Question #151509
3.3.14 Let X and Y be independent, with X ∼ Bernoulli(1/2) and Y ∼ N(0, 1). Let
Z = X +Y and W = X −Y. Compute Var(Z), Var(W), Cov(Z, W), and Corr(Z, W
1
Expert's answer
2020-12-18T14:48:45-0500

Given that X is Bernoulli random variable with parameter 1/2

XBernoulli(12)X\sim Bernoulli(\frac{1}{2}) then E(X)=p=12E(X)=p=\frac{1}{2} and V(X)=pq=1212=14V(X)=pq=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}

We know that V(X)=E(X2)(E(X))2V(X)=E(X^2)-(E(X))^2

than 14=E(X2)(12)2    E(X2)=14+14=12\frac{1}{4}=E(X^2)-(\frac{1}{2})^2 \implies E(X^2)=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}


Y is a standard normal random variable that is

YN(0,1)    E(Y)=0Y\sim N(0,1)\implies E(Y)=0 and V(Y)=1V(Y)=1

V(Y)=E(Y2)(E(Y))2    1=E(Y2)02    E(Y2)=1V(Y)=E(Y^2)-(E(Y))^2 \implies 1=E(Y^2)-0^2\implies E(Y^2)=1


X and Y are independent variables and given that Z=X+YZ = X + Y then

E(Z)=E(X)+E(Y)=12+0=12E(Z)=E(X)+E(Y)=\frac{1}{2}+0=\frac{1}{2}

V(Z)=V(X)+V(Y)+2Cov(X,Y)V(Z)=V(X)+V(Y)+2Cov(X,Y)

Since X and Y are independent Cov(X, Y)=0 then

V(Z)=14+1=54V(Z)=\frac{1}{4}+1=\frac{5}{4}


W=XY so E(W)=E(X)E(Y)=120=12W=X-Y\ so\ E(W)=E(X)-E(Y)=\frac{1}{2}-0=\frac{1}{2}

V(W)=V(X)+V(Y)+2Cov(X,Y)=14+1=54V(W)=V(X)+V(Y)+2Cov(X,Y)=\frac{1}{4}+1=\frac{5}{4}


Cov(Z,W)=E(ZW)E(Z)E(W)=E((X+Y)(XY))E(Z)E(W)Cov(Z,W)=E(ZW)-E(Z)E(W)=E((X+Y)(X-Y))-E(Z)E(W)

Cov(Z,W)=E(X2Y2)1212=E(X2)E(Y2)14=12114=34Cov(Z,W)=E(X^2-Y^2)-\frac{1}{2}\cdot\frac{1}{2}=E(X^2)-E(Y^2)-\frac{1}{4}=\frac{1}{2}-1-\frac{1}{4}=-\frac{3}{4}


Corr(Z,W)=Cov(Z,W)V(Z)V(W)=345454=35Corr(Z,W)=\frac{Cov(Z,W)}{\sqrt{V(Z)}\sqrt{V(W)}}=\frac{-\frac{3}{4}}{\sqrt{\frac{5}{4}}\sqrt{\frac{5}{4}}}=-\frac{3}{5}


Answer: 54,54,34,35\frac{5}{4},\frac{5}{4},-\frac{3}{4}, -\frac{3}{5} .


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Comments

Assignment Expert
21.12.20, 00:04

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Yosef
19.12.20, 06:22

Thank you very much

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