Problem 2 A Markov chain X., X1, X2, ..., X, has binary state space X, E{-1, +1} for all t. The tran- sition probabiliti
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- Problem 2 A Markov Chain X X1 X2 X Has Binary State Space X E 1 1 For All T The Tran Sition Probabiliti 1 (89.36 KiB) Viewed 140 times
Problem 2 A Markov chain X., X1, X2, ..., X, has binary state space X, E{-1, +1} for all t. The tran- sition probabilities are if i j PX:+1 X.Gli) = Pij = \( = p if i = 1 with p < 1/2. Suppose that the chain has run from some initial state for a long time and reached steady state at time n. We try to make an observation of the state Xn, by observing the value of a random variable Y. Given X = 1, Y is distributed according to N(x,02). (a) (5 points) Find the MAP estimate of X, when we observe Y = y. Write the conditional error probabilities PŘn # X Y = y) as a function of y. Hint: Write your answers to problem 2 in terms of the following functions. = 1 h+(y) 4 fy x,(yl + 1) = e 2702 1 h-(y) 4 fyx, (yl-1) = V2πσ2" (+12 e (b) (5 points) Find the MAP prediction of the next state Xn+1 when we observe Y = y (the noisy observation of Xn), write the conditional error probability P(*n+1 # Xn+1 Y = y). (c) (5 points) Recall that p< 1/2. Show that for any possible value of y, the MAP decision rule în+1 is always less reliable than that of MAP decision rule for X, (d) (5 points) Now we would like to estimate în+2 from the observation Y = y. Again, write the conditional error probability. You can use the notation p(2) = 2p(1 - p), which appears in the two step transition probability 1-p(2) if i=; j rij(2) = P xn+2\x.(li) = { 0(2) z|= if i # = p (e) (5 points) Show that as m, observing Y = y does not help to estimate Xn+m. That is, PXq+Y (-ly) approaches a Bernoulli random variable with parameter 1/2 for any value y.