Random Variable


#probability-theory

Definition

Random Variable

Let $(\Omega ,\mathfrak{S},P)$ be a probability space and let $(\mathcal{X},\mathfrak{A})$ be a measurable value space. A random variable is a measurable function

$$X:(\Omega ,\mathfrak{S})\to (\mathcal{X},\mathfrak{A}),$$

meaning that every measurable set of values has a measurable preimage:

$$\forall B\in \mathfrak{A}:X^{-1}(B)=\{\omega \in \Omega \mid X(\omega )\in B\}\in \mathfrak{S}.$$

Each $\omega \in \Omega$ is an underlying outcome, while $x=X(\omega )\in \mathcal{X}$ is a realisation of $X$. The image $X(\Omega )$ is the set of values that $X$ can attain. A real-valued random variable uses $\mathcal{X}=\mathbb{R}$ with its Borel $\sigma$-algebra.

Why Measurability Is Required

The probability measure $P$ accepts only events in $\mathfrak{S}$. A condition $B\in \mathfrak{A}$ on the value of $X$ lives in the value space, so it must first be pulled back to the outcome space:

$$B\stackrel{X^{-1}}{\to }\{\omega \in \Omega \mid X(\omega )\in B\}.$$

Measurability guarantees that this preimage belongs to $\mathfrak{S}$, making its probability well-defined:

$$B\in \mathfrak{A}⟹X^{-1}(B)\in \mathfrak{S}⟹P(X\in B)=P(X^{-1}(B)).$$

The resulting map

$$P_X(B)=P(X^{-1}(B))$$

is the probability distribution induced by $X$. Thus a random variable transfers probability from sets of raw outcomes to sets of values.

Event Notation

Conditions on a random variable denote events in the sample space. For a real-valued random variable,

$$\begin{array}{rl}\{X=x\}& =X^{-1}(\{x\})=\{\omega \in \Omega \mid X(\omega )=x\},\\ \{X<0\}& =X^{-1}((-\infty ,0)),\\ \{X\ge 0\}& =X^{-1}([0,\infty )).\end{array}$$

The braces describe sets of outcomes; they do not mean that $X$ itself is a set.

Realisation

Definition

Random Variable Realisation

The image $x=X(\omega )$ of a random variable $X$ is called one realisation of $X$.

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Example

Maximum of two dice

For two six-sided dice, take

$$\Omega =\{1,2,3,4,5,6{\}}^2$$

and define

$$X(i,j)=\max \{i,j\}.$$

The raw outcome ${\omega }_0=(2,5)$ produces the realisation

$$X({\omega }_0)=X(2,5)=5.$$

Different outcomes may produce the same value: $X(2,5)=X(5,2)=X(5,5)=5$. The value-level condition $X=5$ corresponds to the outcome event

$$\{X=5\}=\{(i,j)\in \Omega \mid \max \{i,j\}=5\}.$$

With fair independent dice, this preimage contains $5^2-4^2=9$ outcomes, so

$$P(X=5)=\frac{9}{36}=\frac{1}{4}.$$