Section 5.2: Expected Values, Covariance, andCorrelation (Only Discrete Case).1Concepts and Formulae: Let p(x, y) be the jointPMF of X and Y , pX(x) be the marginal PMF ofX, and pY(y) be the marginal PMF of Y .• The expected value of h(X, Y ) isE[h(X, Y )] =XxXyh(x, y)p(x, y).• Then,µX= E(X) =XxxpX(x)andµY= E(Y ) =XyypY(y).Similarly, we can calculate V (X) and V (Y ).• The covariance between X and Y isCov(X, Y ) =E[(X − µX)(Y − µY)]=E(XY ) − µXµY.• We can showCov(aX + b, cY + d) = acCov(X, Y )andCov(X, X) = V (X).2• The correlation between X and Y isCorr(X, Y ) =Cov(X, Y )qV (X)V (Y ).• For any real numbers a > 0, b, c > 0 and d,Corr(aX + b, cY + d) = Corr(X, Y ).• We always have|Corr(X, Y )| ≤ 1.• If X and Y are independent,Cov(X, Y ) = Corr(X, Y ) = 0.• If |Corr(X, Y )| = 1, then Y = aX + b for somea 6= 0.3First example of Section 5.2. Under the PMF offirst problem of Section 5.1. We haveE(X) = 100 × 0.5 + 250 × 0.5 = 175E(Y ) = 0 × 0.25 + 100 × 0.25 + 200 × 0.5 = 125E(X2) = 1002× 0.5 + 2502× 0.5 = 36250E(Y2) = 02×0.25+1002×0.25+2002×0.5 = 22500V (X) = 36250 − 1752= 5625V (Y ) = 22500 − 1252= 6875.E(XY )=(100)(0)(0.2) + (100)(100)(0.1)+ (100)(200)(0.2) + (250)(0)(0.05)+ (250)(100)(0.15) + (250)(200)(0.3)=23750.Thus,Cov(X, Y ) = 23750 − 175 × 125 = 1875Corr(X, Y ) =1875√5625 × 6875= 0.3015.Since Corr(X, Y ) 6= 0, X and Y are not
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