HW2_Solutions

# HW2_Solutions - IE:3610 Stochastic Modeling(Fall 2015...

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IE:3610 Stochastic Modeling (Fall 2015) Homework 2 Solutions HW2 (due Thu, Sep 10): 24-4(a,b,c); 24.5; 24-8(a-d); 24-9(a,b,c); 24.13. 24-4 (a, b, c) The random variable X has density function f given by f X ( y ) = 8 > < > : for 0 y K for < y 1 0 elsewhere (a) Determine K in terms of . (b) Find F X ( b ), the CDF of X . (c) Find E ( X ). Solution: (a) By the property of probability density function: Z + 1 -1 f X ( y ) dy = 1 Then, Z + 1 -1 f X ( y ) dy = Z 0 -1 f X ( y ) dy + Z 0 f X ( y ) dy + Z 1 f X ( y ) dy + Z + 1 1 f X ( y ) dy Z + 1 -1 f X ( y ) dy = 0 + Z 0 dy + Z 1 Kdy + 0 = 2 + K (1 - ) 1 = 2 + K (1 - ) K = 1 - 2 1 - = 1 + (b) by the definition F X ( b ) = R b -1 f X ( y ) dy for b < 0 : F X ( b ) = R b -1 0 dy = 0 1

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for 0 b : F X ( b ) = R 0 -1 0 dy + R b 0 dy = b for b 1 : F X ( b ) = R 0 -1 0 dy + R 0 dy + R b Kdy = +( b - ) K = = 2 + b (1 + ) - - 2 = b (1 + ) - for 1 < b : F X ( b ) = R 0 -1 0 dy + R 0 dy + R 1 Kdy + R b 1 0 dy = = 2 + K (1 - ) = 1 F X ( y ) = 8 > > > < > > > : 0 for b < 0 b for 0 b b (1 + ) - for < b 1 1 for1 < b (c) by the definition E ( X ) = R + 1 -1 yf X ( y ) dy E ( X ) = Z 0 -1 y 0 dy + Z 0 y
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