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hw7sol

hw7sol - Math 280A Fall 2011 Homework 7 Solutions 1 Problem...

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Math 280A, Fall 2011 Homework 7 Solutions 1. Problem 1 (page 260). Clearly P [ X k x ] = braceleftbigg 1 exp( ( x a )) , if x a ; 0 , if x < a . Consequently, for x a , P [ Z n > z ] = P [ X 1 > x, X 2 > x, . . . , X n > x ] = n productdisplay k =1 P [ X k > x ] = exp( n ( x a )) , while P [ Z n > x ] = 1 if x < a . Therefore, given ǫ > 0, we have P [ | Z n a | > ǫ ] = P [ Z n > a + ǫ ] = exp( ) 0 , n → ∞ , which means that Z n P a as required. 2. Let x 1 , x 2 , . . . , x n be strictly positive numbers. Show that 1 n n summationdisplay k =1 x k parenleftBigg n productdisplay k =1 x k parenrightBigg 1 /n , with equality only if x 1 = x 2 = · · · = x n . [Hint: The natural logarithm log is a strictly concave function.] Solution. Let X be a random variable that is uniformly distributed over the finite set { x 1 , x 2 , . . . , x n } ; that is, P [ X = x k ] = 1 n , k = 1 , 2 , . . ., n. Clearly X is integrable and E [ X ] = 1 n n summationdisplay k =1 x k . The function log is strictly concave on (0 , ), and so by Jensen’s inequality (restated for concave functions) (2 . 1) E [log( X )] log( E [ X ]) . But E [log( X )] = 1 n n summationdisplay k =1 log( x k ) = 1 n log( x 1 x 2 · · · x n ) = log parenleftBig ( x 1 x 2 · · · x n ) 1 /n parenrightBig .

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hw7sol - Math 280A Fall 2011 Homework 7 Solutions 1 Problem...

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