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Unformatted text preview: 2 Basic Options Problems 1. a) Show Z ∞∞ 1 √ 2 π e x 2 / 2 dx = 1 . b) Show that Z ∞∞ 1 b √ 2 π 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 π 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 π/ 2 Z ∞ e r 2 / 2 rdrdθ = e r 2 / 2 fl fl fl ∞ · π 2 = π 2 , we have 6 2 Basic Options Z ∞ e φ 2 / 2 dφ = p π/ 2 and furthermore get Z ∞∞ e φ 2 / 2 dφ = 2 Z ∞ e φ 2 / 2 dφ = 2 p π/ 2 = √ 2 π, which can be rewritten as Z ∞∞ 1 √ 2 π e x 2 / 2 dx = 1 . b) Let y = x a b , then we have Z ∞∞ 1 b √ 2 π e ( x a ) 2 / 2 b 2 dx = Z ∞∞ 1 √ 2 π e y 2 / 2 dy = 1 ....
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This note was uploaded on 02/11/2010 for the course MATH 6203 taught by Professor Zhu during the Spring '10 term at University of North Carolina Wilmington.
 Spring '10
 Zhu
 Math, Probability

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