# Let t1 and t2 be two points in time and denote x1 x

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Unformatted text preview: general not practical. Let t1 and t2 be two points in time, and denote X1 = X (t1 ) and X2 = X (t2 ) as the random variable samples of X (t) at these times. The PDF of random variable X (t2 ) given (a value of) X (t1 ) is denoted p(x2 /x1 ). This is just the conditional PDF we’ve been employing. Now consider K samples of X (t), Xn ; n = 1, 2, ..., K , taken at times tn n = 1, 2, · · · , K where tn+1 > tn . If, for all integer K and all possible tn n = 1, 2, · · · , K , we have that p(xK /xK −1 , xK −2, · · · , x1 ) = p(xK /xK −1 ) , (1) 2.4 The Viterbi Algorithm then X (t) is a Markov process. This indicates that, given XK −1 , XK is statistically independent of Xn ; n = K − 2, K − 3, · · · , 1. As a result, for a Markov process, p(xK , xK −1, xK −2 , · · · , x1 ) = p(xK /xK −1, xK −2 , · · · , x1 ) (2) . p(xK −1 /xK −2 , xK −3, · · · , x1 ) · · · p(x2 /x1 ) p(x1 ) = p(xK /xK −1) p(xK −1 /xK −2 ) · · · p(x2 /x1 ) p(x1 ) K = p(x1 ) n=2 p(xn /xn−1 ) . Markov processes are much mo...
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## This document was uploaded on 10/12/2009.

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