Lectures 22-23: Conditional Expectations

Transcription

1 Lectures 22-23: Conditional Expectations 1.) Definitions Let X be an integrable random variable defined on a probability space (Ω, F 0, P ) and let F be a sub-σ-algebra of F 0. Then the conditional expectation of X given F is defined to be any random variable Y that satisfies the following two conditions: 1. Y is measurable with respect to F; 2. For all F, Y dp = XdP. ny Y satisfying these conditions is said to be a version of E[X F]. Our first task is to show the existence, uniqueness and integrability of conditional expectations, which we do in reverse order. Integrability: Let = {Y > 0} F. Then Y dp = XdP X dp < ; Y dp = XdP X dp <. c c c Thus, E Y E X < and so Y is integrable. Uniqueness: If Y is also a version of E[X F], then Y dp = Y dp for every F. In particular, taking = {Y Y ɛ} F for some ɛ > 0, we have 0 = (Y Y )dp ɛp() and so P() = 0. Since this holds for every ɛ > 0, it follows that Y Y a.s. Reversing the roles of Y and Y in this argument then shows that Y = Y a.s. Existence: We begin by recalling: Radon-Nikodym Theorem: Let µ and ν be σ-finite measures on (Ω, F). Recall that ν µ if ν() = 0 whenever µ() = 0. In this case, there is a function f L 1 (Ω, F) such that ν() = f dµ for all F. We usually write f = dν/dµ and say that f is the Radon-Nikodym derivative of ν with respect to µ. 1

2 We can use this theorem to prove that conditional expectations exist. First suppose that X 0. Let µ = P and define ν() = X dp for F. Then ν is a finite measure (with ν(ω) = EX < ) and ν µ, so that the Radon-Nikodym theorem gives dν X dp = ν() = dµ dp for any F. This shows that dν/dµ is a version of E[X F]. For the general case, write X = X + X and let Y 1 = E[X + F] and Y 2 = E[X F]. Then Y 1 Y 2 is integrable and X dp = X + dp X dp = Y 1 dp Y 2 dp = (Y 1 Y 2 )dp for any F. This shows that Y 1 Y 2 is a version of E[X F]. 2.) Examples Example 1: If X is F-measurable, then E[X F] = X a.s. Example 2: Suppose that X is independent of F: P ( {X B}, ) = P(X B) P() for all B R and F. We claim that E[X F] = EX. Since constants are measurable with respect to any σ-algebra, condition (1) is satisfied. lso, if F, then X dp = E[X1 ] = EX E1 = EX dp and so condition (2) is also satisfied. Example 3: Let E 1, E 2, be a (countable) partition of Ω into disjoint sets, each with probability P(E i ) > 0, and let F = σ(e 1, E 2, ). Then E[X F](ω) = E[X; E i] P(E i ), ω E i. Indeed, E[X F] is F-measurable, since it is constant on each set E i and these sets generate F. To verify condition (2), it suffices to consider sets = E i, for which we have E[X; E i ] E i P(E i ) dp = E[X; E i] = XdP. E i 2

3 The conditional expectation of X given Y is defined to be E[X Y ] = E[X σ(y )], where σ(y ) is the σ-algebra generated by Y. Example 4: Suppose that X and Y have joint density f(x, y), P((X, Y ) B) = f(x, y)dxdy for B R 2, B and that f(x, y)dx > 0 for every y. If g is a measurable function such that E g(x) <, then E[g(X) Y ] = h(y ) where h(y) = g(x)f(x, y)dx f(x, y)dx. Example 5: Suppose that X and Y are independent with marginal distributions µ and ν. Let φ be a measurable function such that E φ(x, Y ) < and define g(x) = Eφ(x, Y ) = φ(x, y)ν(dy). Then E[φ(X, Y ) X] = g(x). Indeed, we know that g(x) m(σ(x)). If σ(x), then = {X C} for some Borel set C, so φ(x, Y )dp = E[φ(X, Y )1 C (X)] = φ(x, y)1 C (x)ν(dy)µ(dx) = 1 C (x)g(x)µ(dx) = g(x)dp, which verifies condition (2). 3.) Properties of Conditional Expectations (a) Linearity: E[aX + Y F] = ae[x F] + E[Y F]. Proof: We need to check that the RHS is a version of the left. Condition (1) is clearly satisfied since each of the terms in the sum is F-measurable. To check condition (2), let F and observe that ( ) ae[x F] + E[Y F] dp = a = a 3 E[X F]dP + E[Y F]dP XdP + Y dp = (ax + Y )dp.

