# 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 2 Y ) with σ 2 Y = σ 2 Y n . Given a population µ Y = 100 2 Y =43 . 0 ,wehave : (a) n = 100 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 2 Y = σ 2 Y n = 43 64 . 6719 Pr(101 < 103) = Pr( 101 100 0 . 6719 < 103 100 0 . 6719 ) Φ (3 . 6599) Φ (1 . 2200) = 0 . 9999 0 . 8888 = 0 . 1111 (c) n = 165 2 Y = σ 2 Y n = 43 165 . 2606 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 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 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 Var ( Y i 1 n 2 n X i =1 p (1 p p (1 p ) n . The second equality uses the fact that Y 1 ,Y 2 , ...... , Y n are iid draws and Cov ( Y i j )=0 for i 6 = j.
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## This note was uploaded on 04/30/2011 for the course MECHANICAL MSC taught by Professor Dr.rahula during the Spring '10 term at University of Moratuwa.

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

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