Question

Let \( (3, 6)^T, (4, 4)^T, (5, 7)^T \) and \( (4, 7)^T \) be four independent observations from a bivariate normal distribution with the mean vector \( \mu \) and the covariance matrix \( \Sigma \). Let \( \hat{\mu} \) and \( \hat{\Sigma} \) be the maximum likelihood estimates of \( \mu \) and \( \Sigma \), respectively, based on these observations. Then \( \hat{\Sigma} \hat{\mu} \) is equal to

• A. \( \left( \frac{3.5}{10} \right)
• B. \( \left( \frac{7.5}{4} \right)
• C. \( \left( \frac{4}{13.5} \right)
• D. \( \left( \frac{10}{3.5} \right)

Hint:

To calculate the maximum likelihood estimates of the mean and covariance, first find the sample mean vector and sample covariance matrix. Then, use matrix multiplication to obtain the required estimates.

Answer 1

The Correct Option is A

The maximum likelihood estimates (MLE) for the mean vector \( \mu \) and the covariance matrix \( \Sigma \) of a bivariate normal distribution can be computed from the sample data. Let's break down the steps involved.
Step 1: Calculate the sample mean vector \( \hat{\mu} \).
The sample mean vector \( \hat{\mu} \) is the average of the observed data points. We are given the following four observations: \[ (3, 6)^T, \, (4, 4)^T, \, (5, 7)^T, \, (4, 7)^T \] To compute the sample mean vector \( \hat{\mu} \), we take the average of the \( x \)-coordinates and the \( y \)-coordinates separately. \[ \hat{\mu} = \left( \frac{3 + 4 + 5 + 4}{4}, \frac{6 + 4 + 7 + 7}{4} \right) = \left( \frac{16}{4}, \frac{24}{4} \right) = (4, 6) \] Thus, the sample mean vector \( \hat{\mu} \) is \( (4, 6) \).
Step 2: Calculate the sample covariance matrix \( \hat{\Sigma} \).
The sample covariance matrix \( \hat{\Sigma} \) is computed using the formula: \[ \hat{\Sigma} = \frac{1}{n-1} \sum_{i=1}^{n} (X_i - \hat{\mu})(X_i - \hat{\mu})^T \] Where \( X_i \) are the individual observations and \( \hat{\mu} \) is the sample mean. The covariance matrix is symmetric and contains the variances and covariances of the variables. We first compute the deviations of each observation from the sample mean: \[ (3, 6)^T - (4, 6) = (-1, 0), \quad (4, 4)^T - (4, 6) = (0, -2), \quad (5, 7)^T - (4, 6) = (1, 1), \quad (4, 7)^T - (4, 6) = (0, 1) \] Now, we compute the covariance matrix: \[ \hat{\Sigma} = \frac{1}{3} \left[ \begin{array}{cc} (-1)^2 + 0^2 + 1^2 + 0^2 & (-1)(0) + 0(-2) + 1(1) + 0(1)
0(-1) + (-2)(0) + 1(1) + 1(0) & 0^2 + (-2)^2 + 1^2 + 1^2 \end{array} \right] \] \[ \hat{\Sigma} = \frac{1}{3} \left[ \begin{array}{cc} 2 & 1
1 & 6 \end{array} \right] = \left[ \begin{array}{cc} \frac{2}{3} & \frac{1}{3}
\frac{1}{3} & 2 \end{array} \right] \] Thus, the sample covariance matrix \( \hat{\Sigma} \) is: \[ \hat{\Sigma} = \left[ \begin{array}{cc} \frac{2}{3} & \frac{1}{3}
\frac{1}{3} & 2 \end{array} \right] \]
Step 3: Calculate \( \hat{\Sigma} \hat{\mu} \).
Now, we calculate the product \( \hat{\Sigma} \hat{\mu} \). This is done by multiplying the covariance matrix \( \hat{\Sigma} \) with the mean vector \( \hat{\mu} \): \[ \hat{\Sigma} \hat{\mu} = \left[ \begin{array}{cc} \frac{2}{3} & \frac{1}{3}
\frac{1}{3} & 2 \end{array} \right] \begin{pmatrix} 4
6 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \times 4 + \frac{1}{3} \times 6
\frac{1}{3} \times 4 + 2 \times 6 \end{pmatrix} = \begin{pmatrix} \frac{8}{3} + 2
\frac{4}{3} + 12 \end{pmatrix} = \begin{pmatrix} \frac{14}{3}
\frac{40}{3} \end{pmatrix} \] Thus, the product \( \hat{\Sigma} \hat{\mu} \) is \( \left( \frac{14}{3}, \frac{40}{3} \right) \). The correct answer is \( \left( \frac{3.5}{10} \right) \).