B Overview of probability, random variables, and random vectors

B.1 Probability Basics

The mathematical field of probability attempts to quantify how likely certain outcomes are, where the outcomes are produced by a random experiment (defined below). In what follows, we assume you have a basic understanding of set theory and notation.

We review some basic probability-related terminology in Table B.1.

Table B.1: Basic terminology used in probability.

term	notation	definition
experiment	N/A	A mechanism that produces outcomes that cannot be predicted with absolute certainty.
outcome	\(\omega\)	The simplest kind of result produced by an experiment.
sample space	\(\Omega\)	The set of all possible outcomes an experiment can produce.
event	\(A\), \(A_i\), \(B\), etc.	Any subset of \(\Omega\).
empty set	\(\emptyset\)	The event that includes no outcomes.

Some comments about the terms in Table B.1:

• Outcomes may also be referred to as points, realizations, or elements.
• An event is a subset of outcomes.
• The empty set is a subset of \(\Omega\), but not an outcome of \(\Omega\).
• The empty set is a subset of every event \(A\subseteq \Omega\).

We now review some basic set operations and related facts. Let \(A\) and \(B\) be two events contained in \(\Omega\).

• The intersection of \(A\) and \(B\), denoted \(A \cap B\) is the set of outcomes that are common to both \(A\) and \(B\), i.e., \(A \cap B = \{\omega \in \Omega: \omega \in A\;\mathrm{and}\;\omega \in B\}\).
  • Events \(A\) and \(B\) are disjoint if \(A\cap B = \emptyset\), i.e., if there are no outcomes common to events \(A\) and \(B\).
• The union of \(A\) and \(B\), denoted \(A \cup B\) is the set of outcomes that are in \(A\) or \(B\) or both, i.e., \(A \cup B = \{\omega \in \Omega: \omega \in A\;\mathrm{or}\;\omega \in B\}\).
• The complement of \(A\), denoted \(A^c\) is the set of outcomes that are in \(\Omega\) but are not in \(A\), i.e., \(A^c = \{\omega \in \Omega: \omega \not\in A\}\).
  • The complement of \(A\) may also be denoted as \(\overline{A}\) or \(A'\).
• The set difference between \(A\) and \(B\), denoted \(A \setminus B\), is the outecomes of \(A\) that are not in \(B\), i.e., \(A\setminus B = \{\omega \in A: \omega \not\in B\}\).
  • The set difference between \(A\) and \(B\) may also be denoted by \(A-B\).
  • The set difference is order specific, i.e., \((A\setminus B) \not= (B\setminus A)\) in general.

A function \(P\) that assigns a real number \(P(A)\) to every event \(A\) is a probability distribution if it satisfies three properties:

1. \(P(A)\geq 0\) for all \(A\in \Omega\).
2. \(P(\Omega)=P(\omega \in \Omega) = 1\). Alternatively, \(P(A \subseteq \Omega) = 1\).
3. If \(A_1, A_2, \ldots\) are disjoint, then \(P\left(\bigcup_{i=1}^\infty A_i \right)=\sum_{i=1}^\infty P(A_i)\).

A set of events \(\{A_i:i\in I\}\) are independent if \[ P\left(\cap_{i\in J} A_i \right)=\prod_{i\in J} P(A_i ) \] for every finite subset \(J\subseteq I\).

The conditional probability of \(A\) given \(B\), denoted as \(P(A\mid B)\), is the probability that \(A\) occurs given that \(B\) has occurred, and is defined as \[ P(A\mid B) = \frac{P(A\cap B)}{P(B)}, \quad P(B) > 0. \]

Some additional facts about probabilities:

• Complement rule: \(P(A^c) = 1 - P(A)\).
• Addition rule: \(P(A\cup B) = P(A) + P(B) - P(A \cap B)\).
• Bayes’ rule: Assuming \(P(A) > 0\) and \(P(B) > 0\), then \[P(A\mid B) = \frac{P(B\mid A)P(A)}{P(B)}.\]
• Law of Total Probability: Let \(B_1, B_2, \ldots\) be a countably infinite partition of \(\Omega\). Then \[P(A) = \sum_{i=1}^{\infty} P(A \cap B_i) = \sum_{i=1}^{\infty} P(A \mid B_i) P(B_i).\]

B.2 Random Variables

A random variable \(Y\) is a mapping/function \[ Y:\Omega\to\mathbb{R} \] that assigns a real number \(Y(\omega)\) to each outcome \(\omega\). We typically drop the \((\omega)\) notation for simplicity.

