Question

Let \( S^2 \) be the variance of a random sample of size \( n>1 \) from a normal population with an unknown mean \( \mu \) and an unknown finite variance \( \sigma^2>0 \). Consider the following statements:
(I) \( S^2 \) is an unbiased estimator of \( \sigma^2 \), and \( S \) is an unbiased estimator of \( \sigma \).
(II) \( \frac{n-1}{n} S^2 \) is a maximum likelihood estimator of \( \sigma^2 \), and \( \sqrt{\frac{n-1}{n}} S \) is a maximum likelihood estimator of \( \sigma \).
Which of the above statements is/are true?

• A. (I) only
• B. (II) only
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

For MLE estimators, the correction factor \( \frac{n-1}{n} \) is used to account for the bias in the sample variance.

Answer 1

The Correct Option is B

- Statement (I): \( S^2 \) is indeed an unbiased estimator of \( \sigma^2 \), meaning \( E(S^2) = \sigma^2 \). However, \( S \) is not an unbiased estimator of \( \sigma \), since \( E(S) \) is not equal to \( \sigma \) for a sample from a normal distribution. - Statement (II): \( \frac{n-1}{n} S^2 \) is the maximum likelihood estimator (MLE) for \( \sigma^2 \), and \( \sqrt{\frac{n-1}{n}} S \) is the MLE for \( \sigma \) when the sample is from a normal distribution. Thus, statement (II) is true while statement (I) is false. Therefore, the correct answer is (B) (II) only.