REVIEW 1 Review Conditional pdf Let (Y1,··· ,YN) have joint pdf f (y1,··· ,yN). Let f (yJ+1,··· ,yN) be the marginal pdf of (YJ+1,··· ,YN). The conditional pdf of Y1,··· ,YJ given YJ+1,··· ,YN is defined by f (y1,···yJ|yJ+1,···yN) = f (y1,··· ,yJ+1,··· ,yN)f (y J+1,···yN) for f (yJ+1,··· ,yN) > 0. Example 1 Let Y1 and Y2 be discrete r.v.s with bivariate pdf f (y1,y2) = braceleftbigg (y 1 + y2)/9 , y1 = 0,1,2 and y2 = 0,1 0 , otherwise. The marginal pdf of Y1 is f (y1) = braceleftbigg summationtext1 y2=0 f (y1,y2) = y1 9 + y1+1 9 = 2y1+1 9 , y1 = 0,1,2 0 , otherwise. The conditional pdf of Y2 given Y1 = y1 is f (y2|y1) = braceleftBigg f(y1,y2) f(y1) = y1+y2 2y1+1 , y2 = 0,1 0 , otherwise. Example 2 Let Y1 and Y2 be continuous r.v.s with bivariate pdf f (y1,y2) = braceleftbigg 2 , if 0 < y 1 < y2 < 1 0 , otherwise. Then f (y1) = braceleftbigg integraltext1 y1 2dy2 = 2(1 ? y1) , 0 < y1 < 1 0 , otherwise. f (y2|y1) = braceleftbigg 1 1?y1 , if 0 < y1 < y2 < 1 0 , otherwise. (a function of y1) REVIEW 2 Conditional expectation Let Y = (Y1,··· ,YN)prime be an N?dimensional random variable, let g(Y ) be a given function,andlet f (y1,··· ,yJ|yJ+1,··· ,yN) beagivenconditionalprobabilityfunction. Theconditionalexpectationof g(Y ), given YJ+1 = yJ+1,··· ,YN = yN is E [g(y)|YJ+1 = yJ+1,··· ,YN = yN] = integraldisplay ∞ ?∞ ··· integraldisplay ∞ ?∞ g(y1,··· ,yN)f (y1,··· ,yJ|yJ+1,··· ,yN)dy1 ···dyJ. Thisisanonstochasticfunctionof yJ+1,··· ,yN. E [g(Y )|YJ+1,··· ,YN] = h(YJ+1,··· ,YN) isar.v. thattakesonrealizedvalues E [g(Y )|YJ+1 = yJ+1,··· ,YN = yN]. Example 3 (continued) f (y1,y2) = braceleftbigg 2 , if 0 < y 1 < y2 < 1 0 , otherwise. Then E (Y2|Y1 = y1) = integraldisplay 1 y1 y2 11 ? y 1 dy2 = 11 ? y 1 1 2 bracketleftbig1 ? y2 1 bracketrightbig = 1 + y12 (a function of y1) and E (Y2|Y1) = 1 + Y12 (a random variable) Expectationof g(Y ) isgivenby Eg(Y ) = EYJ+1,··· ,EYNEY1,···YJ|YJ+1,···YN [g(Y )|YJ+1,··· ,YN] Example 4 (continued) E (Y2) = EY1 parenleftbigg1 + Y 1 2 parenrightbigg = 12 (1 + EY1 (Y1)) = 12 parenleftbigg 1 + integraldisplay 1 0 y12(1 ? y1)dy1 parenrightbigg = 12 parenleftbigg 1 + 13 parenrightbigg = 23. REVIEW 3 Stochastic convergences Markov’s inequality Assume g(y) ≥ 0 for all y ∈ R P [g(Y ) ≥ c] ≤ g[Y ]c . Proof. Assume that Y is a continuous r.v. with pdf f (·). Define A1 = {y|g(y) ≥ c} and A2 = {y|g(y) < c}. Then E [g(Y )] = integraldisplay A1 g(y)f (y)dy + integraldisplay A2 g(y)f (y)dy ≥ integraldisplay A1 g(y)f (y)dy ≥ integraldisplay A1 cf (y)dy = cP [g(Y ) ≥ c]. The stated result follows from this. Chebyshev’s inequality Let Y be a r.v. with E (Y ) = μ and V ar(Y ) = σ2. Then P [|Y ?μ| ≥ r] ≤ σ 2 r2 . Proof. P [|Y ?μ| ≥ r] = P bracketleftbig(Y ?μ)2 ≥ r2bracketrightbig ≤ E (Y ?μ) 2 r2 = σ 2 r2 . Example 5 Zn = braceleftbigg c with probability 1 ? 1 nn with probability 1 n P [|Zn ?c| ≥ ε] = P [Zn = n] = 1n → 0. Thus Zn P→ c. REVIEW 4 Example 6 Zn = braceleftbigg 1 with probability 1 20 with probability 1 2 Z1,··· ,Zn are independent. Let Zn = 1n nsummationdisplay i=1 Zi P bracketleftbiggvextendsinglevextendsingle vextendsinglevextendsingleZn ? 1 2 vextendsinglevextendsingle vextendsinglevextendsingle > ε bracketrightbigg = P bracketleftbiggvextendsinglevextendsingle vextendsinglevextendsingle1 n summationdisplayparenleftbigg Zi ? 12 parenrightbiggvextendsinglevextendsingle vextendsinglevextendsingle > ε bracketrightbigg ≤ E parenleftbig1 n summationtextparenleftbigZ i ? 12 parenrightbigparenrightbig2 ε2 = 1n2 nsummationdisplay i=1 E parenleftbigg Zi ? 12 parenrightbigg2 = 1n2 ×n× 12 parenleftbigg 1? 12 parenrightbigg → 0. Thus, Zn P→ 12. Example 7 Zn = braceleftbigg 0 with probability 1? 1 nn2 with probability 1 n P (|Zn| > ε) = P parenleftbigZn = n2parenrightbig = 1n → 0. So Zn P→ 0. But E (Zn) = 0× parenleftbigg 1? 1n parenrightbigg + n2 × 1n = n →∞. So we have an example for the statement that convergence in probability does not imply mean square convergence. REVIEW 5 Example 8 Z = braceleftbigg 0 withprobability 1 21 withprobability 1 2 Define Zn = 1?Z forall n. Sinceallthe Zn havethesamedistribution, Zn d→ Z. However, |Zn ?Z| = |1?2Z| = 1 regardlessofthevalueof Z. So, Zn doesnotconvergeinprobabilityto Z. (P (|Zn ?Z| > ε) = P (|Zn ?Z| = 1) = 1)