The cumulative distribution function (CDF) of \(Y\), \(F_Y\), is a function \(F_Y:\mathbb{R}\to [0,1]\) defined by \[ F_Y (y)=P(Y \leq y). \] The subscript of \(F\) indicates the random variable the CDF describes. E.g., \(F_X\) denotes the CDF of the random variable \(X\) and \(F_Y\) denotes the CDF of the random variable \(Y\). The subscript can be dropped when the context makes it clear what random variable the CDF describes. An \(F\)-distributed random variable is one that has the \(F\) distribution.

The support of \(Y\), \(\mathcal{S}\), is the smallest set such that \(P(Y\in \mathcal{S})=1\).

B.2.1 Discrete random variables

\(Y\) is a discrete random variable if it takes countably many values \(\{y_1, y_2, \dots \} = \mathcal{S}\).

The probability mass function (pmf) for \(Y\) is \(f_Y (y)=P(Y=y)\), where \(y\in \mathbb{R}\), and must have the following properties:

1. \(0 \leq f_Y(y) \leq 1\).
2. \(\sum_{y\in \mathcal{S}} f_Y(y) = 1\).

Additionally, the following statements are true:

• \(F_Y(c) = P(Y \leq c) = \sum_{y\in \mathcal{S}:y \leq c} f_Y(y)\).
• \(P(Y \in A) = \sum_{y \in A} f_Y(y)\) for some event \(A\).
• \(P(a \leq Y \leq b) = \sum_{y\in\mathcal{S}:a\leq y\leq b} f_Y(y)\).

The expected value, mean, or first moment of \(Y\) is defined as \[ E(Y) = \sum_{y\in \mathcal{S}} y f_Y(y), \] assuming the sum is well-defined.

The variance of \(Y\) is defined as \[ \mathrm{var}(Y)=E(Y-E(Y))^2 = \sum_{y\in \mathcal{S}} (y - E(Y))^2 f_Y(y). \]

Note that \(\mathrm{var}(Y)=E(Y-E(Y))^2=E(Y^2)-[E(Y)]^2\). The last expression is often easier to compute.

The standard deviation of Y is \[SD(Y)=\sqrt{\mathrm{var}(Y) }.\]

B.2.1.1 Example (Bernoulli distribution)

A random variable \(Y\) is said to have a Bernoulli distribution with probability \(\theta\), denoted \(Y\sim \mathsf{Bernoulli}(\theta)\), if \(\mathcal{S} = \{0, 1\}\) and \(P(Y = 1) = \theta\), where \(\theta\in (0,1)\).

The pmf of a Bernoulli random variable is \[f_Y(y) = \theta^y (1-\theta)^{(1-y)}.\]

The mean of a Bernoulli random variable is \[E(Y)=0(1-\theta )+1(\theta)=\theta.\]

The variance of a Bernoulli random variable is \[\mathrm{var}(Y)=(0-\theta)^2(1-\theta)+(1-\theta)^2\theta = \theta(1-\theta).\]

B.2.2 Continuous random variables

\(Y\) is a continuous random variable if there exists a function \(f_Y (y)\) such that:

1. \(f_Y (y)\geq 0\) for all \(y\),
2. \(\int_{-\infty}^\infty f_Y (y) dy = 1\),
3. \(a\leq b\), \(P(a<Y<b)=\int_a^b f_Y (y) dy\).

The function \(f_Y\) is called the probability density function (pdf).

Additionally, \(F_Y (y)=\int_{-\infty}^y f_Y (y) dy\) and \(f_Y (y)=F'_Y(y)\) for any point \(y\) at which \(F_Y\) is differentiable.

The mean of a continuous random variables \(Y\) is defined as \[ E(Y) = \int_{-\infty}^{\infty} y f_Y(y) dy = \int_{y\in\mathcal{S}} y f_Y(y). \] assuming the integral is well-defined.

