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lecture08 - ECE 5510 Random Processes Lecture Notes Fall...

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ECE 5510: Random Processes Lecture Notes Fall 2008 Lecture 8 Today: (1) Expectation of functions (Y&G 2.7-2.8, 3.3), (2) Conditional Distributions (Y&G 2.9, 3.8) Exam is Thursday Sept 28. Today’s lecture is the last material covered in the exam. Tues- day’s is not. Thu Sept 25 is a review session. Bring questions . HW 3 due today at 5pm. HW 4 due Sept 25 at 10:45 am at start of lecture. Short list: Lectures 1-8, Chapter 1,2,3 What don’t we have to worry about for the exam? 1.10, 1.11 (reliability, Matlab) 2.10 (Matlab) 3.6 (mixed r.v.s), 3.9 (Matlab) 1 Expectation of Functions of a Random Vari- able Expression X is a discrete r.v. X is a continuous r.v. E X [ X ] = x S X xP X ( x ) = integraltext S X xf X ( x ) E X [ g ( X )] = x S X g ( x ) P X ( x ) = integraltext S X g ( x ) f X ( x ) E X [ aX + b ] = aE X [ X ] + b = aE X [ X ] + b E X [ X 2 ] 2 nd moment = x S X x 2 P X ( x ) = integraltext S X x 2 f X ( x ) Var X [ X ] = E X [( X μ X ) 2 ], μ X = E X [ X ] = x S X ( x μ X ) 2 P X ( x ) = integraltext S X ( x μ X ) 2 f X ( x ) Example: Variance of Uniform r.v. Let X be a continous uniform r.v. on ( a, b ), with a, b > 0. 1. What is E X [ X ]? It is integraldisplay b a x b a dx = 1 2( b a ) x 2 vextendsingle vextendsingle b a = b 2 a 2 2( b a ) = b + a 2 .
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ECE 5510 Fall 2008 2 2. What is E X bracketleftbig 1 X bracketrightbig ? integraldisplay b a 1 b a 1 x dx = 1 b a (ln b ln a ) = 1 b a ln b a .
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