Question

Consider the location Model (10.3.35). Assume that the pdf of the random errors, \(f(x)\), is symmetric about \(0 .\) Let \(\widehat{\theta}\) be a location estimator of \(\theta\). Assume that \(E\left(\widehat{\theta}^{4}\right)\) exists. (a) Show that \(\widehat{\theta}\) is an unbiased estimator of \(\theta\). Hint: Assume without loss of generality that \(\theta=0 ;\) start with \(E(\hat{\theta})=\) \(E\left[\widehat{\theta}\left(X_{1}, \ldots, X_{n}\right)\right]\); and use the fact that \(X_{i}\) is symmetrically distributed about \(0 .\) (b) As in Section \(10.3 .4\), suppose we generate \(n_{s}\) independent samples of size \(n\) from the pdf \(f(x)\) which is symmetric about \(0 .\) For the \(i\) th sample, let \(\widehat{\theta}_{i}\) be the estimate of \(\theta\). Show that \(n_{s}^{-1} \sum_{i=1}^{n_{x}} \widehat{\theta}_{i}^{2} \rightarrow V(\hat{\theta})\), in probability.

Short answer

Part (a): By using the fact that the expected value of an unbiased estimator equals the parameter being estimated, it is found that \(\widehat{\theta}\) is an unbiased estimator of \(\theta\). Part (b): With the help of the Law of Large Numbers, it is proven that \(n_{s}^{-1} \sum_{i=1}^{n_{x}} \widehat{\theta}_{i}^{2}\) converges to the variance of \(\widehat{\theta}\), \(V(\widehat{\theta})\), in probability.

Step-by-step solution

Part (a) - Step 1: Recall the Definition of an Unbiased Estimator

By definition, an estimator \(T\) is said to be an unbiased estimator of the parameter \(\theta\) if it fulfills the condition \(E(T) = \theta\). To prove \(\widehat{\theta}\) is an unbiased estimator, it will be necessary to show that \(E(\widehat{\theta}) = \theta\).

Part (a) - Step 2: Calculate the Expected Value

Using the hint, start with \(E(\widehat{\theta}) = E[\widehat{\theta}(X_{1}, \ldots, X_{n})]\). Given that the distribution is symmetric about \(0\) and \(\theta = 0\), we find that \(E(\widehat{\theta})\) equals \(0\) as well, since in a symmetric distribution the mean equals the median, which is \(0\). Therefore, we have \(E(\widehat{\theta}) = \theta = 0\).

Part (a) - Step 3: Conclusion

Since \(E(\widehat{\theta})\) equals \(\theta\), \(\widehat{\theta}\) is an unbiased estimator of \(\theta\).

Part (b) - Step 1: Recall the Law of Large Numbers

According to the Law of Large Numbers, if \(X_1, X_2, \ldots, X_n\) are \(n_s\) independent and identically distributed random variables with finite expected values, the average \(X = n_{s}^{-1} \sum_{i=1}^{n_{s}} X_{i}\) converges in probability to \(E(X)\). This is the principle that will be used to solve part (b).

Part (b) - Step 2: Apply the Law of Large Numbers

Applying the Law of Large Numbers to the given expression \(n_{s}^{-1} \sum_{i=1}^{n_{x}} \widehat{\theta}_{i}^{2}\), it converges in probability to the expected value \(E(\widehat{\theta}^2)\). Since \(E(\widehat{\theta}^2)\) represents the mean square error, which is equal to the variance (\(V(\widehat{\theta})\)) as \(\widehat{\theta}\) is unbiased, we find that the expression converges to \(V(\widehat{\theta})\).

Part (b) - Step 3: Conclusion

With the help of the Law of Large Numbers, we have conclude that \(n_{s}^{-1} \sum_{i=1}^{n_{x}} \widehat{\theta}_{i}^{2} \rightarrow V(\hat{\theta})\) in probability.