hwsol5 - EE 278 Wednesday, August 12, 2009 Statistical...

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Unformatted text preview: EE 278 Wednesday, August 12, 2009 Statistical Signal Processing Handout #15 Homework #5 Solutions 1. (10 points) Absolute value random walk. a. This is a straightforward calculation and we can use results from lecture notes. If k ≥ then P { Y n = k } = P { X n = + k or X n = − k } . If k > 0 then P { Y n = k } = 2P { X n = k } , while P { Y n = 0 } = P { X n = 0 } . Thus P { Y n = k } = ( n ( n + k ) / 2 ) ( 1 2 ) n − 1 k > , n − k is even , n − k ≥ ( n n/ 2 ) ( 1 2 ) n k = 0 , n is even , n ≥ otherwise b. If Y 20 = | X 20 | = 0 then there are only two sample paths with max 1 ≤ i< 20 | X i | = 10. These two paths are shown in Figure 1. Since the total number of sample paths is ( 20 10 ) and all paths are equally likely, P braceleftbig max 1 ≤ i< 20 Y i = 10 | Y 20 = 0 bracerightbig = 2 ( 20 10 ) = 2 184756 = 1 92378 . 10 10 20 n − 10 X n Figure 1: Sample paths for problem 1. 2. (10 points) Random walk with random start. a. We must show that for every sequence of indexes i 1 ,i 2 ,...,i n such that i 1 < i 2 < ... < i n , the increments X i 1 ,X i 2 − X i 1 ,...,X i n − X i n- 1 are independent. This is true by the definition of the { X i } random process; each X i j − X i j- 1 is the sum of a different set of Z i ’s, and the Z i ’s are i.i.d. and independent of X , which appears only in the first increment. b. Starting at an even number (0 or ± 2) can be ruled out, since there is no way that the process could then end up at X 11 = 2. Using Bayes rule for the remaining possibilities, we get P( X = − 1 | X 11 = 2) = P( X 11 = 2 | X = − 1)P( X = − 1) P( X 11 = 2) = ( 1 5 )( 11 7 )( 1 2 ) 7 ( 1 2 ) 4 ( 1 5 )( 11 7 )( 1 2 ) 7 ( 1 2 ) 4 + ( 1 5 )( 11 6 )( 1 2 ) 6 ( 1 2 ) 5 = ( 11 7 ) ( 11 7 ) + ( 11 6 ) = 1 1 + 11!7!4! 11!6!5! = 1 1 + 7 5 = 5 12 Similarly, P( X = 1 | X 11 = 2) = 7 12 . To summarize, P( X = x | X 11 = 2) = 5 12 x = − 1 7 12 x = +1 otherwise 3. (15 points) Markov processes. a. We are given that f ( x n +1 | x 1 ,x 2 ,...,x n ) = f ( x n +1 | x n ). From the chain rule, in general, f ( x 1 ,x 2 ,...,x n ) = f ( x 1 ) f ( x 2 | x 1 ) f ( x 3 | x 1 ,x 2 ) ··· f ( x n | x 1 ,x 2 ,...,x n − 1 ) . Thus, by the definition of Markovity, f ( x 1 ,x 2 ,...,x n ) = f ( x 1 ) f ( x 2 | x 1 ) f ( x 3 | x 2 ) ··· f ( x n | x n − 1 ) . (1) We will need the following to prove the second equality. f ( x i +1 | x k ,x k +1 ,...,x i ) = integraltext ··· integraltext f ( x 1 ,...,x i ,x i +1 ) dx 1 ··· dx k − 1 f ( x 1 ,...,x k ) = integraltext ··· integraltext f ( x 1 ,...,x i ) f ( x i +1 | x i ) dx 1 ··· dx k − 1 f ( x 1 ,...,x k ) = f ( x i +1 | x i ) (2) Now, applying the chain rule in reverse we get f ( x 1 ,x 2 ,...,x n ) = f ( x n ) f ( x n − 1 | x n ) f ( x n − 2 | x n − 1 ,x n ) ··· f ( x 1 | x 2 ,x 3 ,...,x n ) ....
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hwsol5 - EE 278 Wednesday, August 12, 2009 Statistical...

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