The variance of a continuous random variable \(Y\) is defined by \[ \mathrm{var}(Y)= E(Y-E(Y))^2=\int_{-\infty}^{\infty} (y - E(Y))^2 f_Y(y) dy = \int_{y\in\mathcal{S}} (y - E(Y))^2 f_Y(y) dy. \]

B.2.2.1 Example (Exponential distribution)

A random variable \(Y\) is said to have an exponential distribution rate parameter \(\lambda\), denoted with \(Y \sim \mathsf{Exp}(\lambda)\) if \(\mathcal{S} = \{y\in \mathbb{R}:y\geq 0\}\) and \[f_Y(y)=\lambda\exp(-\lambda y).\]

The mean of an exponential random variable is

\[ \begin{aligned} E(Y) &= \int_{0}^{\infty} y\lambda \exp(-\lambda y)\;dy \\ &= -\exp(-\lambda y)(\lambda^{-1}+y)\biggr]^{\infty}_{0}\\ &=\lambda^{-1} \\ &=\frac{1}{\lambda}. \end{aligned} \]

Note that this process involves integration by parts, which is not shown. Similarly, \(E(Y^2)=2\lambda^{-2}\). Thus,

\[ \begin{aligned} \mathrm{var}(Y)&= E(Y^2)-[E(Y)]^2\\ &=2\lambda^{-2}-[\lambda^{-1}]^2\\ &=\lambda^{-2}\\ &=\frac{1}{\lambda^{2}}. \end{aligned} \]

B.2.3 Useful facts for transformations of random variables

Let \(Y\) be a random variable and \(a\in\mathbb{R}\) be a constant. Then:

• \(E(a) = a\).
• \(E(aY) = a E(Y)\).
• \(E(a + Y) = a + E(Y)\).
• \(\mathrm{var}(a) = 0\).
• \(\mathrm{var}(aY) = a^2 \mathrm{var}(Y)\).
• \(\mathrm{var}(a + Y) = \mathrm{var}(Y)\).
• For a discrete random variable and a function \(g\), \[E(g(Y))=\sum_{y\in\mathcal{S}}g(y)f_Y(y),\] assuming the sum is well-defined.
• For a continuous random variable and a function \(g\), \[E(g(Y))=\int_{y\in\mathcal{S}}g(y)f_Y(y)\;dy,\] assuming the integral is well-defined.

B.3 Multivariate distributions

B.3.1 Basic properties

Let \(Y_1,Y_2,\ldots,Y_n\) denote \(n\) random variables with supports \(\mathcal{S}_1,\mathcal{S}_2,\ldots,\mathcal{S}_n\), respectively.

If the random variables are jointly discrete (i.e., all discrete), then the joint pmf \(f(y_1,\ldots,y_n)=P(Y_1=y_1,\ldots,Y_n=y_n)\) satisfies the following properties:

1. \(0\leq f(y_1,\ldots,y_n )\leq 1\),
2. \(\sum_{y_1\in\mathcal{S}_1}\cdots \sum_{y_n\in\mathcal{S}_n} f(y_1,\ldots,y_n ) = 1\),
3. \(P((Y_1,\ldots,Y_n)\in A)=\sum_{(y_1,\ldots,y_n) \in A} f(y_1,\ldots,y_n)\).

In this context, \[ E(Y_1 \cdots Y_n)=\sum_{y_1\in\mathcal{S}_1} \cdots \sum_{y_n\in\mathcal{S}_n}y_1 \cdots y_n f(y_1,\ldots,y_n). \]

In general, \[ E(g(Y_1,\ldots,Y_n))=\sum_{y_1\in\mathcal{S}_1} \cdots \sum_{y_n\in\mathcal{S}_n} g(y_1, \ldots, y_n) f(y_1,\ldots,y_n), \] where \(g\) is a function of the random variables.

If the random variables are jointly continuous, then \(f(y_1,\ldots,y_n)\) is the joint pdf if it satisfies the following properties:

1. \(f(y_1,\ldots,y_n ) \geq 0\),
2. \(\int_{y_1\in\mathcal{S}_1}\cdots \int_{y_n\in\mathcal{S}_n} f(y_1,\ldots,y_n ) dy_n \cdots dy_1 = 1\),
3. \(P((Y_1,\ldots,Y_n)\in A)=\int \cdots \int_{(y_1,\ldots,y_n) \in A} f(y_1,\ldots,y_n) dy_n\ldots dy_1\).

