Question #99287
Problem 2: Let X and Y be two independent random variables with E(X)=1, E(Y)=0, Var(X)=4, Var(Y)=2. Let W=2X +Y+1 and Z=3X+Y.
a) Find the expected value of W, the expected value of Z.
b) Find the variance of W.
c) Find the covariance of Z and W.
1
Expert's answer
2019-11-25T12:00:51-0500

Given that XX and YYare two independent random variables

E(X)=1,E(Y)=0,Var(X)=4,Var(Y)=2E(X)=1, E(Y)=0, Var(X)=4, Var(Y)=2W=2X+Y+1,Z=3X+YW=2X+Y+1, Z=3X+Y

a) Distributive property of expected value: for any rv X,YX,Y and any constants,a,b,ca,b,c


E(aX+bY+c)=aE(X)+b(Y)+cE(aX+bY+c)=aE(X)+b(Y)+c

E(W)=E(2X+Y+1)=2E(X)+E(Y)+1=E(W)=E(2X+Y+1)=2E(X)+E(Y)+1==2(1)+0+1=3=2(1)+0+1=3

E(Z)=E(3X+Y)=3E(X)+E(Y)=E(Z)=E(3X+Y)=3E(X)+E(Y)==3(1)+0=3=3(1)+0=3

b) If XX is any random variable and cc is any constant, then


Var(cX)=c2Var(X)Var(cX)=c^2Var(X)Var(X+c)=Var(X)Var(X+c)=Var(X)

If XXand YYare two independent random variables, then


Var(X+Y)=Var(X)+Var(Y)Var(X+Y)=Var(X)+Var(Y)


Var(W)=Var(2X+Y+1)=4Var(X)+Var(Y)=Var(W)=Var(2X+Y+1)=4Var(X)+Var(Y)==4(4)+2=18=4(4)+2=18


c) For any two random variables XX and Y,Y,


Cov(X,Y)=E(XY)μXμYCov(X, Y)=E(XY)-\mu_X\mu_YCov(X,Y)=Cov(Y,X)Cov(X,Y)=Cov(Y,X)Cov(X,X)=Var(X)Cov(X,X)=Var(X)

If XX and YY are independent, then Cov(X,Y)=0.Cov(X,Y)=0. by observing that E(XY)=E(X)E(Y)E(XY)=E(X)E(Y)


Distributive property of covariance: for any rv X,Y,ZX,Y,Z and any constants, a, b, c,


Cov(aX+bY+c,Z)=aCov(X,Z)+bCov(Y,Z)Cov(aX+bY+c, Z)=aCov(X, Z)+bCov(Y,Z)

Cov(W,Z)=Cov(Z,W)=Cov(W,Z)=Cov(Z,W)==Cov(2X+Y+1,3X+Y)==Cov(2X+Y+1,3X+Y)==2Cov(X,3X+Y)+Cov(Y,3X+Y)==2Cov(X,3X+Y)+Cov(Y,3X+Y)==6Cov(X,X)+2Cov(X,Y)+3Cov(Y,X)+Cov(Y,Y)==6Cov(X,X)+2Cov(X,Y)+3Cov(Y,X)+Cov(Y,Y)==6Var(X)+5Cov(X,Y)+Var(Y)=6Var(X)+5Cov(X,Y)+Var(Y)

Cov(W,Z)=6(4)+5(0)+2=26Cov(W,Z)=6(4)+5(0)+2=26

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