4) = 24 \text{ lbs}$.

• $\text{Var}(S(2)) = (3)(2) E[Y_1^2] = (6)( \text{Var}(Y_1) + E[Y_1]^2 ) = 6(2^2 + 4^2) = 6(20) = 120 \text{ lbs}^2$.

• Standard deviation: $\sqrt{120}$.

4.5 Combining and Thinning

Theorem 4.5.1: Thinning (Splitting)

If each arrival of a Poisson process with rate $\lambda$ is classified as type $j \in \{1, \dots, M\}$ with probability $p_j$, then the processes $N_j(t)$ counting type-$j$ events are independent Poisson processes with intensities $p_j \lambda$.

Example 4.5.2 (Fisherman): Fish bite at rate 2 per hour. 40% are salmon, 60% trout. Salmon process rate is $(0.4)(2) = 0.8$. Trout process rate is $(0.6)(2) = 1.2$. Probability of 1 salmon and 2 trout in 2.5 hours: $P(N_s(2.5)=1) \cdot P(N_t(2.5)=2) = \frac{e^{-2} 2^1}{1!} \cdot \frac{e^{-3} 3^2}{2!} = 9 e^{-5}$.

Theorem 4.5.3: Superposition

The sum of $m$ independent Poisson processes with intensities $\lambda_1, \dots, \lambda_m$ is a Poisson process with intensity $\lambda = \sum_{k=1}^m \lambda_k$.

Example 4.5.4 (Bus Stop): Routes A ($\lambda_A$) and B ($\lambda_B$).

• Overall arrival rate: $\lambda_A + \lambda_B$.

• Expected wait for any bus: $\frac{1}{\lambda_A + \lambda_B}$.

• Probability next bus is A: $\frac{\lambda_A}{\lambda_A + \lambda_B}$.

• Expected A buses before the first B bus: Let $p = \lambda_A/(\lambda_A + \lambda_B)$. The number of A labels before the first B is geometric with parameter $(1-p)$, yielding mean $\frac{p}{1-p} = \frac{\lambda_A}{\lambda_B}$.