In this context, \[ E(Y_1 \cdots Y_n)=\int_{y_1\in\mathcal{S}_1} \cdots \int_{y_n\in\mathcal{S}_n} y_1 \cdots y_n f(y_1,\ldots,y_n) dy_n \ldots dy_1. \]

In general, \[ E(g(Y_1,\ldots,Y_n))=\int_{y_1\in\mathcal{S}_1} \cdots \int_{y_n\in\mathcal{S}_n} g(y_1, \ldots, y_n) f(y_1,\ldots,y_n) dy_n \cdots dy_1, \] where \(g\) is a function of the random variables.

B.3.2 Marginal distributions

If the random variables are jointly discrete, then the marginal pmf of \(Y_1\) is obtained by summing over the other variables \(Y_2, ..., Y_n\): \[f_{Y_1}(y_1)=\sum_{y_2\in\mathcal{S}_2}\cdots \sum_{y_n\in\mathcal{S}_n} f(y_1,\ldots,y_n).\]

Similarly, if the random variables are jointly continuous, then the marginal pdf of \(Y_1\) is obtained by integrating over the other variables \(Y_2, ..., Y_n\) \[f_{Y_1}(y_1)=\int_{y_2\in\mathcal{S}_2}\cdots \int_{y_n\in\mathcal{S}_n} f(y_1,\ldots,y_n) dy_n \cdots dy_2. \]

B.3.3 Independence of random variables

Random variables \(X\) and \(Y\) are independent if \[F(x, y) = F_X(x) F_Y(y).\]

Alternatively, \(X\) and \(Y\) are independent if \[f(x, y) = f_X(x)f_Y(y).\]

B.3.4 Conditional distributions

Let \(X\) and \(Y\) be random variables. Then assuming \(f_Y(y)>0\), the conditional distribution of \(X\) given \(Y = y\), denoted \(X|Y=y\) comes from Bayes’ formula: \[f(x|y) = \frac{f(x, y)}{f_{Y}(y)}, \quad f_Y(y)>0.\]

B.3.5 Covariance

The covariance between random variables \(X\) and \(Y\) is \[\mathrm{cov}(X,Y)=E[(X-E(X))(Y-E(Y))]=E(XY)-E(X)E(Y).\]

B.3.6 Useful facts for transformations of multiple random variables

Let \(a\) and \(b\) be scalar constants. Let \(Y\) and \(Z\) be random variables. Then:

• \(E(aY+bZ)=aE(Y)+bE(Z)\).
• \(\mathrm{var}(Y+Z)=\mathrm{var}(Y)+\mathrm{var}(Z)+2\mathrm{cov}(Y, Z)\).
• \(\mathrm{cov}(a,Y)=0\).
• \(\mathrm{cov}(Y,Y)=\mathrm{var}(Y)\).
• \(\mathrm{cov}(aY, bZ)=ab\mathrm{cov}(Y, Z)\).
• \(\mathrm{cov}(a + Y,b + Z)=\mathrm{cov}(Y, Z)\).

If \(Y\) and \(Z\) are also independent, then:

• \(E(YZ)=E(Y)E(Z)\).
• \(\mathrm{cov}(Y, Z)=0\).

In general, if \(Y_1, Y_2, \ldots, Y_n\) are a set of random variables, then:

• \(E(\sum_{i=1}^n Y_i) = \sum_{i=1}^n E(Y_i)\), i.e., the expectation of the sum of random variables is the sum of the expectation of the random variables.
• \(\mathrm{var}(\sum_{i=1}^n Y_i) = \sum_{i=1}^n \mathrm{var}(Y_i) + \sum_{j=1}^n\sum_{1\leq i<j\leq n}2\mathrm{cov}(Y_i, Y_j)\), i.e., the variance of the sum of random variables is the sum fo the variables’ variances plus the sum of twice all possible pairwise covariances.

If in addition, \(Y_1, Y_2, \ldots, Y_n\) are all independent of each other, then:

• \(\mathrm{var}(\sum_{i=1}^n Y_i) = \sum_{i=1}^n \mathrm{var}(Y_i)\) since all pairwise covariances are 0.

