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# hw1-2 - 2 Basic Options Problems 1 a Show 2 1 ex/2 dx = 1 2...

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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 0 e - x 2 / 2 dx · Z 0 e - y 2 / 2 dy = Z π/ 2 0 Z 0 e - r 2 / 2 rdrdθ = - e - r 2 / 2 fl fl fl 0 · π 2 = π 2 , we have

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6 2 Basic Options Z 0 e - φ 2 / 2 = p π/ 2 and furthermore get Z -∞ e - φ 2 / 2 = 2 Z 0 e - φ 2 / 2 = 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 . c) According to the definition of the expectation and the variance, we have E[ x ] = Z -∞ x · 1 b 2 π e - ( x - a ) 2 / 2 b 2 dx = Z -∞ ( by + a ) · 1 2 π e - y 2 / 2 dy = a.
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