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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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-
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Ask a homework question - tutors are online