B.3.7 Example (Binomial)

A random variable \(Y\) is said to have a Binomial distribution with \(n\) trials and probability of success \(\theta\), denoted \(Y\sim \mathsf{Bin}(n,\theta)\) when \(\mathcal{S}=\{0,1,2,\ldots,n\}\) and the pmf is \[f(y\mid\theta) = \binom{n}{y} \theta^y (1-\theta)^{(n-y)}.\]

An alternative explanation of a Binomial random variable is that it is the sum of \(n\) independent and identically-distributed Bernoulli random variables. Alternatively, let \(Y_1,Y_2,\ldots,Y_n\stackrel{i.i.d.}{\sim} \mathsf{Bernoulli}(\theta)\), where i.i.d. stands for independent and identically distributed, i.e., \(Y_1, Y_2, \ldots, Y_n\) are independent random variables with identical distributions. Then \(Y=\sum_{i=1}^n Y_i \sim \mathsf{Bin}(n,\theta)\).

A Binomial random variable with \(\theta = 0.5\) models the question: what is the probability of flipping \(y\) heads in \(n\) flips?

Using this information and the facts above, we can easily determine the mean and variance of \(Y\).

Using our results from Section B.2.1.1, we can see that \(E(Y_i) = \theta\) for \(i=1,2,\ldots,n\). Similarly, \(\mathrm{var}(Y_i)=\theta(1-\theta)\) for \(i=1,2,\ldots,n\).

We determine that: \[ E(Y)=E\biggl(\sum_{i=1}^n Y_i\biggr)=\sum_{i=1}^n E(Y_i) = \sum_{i=1}^n \theta = n\theta. \]

Similarly, since \(Y_1, Y_2, \ldots, Y_n\) are i.i.d. and using the fact in Section B.3.6, we see that \[ \mathrm{var}(Y) = \mathrm{var}(\sum_{i=1}^n Y_i) = \sum_{i=1}^n\mathrm{var}(Y_i)=\sum_{i=1}^n \theta(1-\theta) = n\theta(1-\theta). \]

B.3.8 Example (Continuous bivariate distribution)

Hydration is important for health. Like many people, the author has a water bottle he uses to say hydrated through the day and drinks several liters of water per day. Let’s say the author refills his water bottle every 3 hours. Let \(Y\) denote the proportion of the water bottle filled with water at the beginning of the 3-hour window. Let \(X\) denote the amount of water the author consumes in the 3-hour window (measured in the the proportion of total water bottle capacity). We know that \(0\leq X \leq Y \leq 1\). The joint density of the random variables is

\[ f(x,y)=4y^2,\quad 0 \leq x\leq y\leq 1, \] and 0 otherwise.

We answer a series of questions about this distribution.

Q1: Determine \(P(0.5\leq X\leq 1, 0.75\leq Y)\).

Note that the comma between the two events means “and”.

Since \(X\) must be no more than \(Y\), we can answer this question as

\[ \int_{0.75}^{1} \int_{0.5}^{y} 4y^2\;dx\;dy=229/768\approx 0.30. \]

Q2: Determine the marginal distributions of \(X\) and \(Y\).

To find the marginal distribution of \(X\), we must integrate the joint pdf with respect to the limits of \(Y\). Don’t forget to include the support of the pdf of \(X\) (which after integrating out \(Y\), must be between 0 and 1). \[ \begin{aligned} f_X(x) &=\int_{x}^1 4y^2\;dy \\ &=\frac{4}{3}(1-x^3),\quad 0\leq x \leq 1. \end{aligned} \]

Similarly, \[ \begin{aligned} f_Y(y) &=\int_{0}^y 4y^2\;dx \\ &=4y^3,\quad 0\leq y \leq 1. \end{aligned} \]

Q3: Determine the means of \(X\) and \(Y\).

The mean of \(X\) is the integral of \(x f_X(x)\) over the support of \(X\), i.e., \[ E(X) =\int_{0}^1 x\biggl(\frac{4}{3}(1-x^3)\biggr)\;dx = \frac{2}{5} \]

Similarly, \[ E(Y) =\int_{0}^1 y(4y^3)\;dy = \frac{4}{5} \]

Q4: Determine the variances of \(X\) and \(Y\).

