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Unformatted text preview: Homework 6 October 17, 2007 Problem 1. Let be a rate parameter. Then notice that P ( X > . 01) = 1 P ( X . 01) = e . 01 . Solving 1 / 2 = e . 01 , we have = 100ln2 . Thus, P ( X > t ) = e 100 t ln2 , so by solving . 9 = e 100 t ln2 , we have t = . 01log 2 . 9 . Problem 2. EX = R  x e  x  dx = R 1 2 xe x + R 1 2 xe x = 1 2 R xde x 1 2 R xde x = 1 2 R e x dx + 1 2 R e x dx = 1 2 + 1 2 = 0 E [ X 2 ] = R  x 2 e  x  dx = R 1 2 x 2 e x + R 1 2 x 2 e x = 1 2 R x 2 de x 1 2 R x 2 de x = R xe x dx + R xe x dx = 1 2 + 1 2 = 2 2 V arX = E [ X 2 ] ( EX ) 2 = 2 2 Problem 3. If X Bin ( n,p ) , then X ( t ) = Ee tX = n X j =0 P [ X = j ] = n X j =0 e tj p j (1 p ) n j n j = n X j =0 ( e t p ) j (1 p ) n j n j = ( e t p + (1 p ) ) n , 1 where the last step invokes the binomial theorem. So X ( t ) = npe t ( e t p + (1 p )...
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This note was uploaded on 11/12/2008 for the course ORIE 360 taught by Professor Ehrlichman during the Fall '07 term at Cornell University (Engineering School).
 Fall '07
 EHRLICHMAN

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