Section 5.5: The Distribution of a LinearCombination1Formulae:• A linear combination of X1, ···, XnisY = a1X1+ a2X2+ ··· + anXn+ b.AssumeE(Xi) = µi; V (Xi) = σ2i• Then, we haveE(Y ) = a1µ1+ a2µ2+ ··· + anµn+ b.• In addition, if X1, ···, Xnare (mutually) inde-pendent, then we haveV (Y ) = a21σ21+ a22σ22+ ··· + a2nσ2n.• In the general case, assumeCorr(Xi, Yi) = ρij; i 6= j.Then, we haveV (Y ) =nXi=1a2iσ2i+X Xi6=jaiajCov(Xi, Xj)=nXi=1a2iσ2i+X Xi6=jaiajσiσjρij.2Special case: Assume X1, ···, Xnare independentand normal. Then, we haveY ∼ N(nXi=1aiµi+ b,nXi=1a2iσ2i).3First example of Section 5.5: example 5.28 and5.30 on textbook. SupposeX1∼ N(1000, 1002)X2∼ N(500, 802)andX3∼ N(300, 502).Assume X1, X2, X3are independent.Let Y = 3.0X1+ 3.2X2+ 3.4X3. ThenE(Y ) =3.0 × 1000 + 3.2 × 500+ 3.4 × 300 = 5620andV (Y ) =3.02× 1002+ 3.22× 802+ 3.42× 502= 184436Thus,Y ∼ N(5620, 184436).P (Y > 4500) = 1 − Φ(4500 − 5620√184436) = 0.9955.4Second example of Section 5.5: example 5.29.Assume X1and X2are independent,X1∼ N(22, 1.22)andX2∼ N(26, 1.52).Then, we haveE(X1− X2) = 22 − 26 = −4andV (X1− X2) =V (X1) + V (X2)=1.22+ 1.52=3.69.Thus,(X1− X2) ∼ N(−4, 3.69).5Third example of Section 5.5. Assume X1, X2,X3, X4, X5, and X6are all N(2, 4) and assumeindependence. LetY =X1+ X22−X3+ X4+ X5+ X64.Compute P (−2 ≤ Y ≤ 3).Answer:E(Y ) =12[E(X1) + E(X2)]−14[E(X3) + E(X4) + E(X5) + E(X6)]=0andV (Y )=(12)2[V (X1) + V (X2)]+ (−14)2[V (X3) + V (X4) + V (X5) + V (X6)]=3.Then, Y ∼ N(0, 3) andP (−2 ≤ Y ≤ 3) =Φ(3 − 0√3) − Φ(−2 − 0√3)=Φ(1.73) − Φ(−1.15)=0.8331.6Fourth example of Section 5.5. X1, X2, X3areindependent. AssumeX1∼ N(1.2, 1.4)X2∼ N(1.3, 1.7)X3∼ N(1.15, 2.0).• ComputeP (−4 ≤ 1.2X1+ 1.1X2− 1.3X3≤ 3)Answer: Let Y = 1.2X1+ 1.1X2− 1.3X3.Then,E(Y ) =1.2 × 1.2 + 1.1 × 1.3 − 1.3 × 1.15=1.375;V (Y )=1.22× 1.4 + 1.12× 1.7 + (−1.3)2× 2.0=7.453.Thus, Y ∼ N(1.375, 7.453) andP (−4 ≤ Y ≤ 3)=Φ(3 − 1.375√7.453) − Φ(−4 − 1.375√7.453)=Φ(0.60) − Φ(−1.97)=0.7013.7• ComputeP (|1.5X1+ 1.4X2− 2.0X3| ≤ 3).Answer: LetY = 1.5X1+ 1.4X2− 2.0X3.Then,E(Y ) =1.5 × 1.2 + 1.4 × 1.3 − 2.0 × 1.15=1.32V (Y )=1.52× 1.4 + 1.42× 1.7 + (−2.0)2× 2.0=13.53,andY ∼ N(1.32, 13.53).Thus,P (|Y | ≤ 3)=P (−3 ≤ Y ≤ 3)=Φ(3 − 1.32√13.53) − Φ(−3 − 1.32√13.53)=Φ(0.46) − Φ(−1.17)=0.5562.8• ComputeP (X1+ X2≥ 2X3).Answer: Note thatP (X1+ X2≥ 2X3) = P (X1+ X2− 2X3≥ 0).LetY = X1+ X2− 2X3.ThenE(Y ) = 0.2andV (Y ) = 11.1.Then,Y ∼ N(0.2, 11.1)Thus,P (Y ≥ 0) =1 − Φ(0 − 0.2√11.1)=1 −
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