351K EE Probability, Statistics, and Random Processes SPRING 2008 Instructor:
Shakkottai/Vishwanath {shakkott,[email protected] 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, fXA (x)
= We get fX (x) P (A) , for 1.5 < X 2. fXA (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. FXZ (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 zx 0 zx 2 e
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This note was uploaded on 02/05/2012 for the course EE 351k taught by Professor Bard during the Spring '07 term at University of Texas.
 Spring '07
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