Definition of Random Variable

DefinitionRandom variable

On a probability space $(\Omega,\mathcal F,P)$ , a real random variable is a function $X:\Omega\to\mathbb R$ such that $\{\omega:X(\omega)\le x\}\in\mathcal F$ for every real $x$ . This measurability condition makes the probabilities below meaningful.

Probability functions

A discrete variable has a countable set $S$ with $P(X\in S)=1$ . Its probability mass function is $p_X(x)=P(X=x)$ . Countable additivity on the disjoint events $\{X=x\}$ gives

p_X(x)\ge0,\qquad \sum_{x\in S}p_X(x)=1.

An absolutely continuous distribution has a nonnegative integrable density $f_X$ such that, for every Borel set $B$ ,

P(X\in B)=\int_B f_X(t)\,dt.

Taking $B=\mathbb R$ gives total integral one; taking a singleton gives $P(X=x)=0$ . Density values at individual points do not affect probabilities. Having infinitely many possible values does not imply existence of a density. Mixed distributions have both atoms and a continuous part; singular continuous distributions also exist, so PMFs and PDFs do not exhaust all distributions. In the examples below, “continuous” means absolutely continuous.

Cumulative distribution function

Every real random variable has a CDF:

F_X(x)=P(X\le x).

For a discrete variable it is the sum of masses at $t\le x$ ; when a density exists it is the integral of that density over $(-\infty,x]$ .

TheoremCDF properties

$F_X$ is nondecreasing and right-continuous, with limits zero at $-\infty$ and one at $+\infty$ . Moreover,

P(a<X\le b)=F_X(b)-F_X(a).

P(X=x)=F_X(x)-F_X(x-).

Proof

First derive continuity of probability from countable additivity. If $A_n\uparrow A$ , partition $A$ into $A_1$ and the disjoint increments $A_n\setminus A_{n-1}$ . The partial sums show $P(A_n)\to P(A)$ . Taking complements proves the analogous result for decreasing events.

The events $\{X\le x\}$ are nested, giving monotonicity. As $n\to\infty$ , the events $\{X\le x+1/n\}$ decrease to $\{X\le x\}$ , proving right-continuity (monotonicity then handles every approach from the right). The events $\{X\le n\}$ increase to $\Omega$ and $\{X\le -n\}$ decrease to the empty set because $X$ is finite-valued. These give the two limits. Subtract the probabilities of nested events to obtain the interval formula. Finally $\{X\le x-1/n\}$ increases to $\{X<x\}$ , so its limiting probability is $F_X(x-)$ ; subtracting from $F_X(x)$ gives the atom formula.

See Expectation and Variance for integrals of random variables.

Some Examples

Below are some examples of random variables in different contexts: discrete, continuous, and mixed.

Discrete Random Variable

ExampleRolling a Die

Consider a simple example of rolling a fair six-sided die. The sample space $\Omega$ consists of the outcomes $\{1, 2, 3, 4, 5, 6\}$ . We can define a random variable $X$ that maps each outcome to its value. For example, if we roll a die and get a 3, then $X(\omega) = 3$ . The probability distribution of this random variable is uniform, meaning each outcome has an equal probability of $\frac{1}{6}$ .

PMF: $p_X(x) = \frac{1}{6}$ for $x \in \{1, 2, 3, 4, 5, 6\}$

For detailed calculations of expected value and variance, see Expectation and Variance.

Discrete PMF and step CDF. For a fair die, the CDF accumulates the six equal probability masses into a staircase.

For a fair die, the CDF accumulates the six equal probability masses into a staircase.

Continuous Random Variable

ExampleRainfall measurement

A normal variable $W\sim N(100,30^2)$ can be negative. To model nonnegative rainfall, let $Y$ have the conditional distribution of $W$ given $W\ge0$ . Write $\phi$ and $\Phi$ for the standard normal density and CDF. Conditional probability gives

f_Y(y)=\begin{cases}\dfrac{\phi((y-100)/30)}{30\Phi(10/3)},&y\ge0,\\0,&y<0,\end{cases}

because $P(W\ge0)=1-\Phi(-10/3)=\Phi(10/3)$ by normal symmetry. Integrating gives

F_Y(y)=\begin{cases}0,&y<0,\\ \dfrac{\Phi((y-100)/30)-\Phi(-10/3)}{\Phi(10/3)},&y\ge0. \end{cases}

The limits are zero and one, verifying normalization. For example,

P(70\le Y\le130)=\frac{\Phi(1)-\Phi(-1)}{\Phi(10/3)}\approx0.6830,

P(40\le Y\le160)=\frac{\Phi(2)-\Phi(-2)}{\Phi(10/3)}\approx0.9549.

