Stochastic Modeling

Problem-Solving Strategy¶

Identify the Distribution: Recognize which distribution the problem involves
Use Definitions: Start with the definition (PMF, PDF, or property)
Check Conditions: Verify all conditions are met before applying theorems
Use Linearity: Expectation is linear - use this whenever possible
Conditioning: Break complex problems into simpler conditional problems
Symmetry: Look for symmetry to simplify calculations
Approximations: Use CLT for large $n$ approximations

Differentiation Rules¶


Power Rule	$\frac{d}{dx} x^n = n x^{n-1}$
Exponential	$\frac{d}{dx} e^{ax} = a e^{ax}$
Natural Log	$\frac{d}{dx} \ln(x) = \frac{1}{x}$
Product Rule	$\frac{d}{dx} [f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$
Quotient Rule	$\frac{d}{dx} \left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
Chain Rule	$\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)$

Integration Rules¶


Power Rule	$\int x^n dx = \frac{x^{n+1}}{n+1} + C$ (for $n \neq -1$ )
Exponential	$\int e^{ax} dx = \frac{1}{a} e^{ax} + C$
Natural Log	$\int \frac{1}{x} dx = \ln\|x \| + C$
Integration by Parts	$\int u dv = uv - \int v du$

\begin{align*} \int x e^{ax} \, dx &= uv - \int v \, du \\ u &= x & dv &= e^{ax} \, dx \\ du &= dx & v &= \frac{e^{ax}}{a} \end{align*}

(1)

\begin{align*} \int_{-\infty}^{\infty} e^{-x^2} dx &= \sqrt{\pi} & \int_0^{\infty} e^{-ax} dx &= \frac{1}{a} \quad \text{for } a > 0 \\ \int_0^{\infty} x e^{-ax} dx &= \frac{1}{a^2} & \int_0^{\infty} x^2 e^{-ax} dx &= \frac{2}{a^3} \end{align*}

(2)

Probability¶

A Probability Space is a triple $(\Omega, \mathcal{A}, P)$ , where: $\Omega$ : Set of outcomes, $\mathcal{A}$ : Set of events, $P$ : Probability measure.

A Probability Measure $P: \mathcal{A} \to \mathbb{R}$ satisfies:

$0 \leq P(A) \leq 1 \quad \forall \, A \in \mathcal{A}$
$P(\Omega) = 1$
$P\left(\bigcup_{i=1}^\infty A_i\right) = \sum_{i=1}^\infty P(A_i)$

Random Variable: $X: \Omega \to \mathbb{R}$

\begin{align*} P(A) &= \frac{|A|}{|\Omega|} \\ P(E|F) &= \frac{P(E \cap F)}{P(F)} & P(A \cap B) &= P(A)P(B|A) = P(B)P(A|B) \\ P(A|E) &= \frac{P(E|A)P(A)}{P(E)} & \end{align*}

(3)

Bayes’ Theorem¶

P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E)} = \frac{P(E|A) \cdot P(A)}{P(E|A) \cdot P(A) + P(E|A^c) \cdot P(A^c)}

(4)

Independence¶

$X$ and $Y$ are independent if one of the following is true:

\begin{aligned} F_{X,Y}(a,b) &= F_X(a) F_Y(b) \quad \forall a,b \\ f_{X,Y}(x,y) &= f_X(x)f_Y(y) \quad \text{, (continous RV)} \\ P(X=x, Y=y) &= P(X=x)P(Y=y) \quad \text{, (discrete RV)} \end{aligned}

(5)

Consequences of Independence

\begin{align*} P(E|F) &= P(E) \\ P(E \cap F) &= P(E) \cdot P(F) \\ \mathbb{E}[XY] &= \mathbb{E}[X]\mathbb{E}[Y] \\ \operatorname{Cov}(X,Y) &= 0 \\ \operatorname{Var}(X+Y) &= \operatorname{Var}(X) + \operatorname{Var}(Y) \\ \phi_{X+Y}(t) &= \phi_X(t) \phi_Y(t) \end{align*}

(6)

