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Unformatted text preview: Math 280A, Fall 2011 Homework 6 1. [Problem 2 (page 153)] (a) Choose r = 2 / 3, and apply H olders inequality with p = 3 / 2 (and conjugate exponent q = 3) to obtain 1 = E [ X 2 ] = E [ | X | 2 / 3 | X | 4 / 3 ] ( E | X | ) 2 / 3 ( E [ X 4 ] ) 1 / 3 . Taking the 3 / 2 power of the extreme terms in this string of inequalities, we have 1 E | X | radicalbig E [ X 4 ] , which immediately implies the desired inequality. (b) More generally, if E [ X 2 ] = 1 and E [ X 2 m ] < for some m 1, then E | X | 1 ( E [ X 2 m ]) 1 / (2 m- 2) . To see this do the same as in part (a) but now with r = (2 m 2) / (2 m 1) and p = 1 /r = (2 m 1) / (2 m 2). 2. [Problem 5 (page 154)] Splitting E [ X ]: E [ X ] = E [ X ; X > E [ X ]] + E [ X ; X E [ X ]] E [ X ; X > E [ X ]] + E [ X ] Moving the extreme right-hand term to the other side of the inequality, we obtain (1 ) E [ X ] E [ X ; X > E [ X ]] . Let us now apply the Cauchy-Schwarz inequality (aka, H olders inequality with p = q = 2) to the right side of this inequality to obtain (1 ) E [ X ] ( E [ X 2 ] ) 1 / 2 ( P [ X > E [ X ]]) 1 / 2 . Squaring this yields (1 ) 2 ( E [ X ]) 2 E [ X 2 ] P [ X > E [ X ]] , which yields the stated inequality upon dividing by E [ X 2 ]. 3. In this problem let X and Y be independent random variables with cumulative distribution functions F and G ; thus, for example, F ( x ) = P [ X x ] for x R . (a) Use Tonellis theorem to show that E [ F ( Y )] = P [ X Y ] and E [ G ( X )] = P [ Y X ] ....
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