The parameters $100$ and $30$ describe the normal variable before truncation; they are not exactly the mean and standard deviation of $Y$ .

Density, area, and cumulative probability. This separate uniform-distribution example shows probability as area under a density, and the CDF as accumulated area.

This separate uniform-distribution example shows probability as area under a density, and the CDF as accumulated area.

Mixed Random Variable

ExampleCustomer waiting time

A customer receives immediate service with probability $0.3$ . Otherwise the waiting time $Z$ has an exponential distribution with rate $\lambda>0$ . For $x\ge0$ , partitioning by the immediate-service event gives

F_Z(x)=0.3+0.7(1-e^{-\lambda x})=1-0.7e^{-\lambda x}.

For $x<0$ , $F_Z(x)=0$ . The jump at zero has size $0.3$ , so $P(Z=0)=0.3$ . On $(0,\infty)$ the continuous part has density $0.7\lambda e^{-\lambda x}$ and total mass $0.7$ . Its integral alone cannot describe the atom; the full distribution is

P(Z\in B)=0.3\mathbf1_{\{0\in B\}}+0.7\int_{B\cap(0,\infty)}\lambda e^{-\lambda x}\,dx.

Comparison: Discrete and Absolutely Continuous Distributions

Aspect	Discrete	Absolutely continuous
Defining property	Probability one on a countable set	Probabilities are integrals of a density
Individual points	Atoms may have positive mass	Every point has probability zero
CDF	Sum of masses at values at most $x$	Integral of density up to $x$
Expectation, when integrable	Sum of value times mass	Integral of value times density

Joint Random Variables

DefinitionJoint random variables

Random variables $X,Y$ on the same probability space define a random vector $(X,Y)$ . Their joint distribution assigns a probability to each Borel subset of $\mathbb R^2$ .

For discrete variables, $p_{X,Y}(x,y)=P(X=x,Y=y)$ . Disjointness of the point events gives total mass one and the marginal formulas

p_X(x)=\sum_y p_{X,Y}(x,y),\qquad p_Y(y)=\sum_x p_{X,Y}(x,y).

If the joint distribution has a density $f_{X,Y}$ , its integral over a set is the probability of that set. Integrating out a coordinate gives

f_X(x)=\int_{\mathbb R}f_{X,Y}(x,y)\,dy,\qquad f_Y(y)=\int_{\mathbb R}f_{X,Y}(x,y)\,dx.

Indeed, integrate either expression over a Borel set $A$ ; Tonelli's theorem for nonnegative integrands identifies the result with the probability of $A\times\mathbb R$ or $\mathbb R\times A$ . Tonelli is a measure-theoretic prerequisite not proved here. Individual densities do not guarantee a joint density: if $X$ is uniform and $Y=X$ , all joint mass lies on the diagonal, whose planar area is zero, contradicting any putative density formula.

Independence means $P(X\in A,Y\in B)=P(X\in A)P(Y\in B)$ for all Borel sets $A,B$ . In the discrete case this is equivalent to $p_{X,Y}(x,y)=p_X(x)p_Y(y)$ : necessity follows using singletons, and sufficiency by summing over $A\times B$ . For joint densities, factorization $f_{X,Y}=f_Xf_Y$ almost everywhere is sufficient by iterated integration. Necessity uses uniqueness of probability measures determined by rectangles and uniqueness of densities almost everywhere; these measure-theoretic results are prerequisites not proved here.

ExampleTwo Dice

Consider rolling two fair six-sided dice. Let $X$ be the outcome of the first die and $Y$ be the outcome of the second die.

