319handout10 - 2 by calculating: V 2 t = 1 n n X i =1 ( X i...

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Econ 319, Fall 2008 TA: Simon Kwok Handout 10 1. The distribution of the weights X of the US population has unknown mean and variance ± 2 (i.e. E ( X ) = and V ar ( X ) = ± 2 : ) Our goal is to estimate the unknown population mean and the variance. To do this, we collect a random sample of n people and measure their weights, which are denoted by X 1 ; X 2 ; : : : ; X n : Obviously, the random variables X 1 ; X 2 ; : : : ; X n are iid and their distribution is the same as that of X . (a) A natural estimator of the population mean is given by X = 1 n n X i =1 X i : (i) Calculate E ( X ) . Determine whether X is a biased estimator of or not. (Note: X is called the sample mean .) (ii) Calculate V ar ( X ) . (iii) What is the mean squared error (MSE) of X ? (b) Now we are interested to estimate the population variance ± 2 . variable X is V ar ( X ) = E [( X ) 2 ] . He then suggested that we could estimate ±
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Unformatted text preview: 2 by calculating: V 2 t = 1 n n X i =1 ( X i & & ) 2 : Explain why the method suggested by Tom is infeasible. (ii) Amy, knowing the problem that V 2 t has, suggested instead to construct the estimator of 2 as follows: V 2 a = 1 n n X i =1 & X i & & X 2 : (A) Is the method suggested by Amy feasible? (B) Is V 2 a biased? If yes, what is the bias of V 2 a ? [Hint: Rewrite Amy&s estimator as V 2 a = 1 n n X i =1 & X i & & + & & & X 2 = = 1 n n X i =1 ( X i & & ) 2 & ( & X & & ) 2 and evaluate its expectation using (a)(1) and (a)(2).] (C) Show that V 2 a is asymptotically unbiased though. (D) Based on the result in (b)(2)(2), show that an (even better) unbiased estimator of 2 is given by: S 2 = 1 n & 1 n X i =1 & X i & & X 2 : (Note: S 2 is called the sample variance .) 1...
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