Answer to Question #151502 in Statistics and Probability for Yosef

Question

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)

Answer 1

Given that X is Bernoulli random variable with parameter \frac{1}{2}

That is, X ∼ Bernoulli $(\frac{1}{2})$

Then, E(X) = p

$= \frac{1}{2}$

and V(X) = pq

$= \frac{1}{2} \times \frac{1}{2} \\ = \frac{1}{4}$

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

$=> \frac{1}{4} = E(X)^2-(\frac{1}{2})^2 \\ => E(X)^2 = \frac{1}{4}+\frac{1}{4}$

$= \frac{1}{2}$

Also, Y is a standard normal random variable.

That is, Y∼N(0,1) => E(Y) =0 and V(Y)=1

$V(Y) = E(Y^2) - (E(Y))^2 \\ => 1 = E(Y^2) - (0)^2 \\ => E(Y^2) = 1+0 \\ =1$

Given that X and Y are independent variables.

Now, given that Z=X+Y

Then,

$E(Z) = E(X)+E(Y) \\ = \frac{1}{2}+0 \\ = \frac{1}{2} \\ V(Z) = V(X)+V(Y)+2Cov(X,Y) \\ =V(X)+V(Y)+2(0)$

since X and Y are independent, Cov (X,Y) = 0

$= \frac{1}{4}+1 \\ = \frac{5}{4}$

Also,

$W=X-Y \\ => E(W)=E(X)-E(Y) \\ = \frac{1}{2}-0 \\ = \frac{1}{2} \\ V(W) =V(X)+V(Y)-2Cov(X,Y) \\ = V(X)+V(Y)-2(0)$

since X and Y are independent,

$Cov (X,Y) = 0 \\ = \frac{1}{4}+1 \\ = \frac{5}{4} \\ Cov(Z,W)=E(ZW)-E(Z)E(W) \\ =E((X+Y)(X-Y))-E(Z)E(W) \\ = E(X^2-Y^2)-\frac{1}{2} \times \frac{1}{2} \\ =E(X^2)-E(Y^2) - \frac{1}{4} \\ = \frac{1}{2}-1- \frac{1}{4} \\ = -0.75$

$Corr(Z,W) = \frac{Cov(Z,W)}{\sqrt{V(Z)}\sqrt{V(W)}} \\ =\frac{-0.75}{\sqrt{\frac{5}{4}}\sqrt{\frac{5}{4}}} \\ = \frac{-0.75 \times 4}{5} \\ = -0.60$