Answer to Question #99287 in Statistics and Probability for Fatima1

Question

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.

Answer 1

Given that $X$ and $Y$are two independent random variables

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

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

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

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

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

b) If $X$ is any random variable and $c$ is any constant, then

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

If $X$and $Y$are two independent random variables, then

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

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

c) For any two random variables $X$ and $Y,$

$$Cov(X, Y)=E(XY)-\mu_X\mu_Y$$ $$Cov(X,Y)=Cov(Y,X)$$ $$Cov(X,X)=Var(X)$$

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

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

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

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

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