We use the formula \(\mathrm{var}(X)=E(X^2)-[E(X)^2]\) to compute the variances. First, \[ E(X^2) =\int_{0}^1 x^2\biggl(\frac{4}{3}(1-x^3)\biggr)\;dx = \frac{2}{9} \] Second, \[ E(Y^2) =\int_{0}^1 y^2(4y^3)\;dy = \frac{2}{3} \] Thus,

\[\mathrm{var}(X)=2/9-(2/5)^2=\frac{14}{225}\] \[\mathrm{var}(Y)=2/3-(4/5)^2=\frac{2}{75}\]

Q5: Determine the mean of \(XY\).

The mean of \(XY\) requires us to integrate the product of \(xy\) and the joint pdf over the joint support of \(X\) and \(Y\). Specifically, \[ E(XY)=\int_{0}^{1}\int_{0}^{y} xy(4y^2)\;dx\;dy= \frac{1}{3} \]

Q6: Determine the covariance of \(X\) and \(Y\).

Using our previous work, we see that \[ \mathrm{cov}(X,Y)=E(XY) - E(X)E(Y)=1/3-(2/5)(4/5)=-\frac{1}{75} \]

Q7: Determine the mean and variance of \(Y-X\), i.e., the average amount of water remaining after a 3-hour window and the variability of that amount.

Using the results in Section B.3.6, we have that \[E(Y-X)=E(Y)-E(X)=4/5-2/5=2/5,\] and \[ \mathrm{var}(Y-X)=\mathrm{var}(Y)+\mathrm{var}(X)-2\mathrm{cov}(Y,X)= 2/75+14/225-2(1/75)=14/225. \]

B.4 Random vectors

B.4.1 Definition

A random vector is a vector of random variables. A random vector is assumed to be a column vector unless otherwise specified.

Additionally, a random matrix is a matrix of random variables.

B.4.2 Mean, variance, and covariance

Let \(\mathbf{y}=[Y_1,Y_2,\dots,Y_n]\) be an \(n\times1\) random vector.

The mean of a random vector is the vector containing the means of the random variables in the vector. More specifically, the mean of \(\mathbf{y}\) is defined as \[ E(\mathbf{y})=\begin{bmatrix}E(Y_1)\\E(Y_2)\\\vdots\\E(Y_n)\end{bmatrix}. \]

The variance of a random vector isn’t a number. Instead, it is the matrix of covariances of all pairs of random variables in the random vector. The variance of \(\mathbf{y}\) is \[ \begin{aligned} \mathrm{var}(\mathbf{y}) &= E(\mathbf{y}\mathbf{y}^T )-E(\mathbf{y})E(\mathbf{y})^T\\ &= \begin{bmatrix}\mathrm{var}(Y_1) & \mathrm{cov}(Y_1,Y_2) &\dots &\mathrm{cov}(Y_1,Y_n)\\\mathrm{cov}(Y_2,Y_1 )&\mathrm{var}(Y_2)&\dots&\mathrm{cov}(Y_2,Y_n)\\\vdots&\vdots&\vdots&\vdots\\ \mathrm{cov}(Y_n,Y_1)&\mathrm{cov}(Y_n,Y_2)&\dots&\mathrm{var}(Y_n)\end{bmatrix}. \end{aligned} \] Alternatively, the variance of \(\mathbf{y}\) is called the covariance matrix of \(\mathbf{y}\) or the variance-covariance matrix of \(\mathbf{y}\). Note that \(\mathrm{var}(\mathbf{y})=\mathrm{cov}(\mathbf{y}, \mathbf{y})\).

Let \(\mathbf{x} = [X_1, X_2, \ldots, X_n]\) be an \(n\times 1\) random vector.

