Question

Let \((X,Y)\) be a random vector with \(Var(X)=Var(Y)=1\) and \(Cov(X,Y)=-0.5\). For which one of the following values of \(b\), the random variables \(U=bX+Y\) and \(V=X+bY\) are uncorrelated?

• A. \(2-\sqrt{3}\)
• B. \(\sqrt{3}-2\)
• C. \(\sqrt{3}\)
• D. \(2\)

Hint:

To check whether two linear combinations are uncorrelated, compute their covariance using covariance linearity properties and set it equal to zero.

Answer 1

The Correct Option is A

Step 1: Expand the covariance.
With $U=bX+Y$ and $V=X+bY$, bilinearity gives \[ \mathrm{Cov}(U,V)=b\,\mathrm{Var}(X)+b^2\mathrm{Cov}(X,Y)+\mathrm{Cov}(Y,X)+b\,\mathrm{Var}(Y). \]

Step 2: Plug in the numbers.
Using $\mathrm{Var}(X)=\mathrm{Var}(Y)=1$ and $\mathrm{Cov}(X,Y)=-0.5$, \[ \mathrm{Cov}(U,V)=2b-\tfrac12 b^2-\tfrac12. \]

Step 3: Set it to zero.
Uncorrelated means this equals $0$. Multiply by $2$: $4b-b^2-1=0$, i.e. $b^2-4b+1=0$.

Step 4: Solve.
\[ b=\frac{4\pm\sqrt{12}}{2}=2\pm\sqrt3. \]

Step 5: Match the option.
Only $2-\sqrt3$ is listed.
\[ \boxed{2-\sqrt3} \]