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Unformatted text preview: ∞ e-λx dx = 1 λ 4. For α > 1, E [ Y ] = ∫ ∞ 1 y α y α +1 dy = ∫ ∞ 1 αy-α dy =-α α-1 y-( α-1) | ∞ 1 = α α-1 5. (a) Since LHS = k n ! ( n-k )! k ! = n ! ( n-k )!( k-1)! 1 ISyE 2027B Probability with Applications Fall 2010 RHS = n ( n-1)! ( n-1-( k-1))!( k-1)! = n ! ( n-k )!( k-1)! So LHS = RHS. Done. (b) E [ Z ] = n ∑ k =0 k ( n k ) p k (1-p ) n-k = n ∑ k =1 k ( n k ) p k (1-p ) n-k = n ∑ k =1 n ( n-1 k-1 ) pp k-1 (1-p ) n-k by (a) = np n ∑ k =1 ( n-1 k-1 ) p k-1 (1-p ) n-k = np n-1 ∑ j =0 ( n-1 j ) p j (1-p ) ( n-1)-j let j = k-1 = np ( p + (1-p )) n-1 by the binomial theorem* = np You can also see that as the sum of a binomial probability mass function. 2...
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This note was uploaded on 02/21/2011 for the course ISYE 2027 taught by Professor Zahrn during the Fall '08 term at Georgia Tech.
- Fall '08