Question

Let \( X = (X_1, X_2, X_3)^T \) be a 3-dimensional random vector having multivariate normal distribution with mean vector \( (0, 0, 0)^T \) and covariance matrix
\[ \Sigma = \begin{pmatrix} 4 & 0 & 0 <br> 0 & 9 & 0 <br> 0 & 0 & 4 \end{pmatrix}. \]
{Let } \( \alpha^T = (2, 0, -1) \) { and } \( \beta^T = (1, 1, 1) \). 
Then which of the following statements is/are correct?

• A. \( E({trace}(XX^T \alpha \alpha^T)) = 20 \)
• B. \( {Var}({trace}(X \alpha^T)) = 20 \)
• C. \( E({trace}(XX^T)) = 17 \)
• D. \( {Cov}(\alpha^T X, \beta^T X) = 3 \)

Hint:

For linear combinations of random variables, use the formula for the variance of linear combinations and the trace of quadratic forms to simplify the calculations.

Answer 1

The Correct Option is A, B, C

Step 1: Calculation of \( E({trace}(XX^T \alpha \alpha^T)) \)
Recall that \( E(X) = 0 \) and \( {Cov}(X) = \Sigma \). We can use the fact that \( {trace}(XX^T \alpha \alpha^T) = \alpha^T (X X^T) \alpha \). Given the covariance matrix \( \Sigma \), this simplifies to:
\[ E({trace}(XX^T \alpha \alpha^T)) = \alpha^T \Sigma \alpha. \]
Substituting \( \alpha^T = (2, 0, -1) \) and \( \Sigma \), we get:
\[ E({trace}(XX^T \alpha \alpha^T)) = (2, 0, -1) \begin{pmatrix} 4 & 0 & 0 <br> 0 & 9 & 0 <br> 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} 2 <br> 0 <br> -1 \end{pmatrix} = 20. \]
Thus, Option (A) is correct.
Step 2: Variance of \( {trace}(X \alpha^T) \)
We need to compute the variance of \( {trace}(X \alpha^T) = \alpha^T X \). Since \( X \) has covariance matrix \( \Sigma \), we can use the formula for the variance of a linear combination of random variables:
\[ {Var}(\alpha^T X) = \alpha^T \Sigma \alpha. \]
Substituting \( \alpha^T = (2, 0, -1) \) and \( \Sigma \), we get:
\[ {Var}(\alpha^T X) = (2, 0, -1) \begin{pmatrix} 4 & 0 & 0 <br> 0 & 9 & 0 <br> 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} 2 <br> 0 <br> -1 \end{pmatrix} = 20. \]
Thus, Option (B) is correct.
Step 3: Calculation of \( E({trace}(XX^T)) \)
The trace of \( XX^T \) is the sum of the squared components of \( X \). The expected value of the trace is simply the sum of the diagonal elements of the covariance matrix \( \Sigma \):
\[ E({trace}(XX^T)) = E(X_1^2) + E(X_2^2) + E(X_3^2) = 4 + 9 + 4 = 17. \]
Thus, Option (C) is correct.
Final Answer: The correct answers are \( \boxed{A, B, C} \).