A real random variable X on (Ω,P) has an expectation if the family (X(ω)P({ω}))ω∈Ω is summable (Chapter 7); its expectation is then
E(X)=ω∈Ω∑X(ω)P({ω}).
Examples
Example 22.10 (Transfer in action)
For X∼P(λ), compute E(1+X1) — the law of 1+X1 itself is awkward, but transfer never asks for it:
E(1+X1)=k≥0∑k+11e−λk!λk=λe−λk≥0∑(k+1)!λk+1=λe−λ(eλ−1)=λ1−e−λ.
Two lessons. Computationally: recognizing a shifted exponential series is the whole work — transfer reduces expectations of f(X) to series manipulation. Structurally: the naive plug-in value would be 1+EX1=1+λ1, and the true answer is larger,
λ1−e−λ≥1+λ1,
exactly as Jensen’s inequality for the convex function t↦1+t1 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 Y is the second die (independent of the first), E(XY)=E(X)E(Y)=3.52=12.25. If instead Y=X (the “product” of a die with itself),
E(X2)=61+4+9+16+25+36=691≈15.17=12.25:
same marginal laws in both scenarios, different joint laws, different product expectations. The moral, worth engraving: E(XY) is a functional of the pair, not of the two marginals — and the gap E(X2)−E(X)2≈2.92 is, by König–Huygens, precisely the variance 1235 of the die.