# hw5_sol - EE 351K Probability, Statistics, and Random...

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Unformatted text preview: EE 351K Probability, Statistics, and Random Processes SPRING 2008 Instructor: Shakkottai/Vishwanath { shakkott,sriram } @ece.utexas.edu Homework 5 - Solutions Problem 1 f X ( x ) = lim P ( x&lt;X x +) = lim P ( X = Y ) P ( x&lt;Y x +)+ P ( X = Z ) P ( x&lt;Z x +) = p lim P ( x&lt;Y x +) + (1- p ) lim P ( x&lt;Z x +) = pf Y ( x ) + (1- p ) f Z ( x ) Problem 2 Part 1: We know that integraltext - f X ( x ) dx = 1 . Hence, integraltext 2 1 c x 2 dx = 1 . c (1- 1 2 ) = 1 . c = 2 . Part 2: P ( A ) = P ( X &gt; 1 . 5) = integraltext 2 1 . 5 2 x 2 dx = 2 ( 1 1 . 5- 1 2 ) = 1 3 . Hence, f X | A ( x ) = f X ( x ) P ( A ) , for 1 . 5 &lt; X 2 . We get f X | A ( x ) = 6 x 2 , for 1 . 5 &lt; X 2 = 0 , otherwise Problem 3 We first find the conditional CDF, then differentiate it to find the conditional PDF. F X | Z ( x | z ) = P ( X x | Z = z ) We cant directly find an expression for this probability, because we are conditioning...
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## hw5_sol - EE 351K Probability, Statistics, and Random...

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