STAT 418HW15 SolutionsSections 7.5-7.839, 41, 51, 55, 56, 65, 68; TE 4, 2539.Cov(Yn, Yn) = V ar(Yn) = 3σ2Cov(Yn, Yn+1) = Cov(Xn+ Xn+1+ Xn+2, Xn+1+ Xn+2+ Xn+3)= Cov(Xn+1+ Xn+2, Xn+1+ Xn+2) = V ar(Xn+1+ Xn+2) = 2σ2Cov(Yn, Yn+2) = Cov(Xn+2, Xn+2) = σ2Cov(Yn, Yn+j) = 0, j ≥ 341. The number of carp is a hypergeometric random variable.E(X) =20 ∗ 30100= 6V ar(X) = 2030100701008099=33699.51.fX|Y(x) =e−y/yRy0e−y/ydx=1y, 0 < x < yE(X3|Y = y) =Zy0x31ydx = y3/4.55. Let N denote the number of ducks. Given N = n, let I1, . . . , Inbe such thatIi=1 if duck i is hit0 otherwiseBecause given N = n, each hunter will independently hit duck i with probability .6/n,E(Number hit |N = n) = E(nXi=1Ii)=nXi=1E(Ii) = n 1 −1 −.6n10!.E(Number hit ) =∞Xn=0n1 −.6n10e−66n/n!.56. Let X be the number that enter on the ground floor and letIi=1 elevator stops at floor i0 otherwise1E(NXi=1Ii|X = k) =NXi=1E(Ii|X = k) = N 1 −N − 1Nk!E(NXi=1Ii) = N − N∞Xk=0N − 1Nke−1010kk!= N − Ne−10/N= N(1 − e10/N).65. Let X be the number of storms, and let G(B) be the events that it is a good (bad)year. ThenE(X) = E(X|G)P (G) + E(X|B)P (B) = 3(.4) + 5(.6) = 4.2.If Y is Poisson with mean λ, then E(Y2) = λ + λ2. Therefore,E(X2) = E(X2|G)P (G) + E(X2|B)P (B) = 12(.4) + 30(.6) = 22.8.Consequently,V ar(X) = 22.8 − (4.2)2= 5.16.68.a..6e−2+ .4e−3b..6e−2233!+ .4e−3333!c.P (3|0) =P (3, 0)P (0)=.6e−2233!+ .4e−3333!.6e−2+ .4e−3TE 4.g(X) = g(µ) + g0(µ)(X − µ) + g00(µ)(X − µ)22+ . . .≈ g(µ) + g0(µ)(X − µ) + g00(µ)(X − µ)22E(g(X)) ≈ g(µ) +g00(µ)2σ2TE 25.a.E(X|Y = y) =XxxP (X = x, Y = y)P (Y = y)=XxxP (X = x)P(Y = y)P (Y = y)=XxxP (X = x)= E(X)2b.E(X|Y = y) =ZxxfXY(x, y)fY(y)dx=ZxxfX(x)fY(y)fY(y)dx=ZxxfX(x)dx=
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