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: Recall the condition for uncorrelated random variables.
Two random variables are uncorrelated if their covariance is zero.
Thus, we require
\[ Cov(U,V)=0 \]
Given,
\[ U=bX+Y \] and
\[ V=X+bY \]

Step 2: Use covariance properties.
Using linearity of covariance,
\[ Cov(U,V) = Cov(bX+Y,\ X+bY) \]
Expand term by term:
\[ Cov(U,V) = bCov(X,X)+b^2Cov(X,Y)+Cov(Y,X)+bCov(Y,Y) \]
Since
\[ Cov(X,X)=Var(X)=1 \] \[ Cov(Y,Y)=Var(Y)=1 \] and
\[ Cov(X,Y)=Cov(Y,X)=-0.5 \]
Substitute these values:
\[ Cov(U,V) = b(1)+b^2(-0.5)+(-0.5)+b(1) \]
\[ Cov(U,V) = 2b-\frac{1}{2}b^2-\frac{1}{2} \]

Step 3: Apply the uncorrelated condition.
Set covariance equal to zero:
\[ 2b-\frac{1}{2}b^2-\frac{1}{2}=0 \]
Multiply throughout by \(2\):
\[ 4b-b^2-1=0 \]
Rearrange:
\[ b^2-4b+1=0 \]

Step 4: Solve the quadratic equation.
Using the quadratic formula,
\[ b= \frac{4\pm\sqrt{16-4}}{2} \]
\[ b= \frac{4\pm\sqrt{12}}{2} \]
\[ b= \frac{4\pm2\sqrt{3}}{2} \]
\[ b= 2\pm\sqrt{3} \]

Step 5: Match with the given options.
The values are
\[ 2+\sqrt{3} \] and
\[ 2-\sqrt{3} \]
Among the options, only
\[ 2-\sqrt{3} \] is present.
Hence, option (A) is correct.

Step 6: Final conclusion.
Therefore, the required value of \(b\) is
\[ \boxed{2-\sqrt{3}} \]