HW9-2 - R and pdf f N ( n ) = 1 2 e- | n | for a positive...

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ECE 302, Homework #9. Due date: 3/30 http://www.ece.purdue.edu/ chihw/ECE302 07.html Prof. Chih-Chun Wang Email: chihw@purdue.edu Office: MSEE354 Office Hours: MWF: 12pm–1pm TA: Kamesh Krishnamurthy, Email: kkrishna@purdue.edu Office: POTR 370 Office Hours: M: 10–11am TTh: 10:30am–12:30pm Question 1: Select three events from Problem 4.1(a) to 4.1(i) by yourself, and sketch their corresponding regions on the two dimensional plane R 2 . Question 2: Problem 4.4. Question 3: Problem 4.9(a), 4.9(b). Find the marginal cdf of X . How do you derive the marginal pdf of X from the marginal pdf of X ? Problem 4.9(c). Question 4: Problem 4.11. Question 5: Problem 4.12(c) Question 6: Problem 4.17. Do not be afraid by the “Laplacian random variable,” which is simply a random variable with sample space being
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Unformatted text preview: R and pdf f N ( n ) = 1 2 e- | n | for a positive constant > 0. For 4.17(c), you are asked to compute P ( X = 1 | Y > 0) and P ( X =-1 | Y > 0). Question 7: Problem 4.38(ac). Hints: 1. For 4.38(a), We notice that Y = N + x is just another Gaussian with mean x and variance 2 N . 2. For 4.38(c), since f X,Y ( x,y ) = f X | Y ( x | y ) f Y ( y ), the conditional f X | Y ( x | y ) is the ratio of the joint pdf f X,Y ( x,y ) divided by the marginal pdf f Y ( y ). So your goal is to nd f Y ( y ). You can use the following formula to nd f Y ( y ): f Y ( y ) = Z x =- f X,Y ( x,y ) dx (1) [Optional] Ask yourself why f Y ( y ) can be obtained from the above integral....
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This note was uploaded on 05/02/2009 for the course ECE 302 taught by Professor Gelfand during the Spring '08 term at Purdue University-West Lafayette.

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HW9-2 - R and pdf f N ( n ) = 1 2 e- | n | for a positive...

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