Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a \(N(0, \theta)\) distribution. We want to estimate the standard deviation \(\sqrt{\theta}\). Find the constant \(c\) so that \(Y=\) \(c \sum_{i=1}^{n}\left|X_{i}\right|\) is an unbiased estimator of \(\sqrt{\theta}\) and determine its efficiency.

Short answer

The constant \(c\) is \(\frac{\sqrt{\pi}}{n\sqrt{2}}\) and the efficiency of the estimator \(Y\) is approximately 0.797.

Step-by-step solution

Determine the constant \(c\)

To find \(c\), we need to set \(E[Y]=\sqrt{\theta}\). Since \(X_{i}\) is a random variable from a \(N(0, \theta)\) distribution, the expected value of \(|X_{i}|\) will just be \(\sqrt{\frac{2\theta}{\pi}}\). Therefore, the expected value of \(Y\) is \(E[Y]=cE\left[\sum_{i=1}^{n}\left|X_{i}\right|\right] = cn\sqrt{\frac{2\theta}{\pi}} = \sqrt{\theta}\). Solving for \(c\), we have \(c = \frac{\sqrt{\pi}}{n\sqrt{2}}\).

Calculate the variance of \(Y\)

To calculate the efficiency, we need to know the variance of \(Y\). Using the property that the variance of a sum of independent random variables is the sum of their variances, we get \(Var(Y) = Var(c\sum_{i=1}^{n}\left|X_{i}\right|)=(cn\sqrt{\frac{2}{\pi}})^{2}Var(|X_{i}|),\)Where \(Var(|X_{i}|) = \Theta - \frac{2\Theta}{\pi}\). Therefore, \(Var(Y) = n\left(\Theta - \frac{2\Theta}{\pi}\right)\)

Determine the efficiency of the estimator

The efficiency of an estimator is determined by the ratio of the variance of the efficient estimator to the variance of the estimator in question. For a normally distributed sample, the efficient estimator of \(\sqrt{\theta}\) is \(c\sqrt{n}\) and its variance is \(\theta\). Therefore, the efficiency of \(Y\) is given by \(\frac{Var(\sqrt{n})}{Var(Y)} = \frac{\theta}{n\left(\Theta - \frac{2\Theta}{\pi}\right)} = \frac{1}{1 - \frac{2}{\pi}} \approx 0.797\).