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100AHW6 - STAT 100A HWVI Solution Problem 1 Suppose we flip...

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Unformatted text preview: STAT 100A HWVI Solution Problem 1: Suppose we flip a fair coin n times independently. Let X be the number of heads. Let k = n/ 2 + z √ n/ 2, or z = ( k- n/ 2) / ( √ n/ 2). Let g ( z ) = P ( X = k ). (1) Using the Stirling formula n ! ∼ √ 2 πnn n e- n , show that g (0) ∼ 1 √ 2 π 2 √ n . a ∼ b means that a/b → 1 as n → ∞ . g (0) = P ( X = n/ 2) = ˆ n n/ 2 ! / 2 n = n ! ( n/ 2)!( n/ 2)!2 n = √ 2 πnn n e- n [ p 2 π ( n/ 2)( n/ 2) n/ 2 e- n/ 2 ] 2 2 n = 1 √ 2 π 2 √ n . (2) Show that g ( z ) /g (0) → e- z 2 / 2 as n → ∞ . A: Let d = z √ n/ 2, g ( z ) g (0) = ( n n/ 2+ d ) ( n n/ 2 ) = n ! / [( n/ 2 + d )!( n/ 2- d )!] n ! / [( n/ 2)!( n/ 2)!] = ( n/ 2)!( n/ 2)! ( n/ 2 + d )!( n/ 2- d )! = ( n/ 2)( n/ 2- 1) ... ( n/ 2- d + 1) ( n/ 2 + 1) ... ( n/ 2 + d ) = (1- δ ) ... (1- ( d- 1) δ ) (1 + δ ) ... (1 + dδ ) ≈ e- ( δ + ... +( d- 1) δ ) e δ + ... + dδ = e- d ( d- 1) δ/ 2- d ( d +1) δ/ 2 = e- d 2 δ/ 2 = e- z 2 / 2 , where δ = 2 /n , and the “ ≈ ” becomes “=” as n → ∞ . So g ( z ) ∼ 1 √ 2 π e- z 2 / 2 2 √ n . (3) For two integers a < b , let a = ( a- n/ 2) / ( √ n/ 2), and b = ( b- n/ 2) / ( √ n/ 2). Show that P ( a ≤ X ≤ b ) → R b a f ( z ) dz , where f ( z ) = 1 √ 2 π e- z 2 / 2 . A: P ( a ≤ X ≤ b ) = b X k = a P ( X = k ) = b X z = a g ( z ) ≈ b X z = a 1 √ 2 π e- z 2 / 2 2 √ n = b X z = a f ( z )Δ z → Z b a f ( z ) dz, where Δ z = 2 / √ n , which is the space between every two consecutive values of z . “ ≈ ” becomes “=” as n → ∞ . (4) Let Z = ( X- n/ 2) / ( √ n/ 2). Show that P ( a ≤ X ≤ b ) = P ( a ≤ Z ≤ b ). Argue that in the limit Z ∼ N(0 , 1)....
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100AHW6 - STAT 100A HWVI Solution Problem 1 Suppose we flip...

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