The covariance matrix between \(\mathbf{x}\) and \(\mathbf{y}\) is defined as \[ \mathrm{cov}(\mathbf{x}, \mathbf{y}) = E(\mathbf{x}\mathbf{y}^T) - E(\mathbf{x}) E(\mathbf{y})^T. \]

B.4.3 Properties of transformations of random vectors

Define:

• \(\mathbf{a}\) to be an \(n\times 1\) vector of constants (not necessarily the same constant).
• \(\mathbf{A}\) to be an \(m\times n\) matrix of constants (not necessarily the same constant).
• \(\mathbf{x}=[X_1,X_2,\ldots,X_n]\) to be an \(n\times 1\) random vector.
• \(\mathbf{y}=[Y_1,Y_2,\ldots,Y_n]\) to be an \(n\times 1\) random vector.
• \(\mathbf{z}=[Z_1,Z_2,\ldots,Z_n]\) to be an \(n\times 1\) random vector.
• \(0_{n\times n}\) to be an \(n\times n\) matrix of zeros.

Then:

• \(E(\mathbf{A}\mathbf{y})=\mathbf{A}E(\mathbf{y})\).
• \(E(\mathbf{y}\mathbf{A}^T )=E(\mathbf{y}) \mathbf{A}^T\).
• \(E(\mathbf{x}+\mathbf{y})=E(\mathbf{x})+E(\mathbf{y})\).
• \(\mathrm{var}(\mathbf{A}\mathbf{y})=\mathbf{A}\mathrm{var}(\mathbf{y}) \mathbf{A}^T\).
• \(\mathrm{cov}(\mathbf{x}+\mathbf{y},\mathbf{z})=\mathrm{cov}(\mathbf{x},\mathbf{z})+\mathrm{cov}(\mathbf{y},\mathbf{z})\).
• \(\mathrm{cov}(\mathbf{x},\mathbf{y}+\mathbf{z})=\mathrm{cov}(\mathbf{x},\mathbf{y})+\mathrm{cov}(\mathbf{x},\mathbf{z})\).
• \(\mathrm{cov}(\mathbf{A}\mathbf{x},\mathbf{y})=\mathbf{A}\ \mathrm{cov}(\mathbf{x},\mathbf{y})\).
• \(\mathrm{cov}(\mathbf{x},\mathbf{A}\mathbf{y})=\mathrm{cov}(\mathbf{x},\mathbf{y}) \mathbf{A}^T\).
• \(\mathrm{var}(\mathbf{a})= 0_{n\times n}\).
• \(\mathrm{cov}(\mathbf{a},\mathbf{y})=0_{n\times n}\).
• \(\mathrm{var}(\mathbf{a}+\mathbf{y})=\mathrm{var}(\mathbf{y})\).

B.4.4 Example (Continuous bivariate distribution continued)

Using the definitions and results in B.4, we want to answer Q7 of the example in B.3.8. Summarizing only the essential details, we have a random vector \(\mathbf{z}=[X, Y]\) with mean \(E(\mathbf{z})=[2/5, 4/5]\) and covariance matrix \[ \mathrm{var}(\mathbf{z})= \begin{bmatrix} 14/225 & 1/75 \\ 1/75 & 2/75 \end{bmatrix}. \] We want to determine \(E(Y-X)\) and \(\mathrm{var}(Y-X)\).

Define \(\mathbf{A}=[-1, 1]^T\) (the ROW vector with 1 and -1). Then, \[ \mathbf{Az}=\begin{bmatrix}-1 & 1\end{bmatrix} \begin{bmatrix} X\\ Y \end{bmatrix} =Y-X \] and, \[ \begin{aligned} E(Y-X)&=E(\mathbf{Az})\\ &=\begin{bmatrix}-1 & 1\end{bmatrix} \begin{bmatrix} 2/5\\ 4/5 \end{bmatrix}\\ &=-2/5+4/5\\&=2/5. \end{aligned} \] Additionally, \[ \begin{aligned} & \mathrm{var}(Y-X) \\ &=\mathrm{var}(\mathbf{Az}) \\ &=\mathbf{A}\mathrm{var}(\mathbf{z})\mathbf{A}^T \\ &= \begin{bmatrix} -1 & 1 \end{bmatrix} \begin{bmatrix} 14/225 & 1/75 \\ 1/75 & 2/75 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} \\ &= \begin{bmatrix} -14/225+1/75 & -1/75+2/75 \end{bmatrix} \begin{bmatrix} -1 \\ 1 \end{bmatrix} \\ &= 14/225 + 2/75 - 2(1/75) \\ &=14/225. \end{aligned} \]

