# hw12_sol - Solutions to HW1 1 Problem 3.1 from Stock Watson...

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Solutions to HW1 1. Problem 3.1 from Stock & Watson. The central limit theorem suggests that when the sample size ( n ) is large, the distribution of the sample average ( Y ) is approximately N ( µ Y , σ 2 Y ) with σ 2 Y = σ 2 Y n . Given a population µ Y = 100 , σ 2 Y = 43 . 0 , we have: (a) n = 100 , σ 2 Y = σ 2 Y n = 43 100 = 0 . 43 , and Pr( Y < 101) = Pr( Y 100 0 . 43 < 101 100 0 . 43 ) Φ (1 . 525) = 0 . 9364 . (b) n = 64 , σ 2 Y = σ 2 Y n = 43 64 = 0 . 6719 , and Pr(101 < Y < 103) = Pr( 101 100 0 . 6719 < Y < 103 100 0 . 6719 ) Φ (3 . 6599) Φ (1 . 2200) = 0 . 9999 0 . 8888 = 0 . 1111 (c) n = 165 , σ 2 Y = σ 2 Y n = 43 165 = 0 . 2606 , and Pr( Y > 98) = 1 Pr( Y 98) = 1 Pr( Y 100 0 . 2606 98 100 0 . 2606 1 Φ ( 3 . 9178) = Φ (3 . 9178) = 1 . 0000 (rounded to four decimal places). 2. Problem 3.2 from Stock & Watson. Each random draw Y i from the Bernoulli distribution takes a value of either zero or one with probability Pr( Y i = 1) = p and Pr( Y i = 0) = 1 p . The mean and the variance of the random variable Y i are as follows. E ( Y i ) = 0 × Pr( Y i = 0) + 1 × Pr( Y i = 1) = p var( Y i ) = E [( Y i µ Y ) 2 ] = (0 p ) 2 × Pr( Y i = 0) + (1 p ) 2 × Pr( Y i = 1) = p 2 (1 p ) + (1 p ) 2 p = p (1 p ) (a) The fraction of successes is b p = #( successes ) n = #( Y i = 1) n = P n i =1 Y i n = Y 1

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(b) E ( b p ) = E ( P n i =1 Y i n ) = 1 n n X i =1 E ( Y i ) = 1 n n X i =1 p = p (c) var( b p ) = var( P n i =1 Y i n ) = 1 n 2 n X i =1 V ar ( Y i ) = 1 n 2 n X i =1 p (1 p ) = p (1 p ) n .
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