Write a program in \(\mathrm{R}\) that simulates 1000 samples of size \(n\) from a \(\mathrm{N}(\theta, 1)\) distribution. For each sample compute the sample mean given by \(\bar{X}_{n}\) and the Hodges super-efficient estimator \(\hat{\theta}_{n}=\bar{X}_{n}+(a-1) \bar{X}_{n} \delta_{n}\) where \(\delta_{n}=\) \(\delta\left\{\left|\bar{X}_{n}ight| ;\left[0, n^{-1 / 4}ight)ight\}\). Using the results of the 1000 simulated samples estimate the standard error of each estimator for each combination of \(n=5,10,25\), 50 , and 100 and \(a=0.25,0.50,1.00\) and 2.00. Repeat the entire experiment once for \(\theta=0\) and once for \(\theta=1\). Compare the estimated standard errors of the two estimators for each combination of parameters given above and comment on the results in terms of the theory presented in Example 10.9.

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Example 10.9. Let X,..., X, be a set of independent and identically dis- tributed random variables following a N(0, 1) distribution. In this case I(0) = 1 so that an asymptotically efficient estimator is one with o(0) = 1. Consider the estimator given by n = Xn+(a-1)Xnon, where n = 8{|xn|; [0, n1/4)} for all n N. We know from Theorem 4.20 (Lindeberg and Lvy) that n1/2 (Xn-0) Z as n where Z is a N(0, 1) random variable. Hence X is asymptotically efficient by Definition 10.4. To establish the asymptotic behav- ior of n we consider two distinct cases. When 0 0 we have that Xn00 as n by Theorem 3.10 and hence it follows that {|X|: [0, n1/4)} 0 as n . See Exercise 12. Therefore, Theorem 3.9 implies that n > as n when 0 0. However, we can demonstrate an even stronger re- sult which will help us establish the weak convergence of n. Consider the sequence of random variables given by n/2 (a - 1)Xnon. Note that Theorem 4.20 (Lindeberg and Lvy) implies that n/2X Z as n where Z is a N(0, 1) random variable. Therefore, Theorem 4.11 (Slutsky) implies that n1/2 (a-1)Xnon 0 as n o. Now note that Theorem 4.11 implies that n1/2Xn+ (a1)n Xn - 0] n1/2(n-0)+n1/2 (a - 1)5nXn, n1/2 (n-0) == =