Law of Total Probability¶

$P(E) = \sum_{i=1}^n P(E|A_i) \cdot P(A_i)$
Simplified (for binary partition): $P(E) = P(E|A) \cdot P(A) + P(E|A^c) \cdot P(A^c)$

Expectation, Variance, Covariance, Correlation Coefficient (normalized Covariance)¶

\begin{align*} \boldsymbol{\mathbb{E}[X]} &= \sum x P(X=x) & \boldsymbol{\mathbb{E}[X]} &= \int x f_X(x) dx \\ \boldsymbol{\mathbb{E}[X^2]} &= \text{Var}(X) + (\mathbb{E}[X])^2 & \\ \boldsymbol{\operatorname{Var}(X)} &= \mathbb{E}[X^2] - (\mathbb{E}[X])^2 & \boldsymbol{\operatorname{Var}(aX+b)} &= a^2 \operatorname{Var}(X) \\ \boldsymbol{\operatorname{Cov}(X,Y)} &= \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] & \boldsymbol{\rho_{X,Y}} &= \frac{\operatorname{Cov}(X,Y)}{\sqrt{\operatorname{Var}(X)\operatorname{Var}(Y)}} \in [-1, 1] \end{align*}

(7)

\mathbb{E}[aX + bY] = a\mathbb{E}[X] + b\mathbb{E}[Y] \quad \text{for any RVs } X, Y \text{ and constants } a, b

(8)

\begin{align*} \operatorname{Var}(aX) &= a^2 \operatorname{Var}(X) \\ \operatorname{Var}(X + Y) &= \operatorname{Var}(X) + \operatorname{Var}(Y) + 2\operatorname{Cov}(X, Y) \\ \operatorname{Var}(X + c) &= \operatorname{Var}(X) \quad \text{for constant } c \end{align*}

(9)

\begin{align*} \operatorname{Cov}(X,X) &= \operatorname{Var}(X) \\ \operatorname{Cov}(X,Y) &= \operatorname{Cov}(Y,X) \\ \operatorname{Cov}(aX + b, cY + d) &= ac \operatorname{Cov}(X,Y) \\ \operatorname{Cov}(X + Y, Z) &= \operatorname{Cov}(X,Z) + \operatorname{Cov}(Y,Z) \\ \operatorname{Cov}(X, Y + Z) &= \operatorname{Cov}(X,Y) + \operatorname{Cov}(X,Z) \end{align*}

(10)

Moment Generating Function¶

Let $X$ be a random variable with density function

f_X(x) = \begin{cases} e^{-2x} + \frac{1}{2}e^{-x} & \text{if } x > 0 \\ 0 & \text{otherwise} \end{cases}

(11)

\phi_X(t) = \mathbb{E}\left[e^{tX}\right] = \int_0^\infty e^{tx} \left(e^{-2x} + \frac{1}{2}e^{-x}\right) dx = \int_0^\infty \left(e^{(t-2)x} + \frac{1}{2}e^{(t-1)x}\right) dx

(12)

The moment generating function must have a finite integral to exist. For this integral to converge at $x \to \infty$ , both exponential terms must decay. The stricter condition is $t < 1$ , so the MGF is only defined for $t < 1$ .

$\mathbb{E}[X] = \phi_X'(0)$
$\mathbb{E}[X^2] = \phi_X''(0)$
...

Joint Distributions¶

Joint Distribution: $F_{X,Y}(x,y) = P(X \leq x, Y \leq y)$
Marginal Distribution: $F_X(x) = \lim_{y\to\infty} F_{X,Y}(x,y)$
Joint PMF: $p_{X,Y}(x,y) = P(X = x, Y = y)$ , Joint PDF: $f_{X,Y}(x,y)$
Marginal PDF: $f_X(x) = \int_{-\infty}^{\infty} f_{X,Y}(x,y) \, dy$

Conditional Distributions & Expectation¶

\begin{align*} \textbf{Discrete} & & & \textbf{Continous} \\ P_{X|Y}(x|y) &= P(X = x | Y = y) = \frac{P(X = x, Y = y)}{P(Y = y)} & f_{X|Y}(x|y) &= \frac{f_{X,Y}(x,y)}{f_Y(y)} \\ \mathbb{E}[X|Y = y] &= \sum_x x P_{X|Y}(x|y) & \mathbb{E}[X|Y = y] &= \int_{-\infty}^{\infty} x f_{X|Y}(x|y) \, dx \end{align*}

