Question

Let \(X_1,X_2,\ldots,X_n\) be a random sample of size \(n\ (n\geq2)\) from a \(N(0,\sigma^2)\) distribution, where \(\sigma>0\) is an unknown parameter. Which one of the following statements is true?

• A. \(\sum_{i=1}^{n}X_i\) is a sufficient statistic for \(\sigma^2\)
• B. \(\sum_{i=1}^{n}X_i^2\) is a sufficient statistic for \(\sigma^2\)
• C. \(\sum_{i=1}^{n}X_i\) is a complete statistic
• D. \(\frac{1}{n}\sum_{i=1}^{n}X_i^2\) is a biased estimator of \(\sigma^2\)

Hint:

For normal distributions with known mean and unknown variance, the sufficient statistic for variance is based on the sum of squared observations.

Answer 1

The Correct Option is B

Step 1: Write the joint density function.
Since
\[ X_i \sim N(0,\sigma^2), \] the density of each observation is
\[ f(x_i;\sigma^2) = \frac{1}{\sqrt{2\pi\sigma^2}} \exp\left( -\frac{x_i^2}{2\sigma^2} \right) \]
Therefore, the joint density of the sample is
\[ f(x_1,\ldots,x_n;\sigma^2) = (2\pi\sigma^2)^{-n/2} \exp\left( -\frac{1}{2\sigma^2}\sum_{i=1}^{n}x_i^2 \right) \]

Step 2: Apply the Factorization Theorem.
The likelihood depends on the sample observations only through
\[ \sum_{i=1}^{n}X_i^2 \]
Hence, by the Neyman--Fisher Factorization Theorem,
\[ \sum_{i=1}^{n}X_i^2 \] is a sufficient statistic for
\[ \sigma^2 \]
Therefore, option (B) is true.

Step 3: Check option (A).
The statistic
\[ \sum_{i=1}^{n}X_i \] does not capture all information about the variance parameter.
The likelihood depends on squared observations rather than their simple sum.
Thus,
\[ \sum X_i \] is not sufficient for
\[ \sigma^2 \]
Hence, option (A) is false.

Step 4: Check option (C).
The statistic
\[ \sum X_i \] is normally distributed with mean \(0\) and variance \(n\sigma^2\).
However, it is not complete for the parameter
\[ \sigma^2 \]
Thus, option (C) is false.

Step 5: Check option (D).
Consider the estimator
\[ \frac{1}{n}\sum_{i=1}^{n}X_i^2 \]
Since
\[ E(X_i^2)=\sigma^2, \] we get
\[ E\left( \frac{1}{n}\sum_{i=1}^{n}X_i^2 \right) = \frac{1}{n}\sum_{i=1}^{n}E(X_i^2) \]
\[ = \frac{1}{n}(n\sigma^2) \] \[ =\sigma^2 \]
Thus, the estimator is unbiased, not biased.
Hence, option (D) is false.

Step 6: Final conclusion.
Therefore, the correct statement is
\[ \boxed{(B)} \]