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Unformatted text preview: 2 Basic Options Problems 1. a) Show Z  1 2 e x 2 / 2 dx = 1 . b) Show that Z  1 b 2 e ( x a ) 2 / 2 b 2 dx = 1 holds for any a and b . (Because this is true and the integrand is always possitive, it can be a probability density function.) c) If the probability density function of a random variable x is 1 b 2 e ( x a ) 2 / 2 b 2 , then it is called a normal random variable. Show E[ x ] = a and Var[ x ] = b 2 . Solution : a) Because Z e x 2 / 2 dx Z e y 2 / 2 dy = Z / 2 Z e r 2 / 2 rdrd = e r 2 / 2 fl fl fl 2 = 2 , we have 6 2 Basic Options Z e 2 / 2 d = p / 2 and furthermore get Z  e 2 / 2 d = 2 Z e 2 / 2 d = 2 p / 2 = 2 , which can be rewritten as Z  1 2 e x 2 / 2 dx = 1 . b) Let y = x a b , then we have Z  1 b 2 e ( x a ) 2 / 2 b 2 dx = Z  1 2 e y 2 / 2 dy = 1 ....
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 Spring '10
 Zhu
 Math, Probability

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