# Hw5 - 351K EE Probability Statistics and Random Processes...

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351K EE Probability, Statistics, and Random Processes SPRING 2008 Instructor: Shakkottai/Vishwanath {shakkott,sriram}@ece.utexas.edu Homework 5 – Solutions Problem 1 fX (x) = lim0 = lim0 P (x<Xx+) P (X=Y )P (x<Y x+)+P (X=Z)P (x<Zx+) P (x<Y x+) = p lim0 + (1 - p) lim0 P (x<Zx+) = pfY (x) + (1 - p)fZ (x) Problem 2 Part 1: 2 We know that - fX (x)dx = 1. Hence, 1 xc2 dx = 1. c (1 - 1 ) = 1. 2 c=2. Part 2: 2 2 1 P (A) = P (X > 1.5) = 1.5 x2 dx = 2 ( 1.5 - 1 ) = 1 . 2 3 Hence, fX|A (x) = We get fX (x) P (A) , for 1.5 < X 2. fX|A (x) 6 = x2 , = 0, for 1.5 < X 2 otherwise Problem 3 We first find the conditional CDF, then differentiate it to find the conditional PDF. FX|Z (x0 |z) = P (X x0 |Z = z) We can't directly find an expression for this probability, because we are conditioning on an event of probability zero ( P (Z = z) = 0). To circumvent this difficulty, we condition on the event Z [z - , z] and then take the limit as 0, as shown in the figure. Hence, P (Z [z - , z]) = P (X + Y [z - , z]) = z z-x 0 z-x- 2 e-
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