extbf {Last Time} extbf {CE for Finite RV's} extbf {Interpretation} extbf {Motivation for CE in the $L^2$ case} extbf {Existence of CE in the $L^2$ case} extbf {Hilbert Space Properties of $L^2$} extbf {Definition of CE in the $L^2$ case} extbf {Example (part of Exercise 2.1.4)}Last Time•Uniform integrability•Convergence of expectations•Independence of events, σ-fields, random variablesToday’s lecture: Sections 2.1, 2.2MATH136/STAT219 Lecture 7, October 6, 2008 – p. 1/8CE for Finite RV’s•Suppose X and Y are RV’s that take finitely many values•Conditional probability that X = x given that Y = y:IP (X = x|Y = y).=IP (X = x, Y = y)IP (Y = y)•Conditional expectation of X given that Y = y:IE(X|Y = y).=XxxIP (X = x|Y = y)•Let f(y) denote IE(X|Y = y) and define the conditionalexpectation of X given Y by the RV f(Y ) ≡ IE(X|Y )•That is, if Y (ω) = y then IE(X|Y )(ω) = IE(X|Y = y)MATH136/STAT219 Lecture 7, October 6, 2008 – p. 2/8Interpretation•E(X|Y ) is best guess of the value of X given informationabout Y•E(X|Y ) should be based on the information in Y•E(X|Y ) should be close to X in some senseMATH136/STAT219 Lecture 7, October 6, 2008 – p. 3/8Motivation for CE in the L2case•Let X and Y be RV’s on (Ω, F, IP ) such thatX ∈ L2(Ω, F, IP )•Define HY.= L2(Ω, σ( Y ), IP ), i.e.HY= {g(Y ) : g is measurable and IE(|g(Y )|2) < ∞}•Goal: choose the RV IE(X|Y ) to be the minimummean-squared error estimateof X based on the informationin Y•That is, want the RV IE(X|Y ) to satisfy:◦IE(X|Y ) ∈ HY◦IE(X|Y ) minimizes, over W ∈ HY,IE|(X − W )2|MATH136/STAT219 Lecture 7, October 6, 2008 – p. 4/8Existence of CE in the L2caseProposition 2.1.2 shows that•There exists a RV W∗∈ HYfor which the minimum MSE isattained, i.e.IE|(X − W∗)2| = infW ∈HYIE|(X − W )2|•The minimizer W∗is unique (in the a.s. sense)•Furthermore, the optimality of W∗is equivalent to theorthogonality property:IE[(X − W∗)V ] = 0, for all V ∈ HYMATH136/STAT219 Lecture 7, October 6, 2008 – p. 5/8Hilbert Space Properties of L2The space L2(Ω, F, IP ) is a Hilbert space•Linear vector space•Norm: ||X||2= (IE|X|2)1/2•Inner product: (X, Y ) = IE(XY )•Schwarz inequality: (IE(XY ))2≤ IE|X|2IE|Y |2•Complete with respect to L2-normExistence of C.E. in L2case is due to the existence oforthogonal projection in Hilbert spacesMATH136/STAT219 Lecture 7, October 6, 2008 – p. 6/8Definition of CE in the L2case•For X ∈ L2(Ω, F, IP ) the conditional expectation of Xgiven Y, denoted IE(X|Y ), is the (a.s.) unique RV whichsatisfies:◦Measurability: IE(X|Y ) ∈ L2(Ω, σ( Y ), IP )◦Orthogonality: For all V ∈ L2(Ω, σ( Y ), IP ),IE[(X − IE(X|Y ))V ] = 0•The orthogonality condition is equivalent toIE|(X − IE(X|Y ))2| = infW ∈L2(Ω,σ(Y ),IP )IE|(X − W )2|•Note that if Y and˜Y are such that σ(Y ) = σ(˜Y ) thenIE(X|Y ) = IE(X|˜Y ). Therefore, IE(X|Y ) ≡ IE(X|σ(Y ))MATH136/STAT219 Lecture 7, October 6, 2008 – p. 7/8Example (part of Exercise 2.1.4)•Let Ω = {a, b, c, d}, F = 2Ω, with probability measure IPgiven by IP ({a}) = 1/2, IP ({b}) = 1/4, IP ({c}) = 1/6,IP ({d}) = 1/12.•Let A = {a, d} and B = {b, c, d}•Find IE(IA|IB)MATH136/STAT219 Lecture 7, October 6, 2008 – p.