MVUE Analysis for Normal Population

The question asks for the Minimum Variance Unbiased Estimator (MVUE) of the parameter $\theta^2$ from a random sample $X_1, \dots, X_n$ drawn from a normal distribution $N(0, \theta^2)$. This specifies a population with mean $\mu = 0$ and variance $\sigma^2 = \theta^2$. We will evaluate the provided options based on unbiasedness and variance.

Estimator Evaluation

Option 1: $\overline{X}^2$

The sample mean is $\overline{X} = \frac{1}{n} \sum_{i=1}^n X_i$. Since the population mean $E[X_i] = 0$, the expected value of the sample mean is $E[\overline{X}] = 0$.

The variance of the sample mean is $Var(\overline{X}) = \frac{\theta^2}{n}$. Using the relationship $Var(\overline{X}) = E[\overline{X}^2] - (E[\overline{X}])^2$, we find the expected value of $\overline{X}^2$:

Since $E[\overline{X}^2] = \frac{\theta^2}{n} \neq \theta^2$, this estimator is biased.

Option 2: $\frac{1}{n} \sum_{i=1}^n X_i^2$

First, we determine the expected value of $X_i^2$. Since $X_i \sim N(0, \theta^2)$, its variance is $Var(X_i) = \theta^2$. We know $Var(X_i) = E[X_i^2] - (E[X_i])^2$. Given $E[X_i] = 0$, we have $\theta^2 = E[X_i^2] - 0^2$, which means $E[X_i^2] = \theta^2$.

Now, we calculate the expected value of the proposed estimator:

This estimator is unbiased for $\theta^2$. It is also known to be the MVUE because the sampling distribution belongs to the exponential family, and this estimator is derived from the sufficient statistic.

Option 3: $\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n - 1}$

This expression represents the sample variance, $s^2$. For normally distributed data, the sample variance $s^2$ is an unbiased estimator of the population variance $\sigma^2$. In this case, $\sigma^2 = \theta^2$, so $E\left[\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n - 1}\right] = \theta^2$. This estimator is unbiased.

To compare it with Option 2, we consider its variance. The variance of this estimator is $Var(s^2) = \frac{2\theta^4}{n-1}$.

Option 4: $\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n}$

This estimator is similar to Option 3 but divided by $n$ instead of $n-1$. Let's find its expected value:

Since $E\left[\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n}\right] \neq \theta^2$, this estimator is biased.

Comparing Unbiased Estimators' Variances

We identified two unbiased estimators: Option 2 and Option 3.

• The variance of the estimator in Option 2 is $\frac{2\theta^4}{n}$.
• The variance of the estimator in Option 3 is $\frac{2\theta^4}{n-1}$.

For $n > 1$, we have $n > n-1$. Therefore, $\frac{1}{n} < \frac{1}{n-1}$, which implies $\frac{2\theta^4}{n} < \frac{2\theta^4}{n-1}$.

The estimator $\frac{1}{n} \sum_{i=1}^n X_i^2$ (Option 2) has a smaller variance than the estimator $\frac{\sum_{i=1}^n (X_i - \overline{X})^2}{n - 1}$ (Option 3).

Conclusion on MVUE

The estimator $\frac{1}{n} \sum_{i=1}^n X_i^2$ is unbiased for $\theta^2$ and possesses the minimum variance among all unbiased estimators. Hence, it is the MVUE.