part22

# part22 - EE 5375/7375 Random Processes Homework #3...

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EE 5375/7375 Random Processes September 16, 2003 Homework #3 Solutions Problem 1. textbook problem 3.53 The exam grades in a certain class have a Gaussian (normal) pdf with mean m and standard deviation σ . Find the constants a and b so that the random variable Y = aX + b has a Gaussian pdf with mean m 0 and standard deviation σ 0 . The linear function Y = aX + b has a simple eﬀect on the mean and variance of X . First, the mean of Y is m 0 = E ( Y ) = E ( aX + b ) = aE ( X ) + b = am + b Second, the variance of Y is σ 0 2 = var ( Y ) = var ( aX + b ) = a 2 var ( X ) = a 2 σ 2 Thus we have 2 equations for 2 unknowns. The second equation gives a = σ 0 σ Substituting into the ﬁrst equation, we get b = m 0 - am = m 0 - σ 0 σ m Problem 2. textbook problem 3.56 Suppose that a voltage X is a zero-mean Gaussian random variable. Find the pdf of the power dissipated by an R-ohm resistor P = RX 2 . The pdf of X is the Gaussian density function: f X ( x ) = 1 2 πσ 2 e - x 2 / 2 σ 2 The PDF of the power P is F P ( y ) = P ( RX 2 y ) = P ( - p y/R X p y/R ) = F X ( p y/R ) - F X ( - p y/R ) The pdf is found by diﬀerentiating with respect to y : f P ( y ) = f X ( p y/R ) 2 p y/R 1 R - f X ( - p y/R ) - 2 p y/R 1 R = f X ( p y/R ) 2 yR + f X ( - p y/R ) 2 yR = 1 2 yR 1 2 πσ 2 e - ( y/R ) / 2 σ 2 + 1 2 yR 1 2 πσ 2 e - ( y/R ) / 2 σ 2 = 1 p 2 πσ 2 Ry e - y/ 2 σ 2 R Problem 3. textbook problem 3.59 1

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Let Y = e X . (a) Find the PDF and pdf of Y in terms of the PDF and pdf of X . (b) Find the pdf of Y when X is a Gaussian random variable. In this case, Y is said to have a lognormal pdf. (a) The PDF of Y is F Y ( y ) = P ( Y y ) = P ( e X y ) = P ( X ln y ) = F X (ln y ) The pdf is found by diﬀerentiating with respect to y : f Y ( y ) = d dy F X (ln y ) = f X (ln y ) d dy ln y = 1 y f X (ln y ) (b) The pdf of X is the Gaussian density function: f X ( x ) = 1 2 πσ 2 e - ( x - μ ) 2 / 2 σ 2 Then the pdf of Y is f Y ( x ) = 1 y 2 πσ 2 e - (ln y - μ ) 2 / 2 σ 2 Problem 4. textbook problem 3.63 An urn contains 90 \$1 bills, 9 \$5 bills, and 1 \$50 bill. Let the random variable X be the denomination of a bill that is selected at random from the urn. Find the mean m = E ( X ). In what sense is the mean the break-even price of a ticket for the right to draw a single bill from the urn? Since each bill is equally likely, the probability of each denomination is the same as its relative frequency. That is, the probability of \$1 is 0.9, probability of \$5 is 0.09, and probability of \$50 is 0.01. The mean is therefore E ( X ) = (1)(0 . 9) + (5)(0 . 09) + (50)(0 . 01) = 1 . 85 The relative frequency interpretation implies that 1.85 is the long-term expected payoﬀ, if the drawing is repeated independently many times. Problem 5. textbook problem 3.75 (a) Suppose a coin is tossed n times. Each coin toss costs d and the reward in obtaining X heads is aX 2 + bX . Find the expected value of the net reward. (b) Suppose that the reward in obtaining X heads is a X , where a > 0. Find the expected value of the reward.
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## This note was uploaded on 11/29/2009 for the course EE 131A taught by Professor Lorenzelli during the Fall '08 term at UCLA.

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part22 - EE 5375/7375 Random Processes Homework #3...

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