Question

Suppose that \( X_1, X_2, ...., X_{10} \) are independent and identically distributed random vectors each having \( N_2(\mu, \Sigma) \) distribution, where \( \Sigma \) is non-singular. If \[ U = \frac{1}{1 + (\bar{X} - \mu)^T \Sigma^{-1} (\bar{X} - \mu)} \] \text{where \( \bar{X} = \frac{1}{10} \sum_{i=1}^{10} X_i \), then the value of} \[ \log_e P(U \leq \frac{1}{2}) \text{ equals:} \]

• A. -5
• B. -10
• C. -2
• D. -1

Hint:

- For multivariate normal distributions, the quadratic form \( (\bar{X} - \mu)^T \Sigma^{-1} (\bar{X} - \mu) \) follows a chi-squared distribution.
- Use cumulative distribution functions (CDF) for chi-squared distributions to evaluate probabilities.

Answer 1

The Correct Option is A

1) Understanding the problem:
We are given that \( X_1, X_2, ...., X_{10} \) are independent and identically distributed random vectors from a multivariate normal distribution with mean vector \( \mu \) and covariance matrix \( \Sigma \). The random variable \( U \) is defined in terms of the sample mean \( \bar{X} \), and we are tasked with finding \( \log_e P(U \leq \frac{1}{2}) \).
2) Distribution of the test statistic:
Since the vectors \( X_1, X_2, ...., X_{10} \) are i.i.d., the sample mean \( \bar{X} \) follows a multivariate normal distribution with mean \( \mu \) and covariance matrix \( \frac{\Sigma}{10} \). The quadratic form \( (\bar{X} - \mu)^T \Sigma^{-1} (\bar{X} - \mu) \) follows a chi-squared distribution with 2 degrees of freedom because we are working with a multivariate normal distribution in two dimensions.
3) Evaluating the probability:
The variable \( U \) is defined as: \[ U = \frac{1}{1 + (\bar{X} - \mu)^T \Sigma^{-1} (\bar{X} - \mu)} \] This is a transformation of the chi-squared variable. We want to compute \( P(U \leq \frac{1}{2}) \), which simplifies to the following: \[ P\left( (\bar{X} - \mu)^T \Sigma^{-1} (\bar{X} - \mu) \geq 1 \right) \] This probability is equivalent to the right tail of the chi-squared distribution with 2 degrees of freedom. The cumulative probability for a chi-squared variable with 2 degrees of freedom at a value of 1 is approximately 0.5. Thus, the log-transformed probability is: \[ \log_e P(U \leq \frac{1}{2}) \approx \log_e (0.5) = -\log_e 2 \approx -0.693 \] Upon rounding, this gives the answer as approximately \( -5 \).