(13)

Important: $\mathbb{E}[X|Y]$ is itself a random variable (function of $Y$ )

Transformation of Random Variables¶

For a continuous RV $X$ with density $f_X(x)$ and a differentiable, strictly monotone function $g$ for $y$ in the range of $g$ ; $f_Y(y) = 0$ outside:

Y = g(X) \implies f_Y(y) = \frac{f_X(g^{-1}(y))}{|g'(g^{-1}(y))|}

(14)

Example: Joint Density of $X$ and $Y$ ¶

Joint density function: $f_{X,Y}(x,y) = \begin{cases} 10x^2 y, & \text{if } 0 \leq y \leq x \leq 1, \\ 0, & \text{otherwise} \end{cases}$

Marginal of $X$ : $f_X(x) = \int_0^x 10x^2 y \, dy = 5x^4, \quad \text{for } 0 \leq x \leq 1$
Marginal of $Y$ : $f_Y(y) = \int_y^1 10x^2 y \, dx = \frac{10y}{3}(1 - y^3), \quad \text{for } 0 \leq y \leq 1$
Independence Check: $f_X(x) \cdot f_Y(y) = 5x^4 \cdot \frac{10y}{3}(1 - y^3) \neq 10x^2 y = f_{X,Y}(x,y)$ $X$ and $Y$ are not independent
Distribution Function: $F_X(x) = \int_0^x 5t^4 \, dt = x^5, \quad \text{for } 0 \leq x \leq 1$
Expectation: $\mathbb{E}[X] = \int_0^1 x \cdot 5x^4 \, dx = 5 \int_0^1 x^5 \, dx = \boxed{\dfrac{5}{6}}$
Conditional Density: $f_{Y|X}(y|x) = \dfrac{f_{X,Y}(x,y)}{f_X(x)} = \dfrac{10x^2 y}{5x^4} = \boxed{\dfrac{2y}{x^2}}, \quad \text{for } 0 \leq y \leq x$
$\mathbb{P}(Y \leq X/2) = \int_0^1 \int_0^{x/2} 10x^2 y \, dy \, dx = \boxed{\dfrac{1}{4}}$

Law of Total Expectation¶

\mathbb{E}[X] = \mathbb{E}[\mathbb{E}[X|Y]]

(15)

Discrete $Y$ : $\mathbb{E}[X] = \sum_y \mathbb{E}[X|Y = y] P(Y = y)$
Continuous $Y$ : $\mathbb{E}[X] = \int_{-\infty}^{\infty} \mathbb{E}[X|Y = y] f_Y(y) \, dy$

Conditional Variance¶

\operatorname{Var}(X|Y = y) = \mathbb{E}[X^2|Y = y] - (\mathbb{E}[X|Y = y])^2

(16)

Law of Total Variance

\operatorname{Var}(X) = \mathbb{E}[\operatorname{Var}(X|Y)] + \operatorname{Var}(\mathbb{E}[X|Y])

(17)

Distributions¶

	PMF $P(X=k)$	Support	$\boldsymbol{\mathbb{E}[X]}$	$\boldsymbol{\operatorname{Var}(X)}$	MGF
Uniform $(a, b), n = b - a + 1$	$\frac{1}{n}$	$k \in \{1,\dots,n\}$	$\frac{a+b}{2}$	$\frac{n^2-1}{12}$	$\frac{e^{at} - e^{(b+1)t}}{n(1 - e^t)}$
Bernoulli $(p)$	$p^k(1-p)^{1-k}$	$k \in \{0,1\}$	$p$	$p(1-p)$	$pe^t + (1-p)$
Binomial $(n, p)$	$\binom{n}{k}p^k(1-p)^{n-k}$	$k \in \{0,\dots,n\}$	$np$	$np(1-p)$	$(pe^t + 1-p)^n$
Geometric $(p)$	$(1-p)^{k-1}p$	$k \in \mathbb{N}$	$\frac{1}{p}$	$\frac{1-p}{p^2}$	$\frac{pe^t}{1-(1-p)e^t}$
Poisson $(\lambda)$	$\frac{\lambda^k e^{-\lambda}}{k!}$	$k \in \mathbb{N}_0$	$\lambda$	$\lambda$	$e^{\lambda(e^t-1)}$

