Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) denote a random sample from a normal distribution with mean zero and variance \(\theta, 0<\theta<\infty\). Show that \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\) and has variance \(2 \theta^{2} / n\).

Short answer

Hence, we have shown that \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\) with variance \(2\theta^{2} / n\).

Step-by-step solution

Calculate the expected value

Let's first calculate the expected value of our estimator \(\sum_{1}^{n} X_{i}^{2} / n\). When \(X_{i}\) follows a normal distribution with mean zero, we know that the expected value \(E[X^{2}]\) equals its variance, \(\theta\). Therefore, \(E[\sum_{1}^{n} X_{i}^{2} / n] = E[X_{1}^{2} + X_{2}^{2} + \ldots + X_{n}^{2}] / n = n\theta / n = \theta\). Thus, \(\sum_{1}^{n} X_{i}^{2} / n\) is an unbiased estimator of \(\theta\).

Calculate the variance

Next, we calculate the variance of the estimator. We know that the variance \(Var(X_{i}^{2})\) of the square of a normally distributed variable equals \(2\theta^{2}\) when the mean is zero. Thus, \(Var(\sum_{1}^{n} X_{i}^{2} / n) = Var(\theta + \theta + \ldots + \theta) / n = n(\theta)^{2} / n^{2} = 2\theta^{2} / n\). Therefore, the variance of \(\sum_{1}^{n} X_{i}^{2} / n\) is \(2\theta^{2} / n\).