What should you pay to play the following game? A fair coin is flipped repeatedly until the first head appears. If the first head appears on flip $n$, you win $2^n$ dollars. The expected winnings, computed by weighting each outcome by its probability, give

\begin{align*} \sum_{n=1}^\infty 2^n \cdot \frac{1}{2^n} = \sum_{n=1}^\infty 1 = \infty. \end{align*}

According to the naive formula, no finite entry fee is too large. Yet no rational person pays even a thousand dollars to play. This is the St. Petersburg paradox, and it reveals something profound: the intuitive notion of "average value" for a random quantity is not yet mathematically complete. We need a framework that tells us when an average exists, when it is finite, and how to compute it even when outcomes are neither discrete nor continuous.

The answer is the Lebesgue integral with respect to a probability measure — what probabilists call **expectation**. On a probability space $(\Omega, \mathcal{F}, \mathbb{P})$, the expectation of a random variable $X: \Omega \to \mathbb{R}$ is the integral $\int_\Omega X \, d\mathbb{P}$, built up from simple functions through a limiting procedure. Writing $\mathcal{L}^1$ for one-dimensional Lebesgue measure, this definition unifies the discrete formula $\mathbb{E}[X] = \sum_x x \, \mathbb{P}(X = x)$ and the continuous formula $\mathbb{E}[X] = \int_{-\infty}^\infty x f(x) \, d\mathcal{L}^1(x)$, and handles every case in between — singular distributions, random variables defined on abstract probability spaces, and limits of sequences of random variables.

Three properties make expectation indispensable. First, linearity: $\mathbb{E}[aX + bY] = a\mathbb{E}[X] + b\mathbb{E}[Y]$ regardless of whether $X$ and $Y$ are independent, enabling calculations that would be intractable by any other method. Second, the probability–expectation bridge: $\mathbb{E}[\mathbb{1}_A] = \mathbb{P}(A)$, so every probability is an expectation of an indicator function. Third, controlled interaction with limits: under appropriate conditions, we may pass limits through the expectation sign — a prerequisite for any serious analysis of sequences of random variables.

[example: The St. Petersburg Paradox] In the St. Petersburg game, let $X$ be the payoff. The sample space can be taken as $\Omega = \mathbb{N}$, where outcome $n$ represents "first head on flip $n$." Setting $\mathbb{P}(X = 2^n) = 2^{-n}$ for $n \in \mathbb{N}$, the sum $\sum_{n=1}^\infty 2^{-n} = 1$ confirms this is a valid probability distribution. The expected payoff is \begin{align*} \mathbb{E}[X] = \sum_{n=1}^\infty 2^n \cdot \frac{1}{2^n} = \sum_{n=1}^\infty 1 = \infty. \end{align*} What does $\mathbb{E}[X] = \infty$ mean precisely? The payoff $X$ is finite almost surely — every play ends in finite time — but the distribution has no finite mean. For any $M > 0$, define the truncated payoff $X_M = \min(X, M)$. Then \begin{align*} \mathbb{E}[X_M] = \sum_{n=1}^{\lfloor \log_2 M \rfloor} 2^n \cdot \frac{1}{2^n} + M \cdot \mathbb{P}(X > M) = \lfloor \log_2 M \rfloor + M \cdot 2^{-\lfloor \log_2 M \rfloor}, \end{align*} which grows without bound as $M \to \infty$. The expectation is $\infty$ in the precise sense $\sup_{M > 0} \mathbb{E}[X_M] = \infty$. The mathematics is unambiguous; the apparent paradox is that human utility for money is concave rather than linear, so utility-weighted expectation is finite even when dollar-weighted expectation is not. [/example]

To define expectation in full generality, we build from the simplest possible random variables — those taking only finitely many values — and extend by approximation. The strategy mirrors the construction of the Lebesgue integral: simple functions first, non-negative functions next via monotone approximation, and finally general functions via the positive-negative decomposition.

