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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.
- Spring '08