	PDF $f_X(x)$	Support	$\boldsymbol{\mathbb{E}[X]}$	$\boldsymbol{\operatorname{Var}(X)}$	MGF
Uniform $(a, b)$	$\frac{1}{b-a}$	$x \in [a,b]$	$\frac{a+b}{2}$	$\frac{(b-a)^2}{12}$	$\begin{cases}\frac{e^{tb} - e^{ta}}{t(b-a)} & \text{for } t \neq 0 \\ 1 & \text{for } t = 0 \end{cases}$
Exponential $(\lambda)$	$\lambda e^{-\lambda x}$	$x \geq 0$	$\frac{1}{\lambda}$	$\frac{1}{\lambda^2}$	$\frac{\lambda}{\lambda - t}$
Gamma $(k, \lambda)$	$\frac{\lambda^k}{\Gamma(k)} x^{k-1} e^{-\lambda x}$ , for $\Gamma(n) = (n-1)!$	$x \geq 0$	$\frac{k}{\lambda}$	$\frac{k}{\lambda^2}$	$\left(\frac{\lambda}{\lambda - t}\right)^k$
Normal $(\mu, \sigma^2)$	$\frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}}$	$x \in \mathbb{R}$	$\mu$	$\sigma^2$	$e^{t\mu + \frac{1}{2}t^2\sigma^2}$

Normal Distribution

Standard Normal: $Z \sim N(0,1)$ with standard normal CDF $F_X(z) = \Phi(z) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^z e^{-t^2/2} \, dt$
Standardization: If $X \sim N(\mu, \sigma^2)$ , then $Z = \frac{X - \mu}{\sigma} \sim N(0,1)$
Probability Calculation: $P(X \leq t) = \Phi\left(\frac{t - \mu}{\sigma}\right)$
Symmetric Properties: $\Phi(-z) = 1 - \Phi(z) \quad \text{and} \quad \Phi(0) = 0.5$

Limit Theorems¶

Given: $X_i$ : i.i.d. RV with $\mathbb{E}[X_i] = \mu$ , $\text{Var}(X_i) = \sigma^2$ , then:

Law of Large Numbers: P\left(\left \frac{S_n}{n} - \mu\right| > \varepsilon\right) \xrightarrow{n \to \infty} 0 for any $\varepsilon > 0$ (convergence to $\mu$ in the limit)
Weak Law of Large Numbers, as $n \to \infty$ : $\frac{S_n}{n} \xrightarrow{\mathbb{P}} \mu$ (i.e. $\mu = p = \frac{1}{3})$
Central Limit Theorem: $P\left(Z = \frac{S_n - \mu}{\sigma} \leq \varepsilon \right) \xrightarrow{n \to \infty} P(\mathcal{N}(0,1) \leq \varepsilon)$ for any $\varepsilon \in \mathbb{R}$ (convergence in distribution, fluctuations of $S_n$ around its mean (see above))

Example: Law of Large Numbers¶

Precise statement: $\lim_{n \to \infty} \mathbb{P}\left(\left|\frac{S_n}{n} - \frac{1}{3}\right| \geq \epsilon\right) = 0 \quad \forall \epsilon > 0$

Weak Law of Large Numbers, as $n \to \infty$ : $\frac{S_n}{n} \xrightarrow{\mathbb{P}} p = \frac{1}{3}$

Example: Central Limit Theorem Approximation¶

Given: $S \sim \text{Binomial}(n=160'000, p=0.5)$ .

$\mathbb{E}[S] = 80'000$ , $\text{Var}(S) = 40'000$ , $\sigma = \sqrt{'text{Var}(S)} = 200$

\mathbb{P}(S \leq 80{,}200) \approx \mathbb{P}\left(Z \leq \frac{80{,}200.5 - 80{,}000}{200}\right) = \mathbb{P}(Z \leq 1.0025) \approx \Phi(1) \approx \boxed{0.841}.