The first step is to isolate the finite-valued random variables for which expectation can be computed by a direct weighted sum. Naming this class matters because these variables will serve as the approximating building blocks for arbitrary non-negative random variables.

Every random variable taking finitely many values can be written as a linear combination of indicator functions. This is the starting point because indicators are the atoms of the theory: knowing how to integrate them determines everything else.

The obstruction is that an arbitrary random variable may take infinitely many values, so a direct probability-weighted sum over its values need not be available. We first need a finite-valued class whose level sets partition the sample space, because that structure makes expectation a finite sum and later gives the approximants used for general random variables.

[definition: Simple Random Variable] Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. A random variable $X: \Omega \to \mathbb{R}$ is **simple** if it takes only finitely many values. Any such $X$ can be written in **standard form** as \begin{align*} X = \sum_{i=1}^n a_i \mathbb{1}_{A_i}, \end{align*} where $a_1, \ldots, a_n \in \mathbb{R}$ are the distinct values of $X$ and $A_i = \{X = a_i\} \in \mathcal{F}$ are pairwise disjoint events with $\bigcup_{i=1}^n A_i = \Omega$. [/definition]

For a simple random variable, expectation requires no limiting procedure and no approximation. The definition is the probability-weighted average of the values.

[definition: Expectation of a Simple Random Variable] Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $X = \sum_{i=1}^n a_i \mathbb{1}_{A_i}$ be a simple random variable in standard form. The **expectation** of $X$ is \begin{align*} \mathbb{E}[X] = \sum_{i=1}^n a_i \, \mathbb{P}(A_i). \end{align*} [/definition]

One must check that this formula does not depend on how $X$ is written as a sum of indicators. The standard form is unique, but a random variable can also be represented as $X = \sum_j b_j \mathbb{1}_{B_j}$ with the $B_j$ not necessarily disjoint. In all such representations, the sum $\sum_j b_j \mathbb{P}(B_j)$ gives the same value, which follows from finite additivity of $\mathbb{P}$.

The extension to non-negative random variables uses the fact that every non-negative measurable function is the pointwise supremum of all simple functions lying below it. This suggests defining the expectation as the supremum of expectations over all such simple approximants — and this is exactly the right definition.

[definition: Expectation of a Non-Negative Random Variable] Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $X: \Omega \to [0, \infty]$ be a non-negative random variable. The **expectation** of $X$ is \begin{align*} \mathbb{E}[X] = \sup \bigl\{ \mathbb{E}[Z] : Z \text{ is simple and } 0 \leq Z \leq X \bigr\}. \end{align*} The expectation takes values in $[0, \infty]$; we allow $\mathbb{E}[X] = \infty$. [/definition]

For a general random variable, the positive and negative parts may both contribute infinite expectations, making $\mathbb{E}[X^+] - \mathbb{E}[X^-]$ of the form $\infty - \infty$, which is undefined. The expectation is therefore defined only when at least one of $\mathbb{E}[X^+]$ or $\mathbb{E}[X^-]$ is finite. This three-step construction culminates in the primary definition.

## Definition

The positive-negative decomposition is the point where the construction becomes a definition for real-valued random variables. The only obstruction is the indeterminate expression \(\infty - \infty\), so the definition must state exactly when the two one-sided expectations can be combined.

[definition: Expectation] Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space and let $X: \Omega \to \mathbb{R}$ be a random variable. Write $X^+ = \max(X, 0)$ and $X^- = \max(-X, 0)$, so that $X = X^+ - X^-$ and $|X| = X^+ + X^-$. If $\min(\mathbb{E}[X^+], \mathbb{E}[X^-]) < \infty$, the **expectation** of $X$ is \begin{align*} \mathbb{E}[X] = \mathbb{E}[X^+] - \mathbb{E}[X^-] \in [-\infty, \infty]. \end{align*} The random variable $X$ is called **integrable** if $\mathbb{E}[|X|] = \mathbb{E}[X^+] + \mathbb{E}[X^-] < \infty$, in which case $\mathbb{E}[X] \in \mathbb{R}$. [/definition]