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Suppose S = { X 1 ,...,X n } is a simple random sample from a population with and finite variance 2 < . Show that the sample mean X is an unbiased estimator for , so is sample variance s 2 for 2 . 1. Before moving to formal proofs, there are several properties regarding expectation and variance as we metioned in early chapters: E ( aX + b ) = aE ( X ) + b. (1) This property is called linearity of expectation, which naturally implies E ( X 1 + ... + X n ) = E ( X 1 ) + E ( X 2 + ... + X n ) = E ( X 1 ) + E ( X 2 ) + E ( X 3 + ... + X n ) = n X i =1 E ( X i ) . The other properties are regarding variance, V ar ( aX + b ) = a 2 V ar ( X ) , V ar ( X 1 + X 2 ) = V ar ( X 1 ) + V ar ( X 2 ) + 2 Cov ( X 1 ,X 2 ) . (2) The property given in (2) implies: (a) If X 1 ,...,X n are independent, then they are uncorrelated, i.e. Cov ( X i ,X j ) , i 6 = j . Therefore, V ar { n X i =1 X i } = n X i =1 V ar ( X i ) + X i 6 = j Cov ( X i ,X j ) = n X i =1 V ar ( X i ) . (b) If X 1 ,...,X n are independent and identically distributed, V ar ( X ) = V ar { n- 1 n X i =1 X i } = n- 2 V ar { n X i =1 X i } = n- 2 nV ar ( X 1 ) = V ar ( X 1 ) n . 2. The following equality is helpful in many derivations, n X i =1 ( X i- X ) 2 = n X i =1 ( X i- a ) 2- n { X- a } 2 , a. (3) This is because X i =1 ( X i- X ) 2 = n X i =1 { ( X i- a )- ( X- a ) } 2 = n X i =1 [( X i- a ) 2- 2( X i- a )( X- a ) + ( X- a ) 2 ] . Note that n i =1 ( X i- a )( X- a ) = n ( X- a )( X- a ), the result in equation (3) follows. 3. There are two equivalent ways of expressiong variance, they are V ar ( X 1 ) = E { X- E ( X ) } 2 = E ( X 2 )- { E ( X ) } 2 . Now, E ( X ) = n- 1 n X i =1 E ( X i ) = n- 1 n = . E ( s 2 ) = E n i =1 ( X i- X ) 2 n- 1 = 1 n- 1 E { n X i =1 ( X i- X ) 2 } = n i =1 E ( X i- ) 2- nE ( X- ) 2 n- 1 . Note that E ( X i- ) 2 = V ar ( X i ) = 2 and E ( X- ) 2 = V ar ( X ) = n- 1 V ar ( X i ) = n- 1 2 , we then have E ( s 2 ) = n 2- n n- 1 2 n- 1 = 2 . 1... View Full Document

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