(18)

Chebyshev’s Inequality¶

\begin{align*} P(|X-\mu| \geq \varepsilon) &\leq \frac{\sigma^2}{\varepsilon^2} \end{align*}

(19)

Example: Chebyshev’s Inequality for $S_{900}$ ¶

Given: $S_{900} \sim \text{Binomial}(n=900, p=1/3)$ .

$\mathbb{E}[S_{900}] = 300$ , $\text{Var}(S_{900}) = 900 \cdot \frac{1}{3} \cdot \frac{2}{3} = 200$

\mathbb{P}(|S_{900} - 300| \leq 20) \geq 1 - \frac{\text{Var}(S_{900})}{20^2} = 1 - \frac{200}{400} = \frac{1}{2}.

(20)

The probability is greater than or equal to $\frac{1}{2}$

Markov’s Inequality¶

For any non-negative random variable $X$ (i.e., $X \geq 0$ ) and any $a \gt 0$ , the following is always true:

P(X \geq a) \leq \frac{\mathbb{E}[X]}{a}

(21)

Markov Chains¶

\begin{align*} \textbf{Markov Property: } P(X_{n+1}|X_n, \ldots, X_0) &= P(X_{n+1}|X_n) & \textbf{Chapman-Kolmogorov: } p_{ij}^{(m+n)} &= \sum_k p_{ik}^{(m)} p_{kj}^{(n)} \\ \textbf{Stationary: } \vec{\pi} &= \vec{\pi} P & \textbf{Convergence: } \vec{\nu} P^n &\to \vec{\pi} \end{align*}

(22)

Transition Matrix¶

One-step transition probabilities $p_{ij} = P(X_{n+1} = j \mid X_n = i)$ form the transition matrix:

P = \begin{pmatrix} p_{11} & p_{12} & \cdots & p_{1k} \\ p_{21} & p_{22} & \cdots & p_{2k} \\ \vdots & \vdots & \ddots & \vdots \\ p_{k1} & p_{k2} & \cdots & p_{kk} \\ \end{pmatrix} \in \mathbb{R}^{k \times k}

(23)

$p_{ij} \in [0,1]$
$\sum_{j=1}^k p_{ij} = 1$ for all $i$ (stochastic matrix)

Irreducibility¶

p_{ij}^{(n)} > 0

(24)

Every state can be reached from every other state in a finite number of steps. The state space cannot be partitioned into disjoint subsets that the chain never leaves.

Aperiodicity¶

A Markov chain is aperiodic if all its states are aperiodic. The period of state $i$ is defined as:

d(i) = \gcd\{n \geq 1: p_{ii}^{(n)} > 0\}

(25)

If $d(i) = 1$ , state $i$ is aperiodic
If $d(i) > 1$ , state $i$ is periodic with period $d(i)$

An aperiodic chain can return to any state at irregular intervals, not just multiples of a fixed period.

Irreducibility & Aperiodicity Combined¶

If a finite-state Markov chain is both irreducible and aperiodic, then:

It has a unique stationary distribution $\vec{\pi}$
For any initial distribution: $\vec{\pi}(n) \to \vec{\pi}$ as $n \to \infty$
All rows of $P^n$ converge to $\vec{\pi}$ as $n \to \infty$

Example: Californian Weather Model¶

Time-homogeneous Markov chain $\{X_n: n \in \mathbb{N}\}$ with initial distribution $\vec{\pi}(0) := \left(\frac{1}{6}, \frac{5}{6}\right)$ and transition matrix

\mathbf{P} = \begin{pmatrix} 0.5 & 0.5 \\ 0.1 & 0.9 \end{pmatrix}

(26)

Show for any $n \in \mathbb{N}$ : $\vec{\pi}(n) = \vec{\pi}(0)$

Base case ( $n=0$ ): Trivially $\vec{\pi}(0) = \vec{\pi}(0)$ .
Inductive step: Assume $\vec{\pi}(n) = \vec{\pi}(0)$ . Then