Joint PMF: $p_{X,Y}(x,y) = \frac{1}{36}$ for $x,y \in \{1, 2, 3, 4, 5, 6\}$

Marginal PMFs:

$p_X(x) = \sum_{y=1}^{6} p_{X,Y}(x,y) = \frac{1}{6}$
$p_Y(y) = \sum_{x=1}^{6} p_{X,Y}(x,y) = \frac{1}{6}$

Since $p_{X,Y}(x,y) = p_X(x) \cdot p_Y(y)$ , the dice rolls are independent.

ExampleHeight and Weight

Consider the relationship between height $H$ and weight $W$ of adults. These are typically not independent.

In a model admitting a joint density, $f_{H,W}(h,w)$ describes how height and weight are distributed together in the population.

The marginal density $f_H(h) = \int_{0}^{\infty} f_{H,W}(h,w) dw$ gives the distribution of heights regardless of weight
The marginal density $f_W(w) = \int_{0}^{\infty} f_{H,W}(h,w) dh$ gives the distribution of weights regardless of height

Since height and weight are correlated, $f_{H,W}(h,w) \neq f_H(h) \cdot f_W(w)$ .

Exercises

ExerciseSubset Cardinality and Minimum Element

One of the 256 subsets of $\{1,2,3,4,5,6,7,8\}$ is chosen uniformly at random. Let $X$ be the number of elements in this subset. Let $Y$ be 1 if the subset is empty and the least element of the subset otherwise.

Are the events " $X = 1$ " and " $Y = 3$ " independent?
Are the random variables $X$ and $Y$ independent?
What is $\Pr(X \leq 2 \mid Y \leq 5)$ ?

Solution

Since each of the $2^8 = 256$ subsets is equally likely, each is chosen with probability $\frac{1}{256}$ .

Independence of Events $X = 1$ and $Y = 3$ : We check whether $\Pr(X = 1 \land Y = 3) = \Pr(X = 1)\Pr(Y = 3)$ $Pr (X = 1 \land Y = 3) = Pr (X = 1) Pr (Y = 3)$ .
- There are $\binom{8}{1} = 8$ subsets with exactly 1 element, so $\Pr(X = 1) = \frac{8}{256} = \frac{1}{32}$ .
- A subset has least element 3 if and only if it contains 3 and a subset of $\{4,5,6,7,8\}$ , giving $2^5 = 32$ subsets. Thus $\Pr(Y = 3) = \frac{32}{256} = \frac{1}{8}$ .
- The event " $X = 1$ and $Y = 3$ " corresponds to the unique subset $\{3\}$ , so $\Pr(X = 1 \land Y = 3) = \frac{1}{256}$ . Since

\Pr(X = 1)\Pr(Y = 3) = \frac{1}{32} \cdot \frac{1}{8} = \frac{1}{256} = \Pr(X = 1 \land Y = 3)

, the events are independent.

Independence of Random Variables $X$ and $Y$ : Consider $X = 8$ and $Y = 8$ .
- $\Pr(X = 8) = \frac{1}{256}$ (only the full set $\{1,2,3,4,5,6,7,8\}$ ).
- $\Pr(Y = 8) = \frac{1}{256}$ (only the singleton $\{8\}$ ).
- However, a subset cannot simultaneously have 8 elements and have 8 as its minimum element, so $\Pr(X = 8 \land Y = 8) = 0 \neq \Pr(X = 8)\Pr(Y = 8)$ . Therefore, $X$ and $Y$ are not independent random variables.
Conditional Probability $\Pr(X \leq 2 \mid Y \leq 5)$ :
- The subsets with minimum element $\geq 6$ are subsets of $\{6,7,8\}$ (excluding $\emptyset$ , which has $Y=1$ ). There are $2^3 - 1 = 7$ such subsets: $\{6\}, \{7\}, \{8\}, \{6,7\}, \{6,8\}, \{7,8\}, \{6,7,8\}$ . Thus $\Pr(Y \leq 5) = 1 - \frac{7}{256} = \frac{249}{256}$ .
- The total number of subsets with $X \leq 2$ is $\binom{8}{0} + \binom{8}{1} + \binom{8}{2} = 1 + 8 + 28 = 37$ . Among these 37 subsets, those with $Y \geq 6$ are the 6 non-empty subsets with at most 2 elements from $\{6,7,8\}$ : $\{6\}, \{7\}, \{8\}, \{6,7\}, \{6,8\}, \{7,8\}$ . Therefore, the number of subsets with $X \leq 2$ and $Y \leq 5$ is $37 - 6 = 31$ .

