Question

Let \( X_1, X_2, \dots, X_n \), where \( n \geq 2 \), be a random sample from a \( N(-\theta, \theta) \) distribution, where \( \theta>0 \) is an unknown parameter. Then which one of the following options is correct?

• A. \( \sum_{i=1}^{n} X_i \) is a minimal sufficient statistic
• B. \( \sum_{i=1}^{n} X_i^2 \) is a minimal sufficient statistic
• C. \( \left( \frac{1}{n} \sum_{i=1}^{n} X_i, \frac{1}{n-1} \sum_{j=1}^{n} (X_j - \frac{1}{n} \sum_{i=1}^{n} X_i)^2 \right) \) is a complete statistic
• D. \( -\frac{1}{n} \sum_{i=1}^{n} X_i \) is a uniformly minimum variance unbiased estimator of \( \theta \)

Hint:

For normal distributions, the sum of squares of the observations is often the minimal sufficient statistic because it captures both the mean and the variance of the distribution.

Answer 1

The Correct Option is B

We are given a random sample \( X_1, X_2, \dots, X_n \) from a normal distribution \( N(-\theta, \theta) \), where \( \theta > 0 \) is the unknown parameter. We need to determine which of the following options is correct. 
Step 1: Understanding the Distribution 
The random variables \( X_1, X_2, \dots, X_n \) are from the normal distribution \( N(-\theta, \theta) \), which means:
The mean of the distribution is \( -\theta \),
The variance of the distribution is \( \theta \).
Step 2: Factorization Theorem 
To determine which statistic is minimal sufficient, we apply the Factorization Theorem, which states that a statistic is sufficient if the likelihood function can be factored into two parts: one depending on the data through the statistic, and the other not depending on the parameter. The likelihood function for the normal distribution is given by: \[ L(\theta) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi \theta}} \exp \left( -\frac{(X_i + \theta)^2}{2\theta} \right). \] This can be written as: \[ L(\theta) = \frac{1}{(\sqrt{2\pi \theta})^n} \exp \left( -\sum_{i=1}^{n} \frac{(X_i + \theta)^2}{2\theta} \right), \] which depends on the sum \( \sum_{i=1}^{n} X_i^2 \), making it crucial in determining the likelihood. 
Step 3: Identifying the Minimal Sufficient Statistic 
The statistic \( \sum_{i=1}^{n} X_i^2 \) is sufficient for \( \theta \) because it contains all the information needed to estimate \( \theta \). Additionally, it is minimal because there are no redundant pieces of information: it captures both the mean and variance of the distribution. Thus, \( \sum_{i=1}^{n} X_i^2 \) is the minimal sufficient statistic. 
Step 4: Re-evaluating the Other Options
(A) \( \sum_{i=1}^{n} X_i \): This is a sufficient statistic, but not minimal, because it does not capture information about the variance of the distribution.
(C) \( \left( \frac{1}{n} \sum_{i=1}^{n} X_i, \frac{1}{n-1} \sum_{j=1}^{n} (X_j - \frac{1}{n} \sum_{i=1}^{n} X_i)^2 \right) \): While this is a complete statistic, it is not minimal since the sum of squares \( \sum_{i=1}^{n} X_i^2 \) alone is already sufficient and minimal.
(D) \( -\frac{1}{n} \sum_{i=1}^{n} X_i \): This is the unbiased estimator of \( \theta \), but not minimal or sufficient for the variance in this case. Thus, the correct answer is (B) \( \sum_{i=1}^{n} X_i^2 \).