Mathematics · Glossary

What is Expectation?

Definition 18.8 High School Mathematics · Chapter 18 — Probability and Random Variables

The expectation (or mean) of XX is the average of its values weighted by their probabilities:

E(X)=i=1kP(X=xi)  xi.\E(X) = \sum_{i=1}^{k} \P(X = x_i)\; x_i .

Examples

Example 18.10

For the game of Example 18.6:

E(G)=46×(2)+16×3+16×4=8+3+46=160.17.\E(G) = \frac46 \times (-2) + \frac16 \times 3 + \frac16 \times 4 = \frac{-8 + 3 + 4}{6} = -\frac16 \approx -0.17 .

On average the player loses about 1717 cents per game: the game is unfavorable. A game is called fair when the expected gain is 00.

Example 18.15

For the game of Example 18.6: E(G2)=46×4+16×9+16×16=16+9+166=416\E(G^2) = \frac46 \times 4 + \frac16 \times 9 + \frac16 \times 16 = \frac{16 + 9 + 16}{6} = \frac{41}{6}, so

V(G)=416(16)2=416136=245366.8,σ(G)2.6.\V(G) = \frac{41}{6} - \left(-\frac16\right)^2 = \frac{41}{6} - \frac{1}{36} = \frac{245}{36} \approx 6.8, \qquad \sigma(G) \approx 2.6 .

The loss of 0.170.17 per game on average comes with swings of typical size 2.62.6 euros — expectation without standard deviation tells only half the story.

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Definition 22.6 University Mathematics — Year 2 · Chapter 22 — Discrete Random Variables

A real random variable XX on (Ω,P)(\Omega, \P) has an expectation if the family (X(ω)P({ω}))ωΩ\bigl(X(\omega)\,\P(\{\omega\})\bigr)_{\omega \in \Omega} is summable (Chapter 7); its expectation is then

E(X)=ωΩX(ω)P({ω}).\E(X) = \sum_{\omega \in \Omega} X(\omega)\,\P(\{\omega\}) .

Examples

Example 22.10 (Transfer in action)

For XP(λ)X \sim \mathcal P(\lambda), compute E(11+X)\E\bigl(\frac1{1+X}\bigr) — the law of 11+X\frac1{1+X} itself is awkward, but transfer never asks for it:

E(11+X)=k01k+1eλλkk!=eλλk0λk+1(k+1)!=eλλ(eλ1)=1eλλ.\E\Bigl(\frac1{1+X}\Bigr) = \sum_{k\geq0}\frac{1}{k+1}\,\eu^{-\lambda} \frac{\lambda^k}{k!} = \frac{\eu^{-\lambda}}{\lambda}\sum_{k\geq0} \frac{\lambda^{k+1}}{(k+1)!} = \frac{\eu^{-\lambda}}{\lambda}\bigl(\eu^\lambda - 1\bigr) = \frac{1 - \eu^{-\lambda}}{\lambda} .

Two lessons. Computationally: recognizing a shifted exponential series is the whole work — transfer reduces expectations of f(X)f(X) to series manipulation. Structurally: the naive plug-in value would be 11+EX=11+λ\frac1{1 + \E X} = \frac1{1 + \lambda}, and the true answer is larger,

1eλλ11+λ,\frac{1 - \eu^{-\lambda}}{\lambda} \geq \frac{1}{1 + \lambda},

exactly as Jensen’s inequality for the convex function t11+tt \mapsto \frac1{1+t} demands. Expectations of convex images sit above the naive plug-in value, and transfer plus a series check makes the abstract inequality concrete.

Example 22.12 (Products, with and without independence)

Roll two fair dice. If YY is the second die (independent of the first), E(XY)=E(X)E(Y)=3.52=12.25\E(XY) = \E(X)\E(Y) = 3.5^2 = 12.25. If instead Y=XY = X (the “product” of a die with itself),

E(X2)=1+4+9+16+25+366=91615.1712.25:\E(X^2) = \frac{1 + 4 + 9 + 16 + 25 + 36}{6} = \frac{91}{6} \approx 15.17 \neq 12.25 :

same marginal laws in both scenarios, different joint laws, different product expectations. The moral, worth engraving: E(XY)\E(XY) is a functional of the pair, not of the two marginals — and the gap E(X2)E(X)22.92\E(X^2) - \E(X)^2 \approx 2.92 is, by König–Huygens, precisely the variance 3512\frac{35}{12} of the die.

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