\Pr(X \leq 2 \mid Y \leq 5) = \frac{\Pr(X \leq 2 \land Y \leq 5)}{\Pr(Y \leq 5)} = \frac{31/256}{249/256} = \frac{31}{249}.

Remark

A common pitfall in part 3 is attempting to count subsets with $X = 2$ and $Y \leq 5$ by multiplying $\binom{5}{1} \times \binom{7}{1} = 35$ . This overcounts sets where both elements are $\leq 5$ , because pairs like $\{1, 2\}$ would be counted twice depending on which element is picked first.

ExerciseProduct Distribution of Two Dice

Two fair six-sided dice are rolled independently. Let $X$ denote the product of the two face values. Determine the probability mass function $P(X = i)$ for all possible values of $i$ .

Solution

There are $6 \times 6 = 36$ equally likely outcomes $(a, b)$ with $1 \leq a, b \leq 6$ . Counting the factor pairs for each product $i = a \cdot b$ :

\begin{aligned} P(X = 1) &= \frac{1}{36} & P(X = 6) &= \frac{4}{36} & P(X = 15) &= \frac{2}{36} & P(X = 25) &= \frac{1}{36} \\ P(X = 2) &= \frac{2}{36} & P(X = 8) &= \frac{2}{36} & P(X = 16) &= \frac{1}{36} & P(X = 30) &= \frac{2}{36} \\ P(X = 3) &= \frac{2}{36} & P(X = 9) &= \frac{1}{36} & P(X = 18) &= \frac{2}{36} & P(X = 36) &= \frac{1}{36} \\ P(X = 4) &= \frac{3}{36} & P(X = 10) &= \frac{2}{36} & P(X = 20) &= \frac{2}{36} & P(X = \text{other}) &= 0 \\ P(X = 5) &= \frac{2}{36} & P(X = 12) &= \frac{4}{36} & P(X = 24) &= \frac{2}{36} & & \end{aligned}

ExerciseHighest Rank of a Group

Five men and five women are ranked according to their examination scores, with all $10!$ possible rankings equally likely and no tied scores. Let $X$ denote the highest rank achieved by a woman (e.g. $X=1$ if a woman finishes first). Find the PMF of $X$ .

Solution

The woman with the highest rank can finish anywhere from rank 1 to rank 6 (since if all 5 men take the top 5 spots, the top woman finishes 6th).

For $X = i$ ( $1 \leq i \leq 6$ ), the first $i - 1$ positions must be occupied by men, position $i$ must be occupied by a woman, and the remaining positions are filled by the remaining 4 women and $5 - (i - 1)$ men.

Equivalently, choosing the 5 positions for women out of 10 total positions: the top woman at position $i$ means the remaining 4 women must be chosen from the $10 - i$ positions below rank $i$ . Therefore:

P(X = i) = \frac{\binom{10 - i}{4}}{\binom{10}{5}}, \quad i \in \{1, 2, 3, 4, 5, 6\}

Explicit probabilities:

$P(X = 1) = \frac{\binom{9}{4}}{\binom{10}{5}} = \frac{126}{252} = \frac{1}{2}$
$P(X = 2) = \frac{\binom{8}{4}}{252} = \frac{70}{252} = \frac{5}{18}$
$P(X = 3) = \frac{\binom{7}{4}}{252} = \frac{35}{252} = \frac{5}{36}$
$P(X = 4) = \frac{\binom{6}{4}}{252} = \frac{15}{252} = \frac{5}{84}$
$P(X = 5) = \frac{\binom{5}{4}}{252} = \frac{5}{252}$
$P(X = 6) = \frac{\binom{4}{4}}{252} = \frac{1}{252}$

Definition of Random Variable

Probability functions

Cumulative distribution function

Some Examples

Discrete Random Variable

Continuous Random Variable

Mixed Random Variable

Comparison: Discrete and Absolutely Continuous Distributions

Joint Random Variables

Exercises

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