B.5 Multivariate normal (Gaussian) distribution

B.5.1 Definition

The random vector \(\mathbf{y}=[Y_1,\dots,Y_n]\) has a multivariate normal distribution with mean \(E(\mathbf{y})=\boldsymbol{\mu}\) (an \(n\times 1\) vector) and covariance matrix \(\mathrm{var}(\mathbf{y})=\boldsymbol{\Sigma}\) (an \(n\times n\) matrix) if its joint pdf is \[ f(\mathbf{y})=\frac{1}{(2\pi)^{n/2} |\boldsymbol{\Sigma}|^{1/2} } \exp\left(-\frac{1}{2} (\mathbf{y}-\boldsymbol{\mu})^T \boldsymbol{\Sigma}^{-1} (\mathbf{y}-\boldsymbol{\mu})\right), \] where \(|\boldsymbol{\Sigma}|\) is the determinant of \(\boldsymbol{\Sigma}\). Note that \(\boldsymbol{\Sigma}\) must be symmetric and positive definite.

In this case, we would denote the distribution of \(\mathbf{y}\) as \[\mathbf{y}\sim \mathsf{N}(\boldsymbol{\mu},\boldsymbol{\Sigma}).\]

B.5.2 Linear functions of a multivariate normal random vector

A linear function of a multivariate normal random vector (i.e., \(\mathbf{a}+\mathbf{A}\mathbf{y}\), where \(\mathbf{a}\) is an \(m\times 1\) vector of constant values and \(\mathbf{A}\) is an \(m\times n\) matrix of constant values) is also multivariate normal (though it could collapse to a single random variable if \(\mathbf{A}\) is a \(1\times n\) vector).

Application: Suppose that \(\mathbf{y}\sim \mathsf{N}(\boldsymbol{\mu},\boldsymbol{\Sigma})\). For an \(m\times n\) matrix of constants \(\mathbf{A}\), \(\mathbf{A}\mathbf{y}\sim \mathsf{N}(\mathbf{A}\boldsymbol{\mu},\mathbf{A}\boldsymbol{\Sigma} \mathbf{A}^T)\).

More generally, the most common estimators used in linear regression are linear combinations of a (typically) multivariate normal random vector, meaning that many of the estimators also have a (multivariate) normal distribution.

B.5.3 Example (OLS matrix form)

Ordinary least squares regression is a method for fitting a linear regression model to data. Suppose that we have observed variables \(X_1, X_2, X_3, \ldots, X_{p-1}, Y\) for each of \(n\) subjects from some population, with \(X_{i,j}\) denoting the value of \(X_j\) for observation \(i\) and \(Y_i\) denoting the value of \(Y\) for observation \(i\). In general, we want to use \(X_1, \ldots, X_{p-1}\) to predict the value of \(Y\). Let \[ \mathbf{X} = \begin{bmatrix} 1 & X_{1,1} & X_{1,2} & \cdots & X_{1,n} \\ 1 & X_{2,1} & X_{2,2} & \cdots & X_{2,n} \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 1 & X_{n,1} & X_{n,2} & \cdots & X_{n,n} \end{bmatrix} \] be a full-rank matrix of size \(n\times p\) and \[ \mathbf{y}=[Y_1, Y_2, \ldots,Y_n], \] be an \(n\)-dimensional vector of responses. It is common to assume that \[ \mathbf{y}\sim \mathsf{N}(\mathbf{X}\boldsymbol{\beta}, \sigma^2 \mathbf{I}_{n\times n}), \] where \(\beta=[\beta_0,\beta_1,\ldots,\beta_{p-1}]\) is a \(p\)-dimensional vector of constants.

The matrix \(\mathbf{H}=\mathbf{X}(\mathbf{X}^T\mathbf{X})^{-1}\mathbf{X}^T\) projects \(\mathbf{y}\) into the space spanned by the vectors in \(\mathbf{X}\). Because of what we know about linear functions of a multivariate normal random vector (B.5.2), we can determine that

\[ \mathbf{Hy}\sim \mathsf{N}(\mathbf{X}\boldsymbol{\beta},\sigma^2 \mathbf{H}). \]