Mth562-HW8-2

Mth562-HW8-2 - MTH/STA 562 Exercise 8.1 . Write b...

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Unformatted text preview: MTH/STA 562 Exercise 8.1 . Write b & & & = h b & & E & b & i + h E & b & & & i = h b & & E & b & i + B: Then & b & & & 2 = h b & & E & b & i 2 + 2 B h b & & E & b & i + B 2 : Thus, MSE & b & = E & b & & & 2 = E h b & & E & b & i 2 + 2 BE h b & & E & b & i + B 2 Note that V ar & b & = E h b & & E & b & i 2 and E h b & & E & b & i = E & b & & E & b & = 0 : Hence, MSE & b & = E & b & & & 2 = V ar & b & + B 2 : Exercise 8.2 . It is given that E & b & 1 = E & b & 2 = & , V ar & b & 1 = 2 1 , and V ar & b & 2 = 2 2 . Also, b & 3 = a b & 1 + (1 & a ) b & 2 . ( a ) Since E & b & 3 = E h a b & 1 + (1 & a ) b & 2 i = aE & b & 1 + (1 & a ) E & b & 2 = a& + (1 & a ) & = &; b & 3 is an unbiased estimator for & . ( b ) Assuming that b & 1 and b & 2 are independent, the variance of the new estimator, b & 3 , will be V ar & b & 3 = V ar h a b & 1 + (1 & a ) b & 2 i = a 2 V ar & b & 1 + (1 & a ) 2 V ar & b & 2 = a 2 2 1 + (1 & a ) 2 2 2 : In order to choose a value of a such that V ar & b & 3 is minimized, look at d da V ar & b & 3 = 2 a 2 1 & 2 (1 & a ) 2 2 : Setting the derivative equal to zero, we obtain 2 a 2 1 & 2 (1 & a ) 2 2 = 0 or a = 2 2 2 1 + 2 2 : Note that d 2 da 2 V ar & b & 3 = 2 2 1 + 2 2 2 > so that the value is in fact a minimum. Exercise 8.3 . It is given that E & b & 1 = E & b & 2 = & , V & b & 1 = 2 1 , V ar & b & 2 = 2 2 , and Cov & b & 1 ; b & 2 = c . Also, b & 3 = a b & 1 + (1 & a ) b & 2 . ( b ) The variance of the new estimator, b & 3 , will be V ar & b & 3 = V ar h a b & 1 + (1 & a ) b & 2 i = a 2 V ar & b & 1 + (1 & a ) 2 V ar & b & 2 +2 a (1 & a ) Cov & b & 1 ; b & 2 = a 2 2 1 + (1 & a ) 2 2 2 + 2 a (1 & a ) c: In order to choose a value of a such that V ar & b & 3 is minimized, consider d da V ar & b & 3 = 2 a 2 1 & 2 (1 & a ) 2 2 + 2 c (1 & 2 a ) : Setting the derivative equal to zero, we obtain 2 a 2 1 & 2 (1 & a ) 2 2 + 2 c (1 & 2 a ) = 0 a 2 1 & (1 & a ) 2 2 + c (1 & 2 a ) = 0 a 2 1 & 2 2 + a 2 2 + 2 c & 2 ca = 0 a & 2 1 + 2 2 & 2 c = 2 2 & c Hence, a = 2 2 & c 2 1 + 2 2 & 2 c : Note that d 2 da 2 V ar & b & 3 = 2 2 1 + 2 2 2 & 4 c = 2 & 2 1 + 2 2 & 2 c = 2 h V ar & b & 1 + V ar & b & 2 & 2 Cov & b & 1 ; b & 2 i = 2 V ar & b & 1 & b & 2 > so that the value is in fact a minimum....
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Mth562-HW8-2 - MTH/STA 562 Exercise 8.1 . Write b...

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