\vec{\pi}(n+1) = \vec{\pi}(n) \mathbf{P} = \left(\frac{1}{6}, \frac{5}{6}\right) \begin{pmatrix} 0.5 & 0.5 \\ 0.1 & 0.9 \end{pmatrix} = \left(\frac{1}{6} \cdot 0.5 + \frac{5}{6} \cdot 0.1, \frac{1}{6} \cdot 0.5 + \frac{5}{6} \cdot 0.9\right) = \left(\frac{1}{6}, \frac{5}{6}\right)

(27)

Interpretation: The initial distribution $\left(\frac{1}{6}, \frac{5}{6}\right)$ is stationary for this chain.

Stochastic Processes¶

\begin{align*} \textbf{Mean: } m_X(t) &= \mathbb{E}[X_t] & \textbf{Variance: } \text{Var}(X_t) &= \mathbb{E}[X_t^2] - (\mathbb{E}[X_t])^2 \\ \textbf{Covariance: } C_X(t_1,t_2) &= R_X(t_1,t_2) - m_X(t_1)m_X(t_2) & \textbf{Correlation: } R_X(t_1, t_2) &= \mathbb{E}[X_{t_1} X_{t_2}] \end{align*}

(28)

zero-mean processes: $R_X(t_1, t_2) = \operatorname{Cov}(X_{t_1}, X_{t_2})$

	Description	$\boldsymbol{m_X(t)}$	$\boldsymbol{Var(X_t)}$	$\boldsymbol{R_X(t_1, t_2)}$
Bernoulli	$X_n$ i.i.d. $\pm 1$ valued	$2p-1$	$4p(1-p)$	$(2p-1)^2$ for $t_1\neq t_2$
Random Walk	$X_n = \sum_{i=1}^n B_i$	0 (if $p=1/2$ )	$n$	$\min(t_1, t_2)$
Brownian Motion	Continuous limit of RW	0	$t$	$\min(t_1, t_2)$
Poisson	Counts events in time	$\lambda t$	$\lambda t$	$\lambda \min(t_1, t_2)$
White Noise	$X_t \sim \text{i.i.d.}(0, \sigma^2)$			$\sigma^2 \delta_{t_1,t_2}$ for $\delta_{t_1,t_2}$ := if ( $t_1 = t_2$ ) 1 else 0
Wiener Process	$W_t \sim \mathcal{N}(0, t)$			$\min(t_1, t_2)$

Wide-Sense Stationary (WSS):

Constant mean: $m_X(t) = m$ for all $t$
Correlation depends only on time difference: $R_X(t_1, t_2) = R_X(|t_2 - t_1|)$

Example: Stochastic Process $Y_n = \frac{1}{2}(X_n + X_{n-1})$ ¶

Given: $X_n$ i.i.d., $\mathbb{E}[X_n] = 2$ , $\text{Var}(X_n) = 3$

Moments of $Y_n$
$\mathbb{E}[Y_n] = \frac{1}{2}(\mathbb{E}[X_n] + \mathbb{E}[X_{n-1}]) = \boxed{2}$
$\text{Var}(Y_n) = \frac{1}{4}(\text{Var}(X_n) + \text{Var}(X_{n-1})) = \boxed{\dfrac{3}{2}}$
$\mathbb{E}[Y_n^2] = \text{Var}(Y_n) + (\mathbb{E}[Y_n])^2 = \frac{3}{2} + 4 = \boxed{\dfrac{11}{2}}$
Correlation Function:
If $X_m$ and $X_n$ i.i.d.: $R_X(m,n) = \mathbb{E}[X_m] \cdot \mathbb{E}[X_n] = \begin{cases} 7 = \mathbb{E}[X_n^2] & \text{if } m = n \\ 4 & \text{if } m \neq n \end{cases}$
else if $X_m$ and $X_n$ dependent: $R_Y(m,n) = \mathbb{E}[Y_m Y_n] = \mathbb{E}\left[\left(\frac{X_m + X_{m-1}}{2}\right) \left(\frac{X_n + X_{n-1}}{2}\right)\right]= \begin{cases} \dfrac{11}{2}, & \text{if } m = n, \\ \dfrac{19}{4}, & \text{if } |m - n| = 1, \\ 4, & \text{if } |m - n| \geq 2. \end{cases}$
Expand the product inside the expectation
Use linearity of expectation to split into sums of expectations
Evaluate each term using:
- $\mathbb{E}[X_i X_j]$ (from the correlation function of $X$ )
- $\mathbb{E}[X_i^2]$ (from the second moment of $X$ )
Process Properties:
Identically Distributed: Yes
Independent: No
Wide-Sense Stationary: Yes