4 (b) Monotonicity: If X Y, then E[X F] E[Y F] a.s. Proof: Let = {E[X F] E[Y F] ɛ} for ɛ > 0. Then ( ) ɛp() E[X F] E[Y F] dp = ( X Y ) dp 0, and so P() = 0. Since this holds for every ɛ > 0, the result follows. (c) Conditional Monotone Convergence Theorem: If X n 0 and X n X with EX <, then E[X n F] E[X F]. Proof: Let Y n = X X n 0. Since Y n is decreasing, it follows from the monotonicity property of conditional expectations that Z n E[Y n F] is also decreasing. Furthermore, since each Z n 0 a.s., it follows that there is a variable Z such that Z n Z. If F, then Z n dp = Y n dp. Then, since Y 1 Y n 0 and Z 1 Z n Z, the dominated convergence theorem implies that Z dp = lim Z n dp = lim Y n dp = 0. Since this holds for every F, it follows that Z = 0 a.s. and so E[Y n F] 0 a.s. (d) Conditional Jensen s Inequality: If φ is convex and E X + E φ(x) <, then φ ( E[X F] ) E [ φ(x) F ]. Proof: We may assume wlog that φ is non-linear. In this case, if we let then S { (a, b) : a, b Q, ax + b φ(x) for all x }, φ(x) = sup{ax + b : (a, b) S}. If φ(x) ax + b, then monotonicity and linearity imply that E[φ(X) F] ae[x F] + b a.s., and taking the sup over (a, b) S gives Jensen s inequality. (e) E[E[X F]] = EX. Proof: This follows by taking = Ω in condition (2). 4

5 (f) Conditional expectation is a contraction in L p, p 1. Proof: Since φ(x) = x p is convex if p 1, the conditional Jensen s inequality implies that E[X F] p E [ X p F ]. Taking expectations and using (e) gives E[X F] p p = E[ E[X F] p ] E[E[ X p F]] = X p p. Theorem (4.1.2): The Tower Property. Suppose that F 1 F 2. (i) E [ E[X F 1 ] F 2 ] = E[X F1 ]. (ii) E [ E[X F 2 ] F 1 ] = E[X F1 ]. Proof: (i) follows from the observation that E[X F 1 ] is F 2 -measurable. To prove (ii), notice that E[X F 1 ] is F 1 -measurable. lso, if F 1 F 2, then E[X F 1 ]dp = XdP = E[X F 2 ]dp. Theorem (4.1.3): If X m(f) and E Y + E XY <, then E[XY F] = XE[Y F]. Proof: Since XE[Y F] is F-measurable, we only need to check condition (2) which we can do using the standard machine. First, suppose that X = 1 B for some set B F. If F, then 1 B E[Y F]dP = E[Y F]dP = Y dp = 1 B Y dp B and so (2) holds for indicator functions. This result then extends to X that are simple functions by linearity of expectations. If X, Y 0, then we can choose a sequence of simple functions 0 X n X and use the monotone convergence theorem to conclude that XE[Y F]dP = lim X n E[Y F]dP = lim X n Y dp = XY dp. To prove the result in general, we then split X and Y into their positive and negative parts. Theorem (4.1.4): Suppose that EX 2 <. Then E[X F] is the orthogonal projection of X L 2 (F 0 ) onto L 2 (F): E [ { } (X E[X F]) 2 ] = inf E[(X Y ) 2 ] : Y L 2 (F). Proof: If Z L 2 (F), then E XZ < and B ZE[X F] = E[ZX F]. 5

6 This shows that E[Z(X E[X F])] = E[ZX] E[Z E[X F]] = E[ZX] E[E[ZX F]] = 0 whenever Z L 2 (F). If Y = E[X F] + Z L 2 (F), then E[(X Y ) 2 ] = E[(X E[X F] Z) 2 ] = E[(X E[X F]) 2 ] 2E[Z(X E[X F])] + EZ 2 = E[(X E[X F]) 2 ] + EZ 2, which is minimized when Z = 0, i.e., when Y = E[X F]. 6