Question

Let \( X_1, X_2, \dots, X_5 \) be i.i.d. random vectors following the bivariate normal distribution with zero mean vector and identity covariance matrix. Define the \( 5 \times 2 \) matrix \( X = (X_1, X_2, \dots, X_5)^T \). Further, let \( W = (W_{ij}) = X^T X \), and \[ Z = W_{11} + 4W_{12} + 4W_{22}. \] Then \( {Var}(Z) \) equals:

• A. 150
• B. 200
• C. 250
• D. 300

Hint:

When dealing with quadratic forms in normal variables, the variance of the sum of these forms can be calculated by computing the variances and covariances of the individual components.

Answer 1

The Correct Option is C

Step 1: Understanding the Components
The matrix \( X = (X_1, X_2, \dots, X_5)^T \) consists of independent standard normal random variables. Each \( X_i \) is a \( 2 \times 1 \) vector, and we are interested in the variance of \( Z = W_{11} + 4W_{12} + 4W_{22} \), where \( W = X^T X \) is a \( 2 \times 2 \) matrix.
Step 2: Matrix Formulation
The matrix \( X \) is a \( 5 \times 2 \) matrix, and \( W = X^T X \) is a \( 2 \times 2 \) matrix, where the elements of \( W \) are given by: \[ W = \begin{pmatrix} \sum_{i=1}^{5} X_{i1}^2 & \sum_{i=1}^{5} X_{i1} X_{i2} \\ \sum_{i=1}^{5} X_{i1} X_{i2} & \sum_{i=1}^{5} X_{i2}^2 \end{pmatrix}. \] The statistic \( Z \) is a linear combination of the elements of \( W \): \[ Z = W_{11} + 4W_{12} + 4W_{22}. \] Step 3: Variance Calculation
To calculate \( \text{Var}(Z) \), we use the fact that the elements of \( X \) are i.i.d. normal random variables with unit variance. The variance of \( Z \) is computed by finding the variances and covariances of the elements \( W_{11} \), \( W_{12} \), and \( W_{22} \). By applying the formulas for the variance of quadratic forms in normal random variables, we find: \[ \text{Var}(Z) = 250. \] Thus, the correct answer is \( \boxed{250} \).