Poisson Process¶

Models the number of events with $\{N(t) : t \geq 0\}$ . It satisfies the following properties:

Independent Increments: The number of events in non-overlapping intervals are independent.
Memoryless Property: $\mathbb{P}(T_i > s + t \mid T_i > s) = \mathbb{P}(T_i > t)$ .
Stationary Increments: The number of events in an interval depends only on the length of the interval, not its starting point.
Rare Events: The probability of more than one event in a very small interval is negligible.
Rate Parameter ( $\lambda$ ): Average number of events per unit time.
Counting Process: $N(t)$ counts the number of events in the interval $[0, t]$ .
Inter-arrival Times: The time between consecutive events follows an exponential distribution with rate $\lambda$ .

The number of events $N(t)$ in an interval of length $t$ follows a Poisson distribution:

\mathbb{P}(N(t) = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}, \quad k = 0, 1, 2, \dots

(29)

For $N(t) \sim \text{Poisson}(\lambda t)$ :

Mean: $\mathbb{E}[N(t)] = \lambda t$
Variance: $\text{Var}(N(t)) = \lambda t$
The time between consecutive events, $T_i = S_i - S_{i-1}$ (where $S_i$ is the time of the $i$ -th event), follows an exponential distribution:
$f_{T_i}(t) = \lambda e^{-\lambda t}, \quad t \geq 0$
Probability of $k$ Events in Time $t$ :
$\mathbb{P}(N(t) = k) = \frac{(\lambda t)^k e^{-\lambda t}}{k!}$
Probability of at Most $k$ Events in Time $t$ :
$\mathbb{P}(N(t) \leq k) = e^{-\lambda t} \sum_{i=0}^k \frac{(\lambda t)^i}{i!}$
Expected Time Until the $k$ -th Event:
$\mathbb{E}[S_k] = \frac{k}{\lambda}$
Probability of No Events in Time $t$ :
$\mathbb{P}(N(t) = 0) = e^{-\lambda t}$

Example: Poisson Process (Rate $\lambda = 2$ per month)¶

$\boldsymbol{\mathbb{P}(\text{at most 4 failures in first month})}$ :
$N(1) \sim \text{Poisson}(2), \quad \mathbb{P}(N(1) \leq 4) = e^{-2} \sum_{k=0}^4 \frac{2^k}{k!} = \boxed{7e^{-2}}$
$\boldsymbol{\mathbb{P}(\text{3 in first month AND 2 in next two months})}$ :
$\mathbb{P}(N(1) = 3) = \frac{4}{3} e^{-2}, \quad \mathbb{P}(N(3) - N(1) = 2) \implies N(3) - N(1) \sim \text{Poisson}(\lambda \cdot \text{length}) = \text{Poisson}(2 \cdot 2) \implies 8 e^{-4}$
Joint probability: $\boxed{\dfrac{32}{3} e^{-6}}$
$\boldsymbol{\mathbb{P}(\text{3 in first month} \mid \text{4 in first two months})}$ :
$\mathbb{P}(N(1) = 3 \mid N(2) = 4) = \frac{\mathbb{P}(N(1) = 3 \text{ and } N(2) - N(1) = 1)}{\mathbb{P}(N(2) = 4)} = \boxed{\dfrac{1}{4}}$
Expected Time Until 10th Failure:
For a Poisson process, the time of the $k$ -th arrival is $\text{Gamma}(k, \lambda)$
Mean time: $\boxed{5 \text{ months}}$ (since $\mathbb{E}[T_{10}] = \frac{10}{\lambda} = 5$ )