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100AHW6S - STAT 100A HWVI Solution Problem 1 Suppose we ip...

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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 → ∞ . A: 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 + ) e - ( δ + ... +( d - 1) δ ) e δ + ... + = 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 0 = ( a - n/ 2) / ( n/ 2), and b 0 = ( b - n/ 2) / ( n/ 2). Show that P ( a X b ) R b 0 a 0 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 0 X z = a 0 g ( z ) b 0 X z = a 0 1 2 π e - z 2 / 2 2 n = b 0 X z = a 0 f ( z z Z
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