where the expression on the right hand side is defined by the left hand side Definition. Let Y E L(Q, F, P) and let X be a stochastic variable. Then the function EYIX=z is defined as any Borel function f: R+R with the propert that f(X)is a version of E Y X. Note that E Y X=r is not always uniquely defined, but that this does not matter in practice Having defined the conditional expectation, we now note some of its properties. Let the given probability space be( Q, F, P) Proposition. Let g=1Q,0. Then E[XIG=EX Proof. Exercise.■ Proposition. Let X and Y be integrable random variables, let g be a o algebra and let a, B be scalars. Then E[ax+BYg=aEXI9+BEYIg Proof. Exercise.■ Proposition [ Law of iterated expectations]. Let X EC(Q2, F, P)and letgCHCF beσ- - algebras.Then EEIXH IG=E[XIg (5) Proof. We check that the left hand side satisfies the conditions for being the con- ditional expectation of X with respect to g. Clearly it is g-measurable. Now let G Eg and we have, since g C H and consequently G E H E[E[XIMIC]dP=/ E(X1H]dwhere the expression on the right hand side is defined by the left hand side. Definition. Let Y ∈ L 1 (Ω, F, P) and let X be a stochastic variable. Then the function E [Y |X = x] is defined as any Borel function f : R → R with the property that f (X) is a version of E [Y |X] . Note that E [Y |X = x] is not always uniquely defined, but that this does not matter in practice. Having defined the conditional expectation, we now note some of its properties. Let the given probability space be (Ω, F,P). Proposition. Let G = {Ω, ∅}. Then E [X|G] = E [X]. Proof. Exercise. Proposition. Let X and Y be integrable random variables, let G ⊂ F be a σ- algebra and let α, β be scalars. Then E [αX + βY |G] = αE [X|G] + βE [Y |G] (4) Proof. Exercise. Proposition [Law of iterated expectations]. Let X ∈ L1 (Ω, F,P) and let G ⊂ H ⊂ F be σ-algebras. Then E [E [X|H] |G] = E [X|G] (5) Proof. We check that the left hand side satisfies the conditions for being the con￾ditional expectation of X with respect to G. Clearly it is G-measurable. Now let G ∈ G and we have, since G ⊂ H and consequently G ∈ H, Z G E [E [X|H] |G] dP = Z G E [X|H] dP = Z G XdP. (6) 8