Hi, this was a midterm problem in a probability class.(adsbygoogle = window.adsbygoogle || []).push({});

1. The problem statement, all variables and given/known data

A fair coin is tossed repeatedly with results Y(0), Y(1)... that are 1 or 0 for heads or tails. For n>0, define a new sequence X(n) = Y(n)+Y(n-1), i.e. the number of 1's in the last two tosses.

Is Xn a markov sequence?

2. Relevant equations

3. The attempt at a solution

I thought, when X(n) = 1, we cannot determine P( X(n+1) | X(n) ), due to the fact that the values X(n+1) can take depends on whether the last two flips were 0 and 1 or 1 and 0. In the former case we can have X(n+1)= 0 or 1, whereas in the latter we can have X(n+1)=1 or 2. The probabilities of all these are 1/2, but the values are different. So i said X(n) is not markov.

In the posted solutions, X(n) is given as markov. Probabilities in the above case are calculated as

P(X(n+1)=0 | X(n) = 1) = P(X(n+1) =0 | Y(n)=0, Y(n-1)=1)*P(Y(n)=0, Y(n-1)=1) + P(X(n+1) =0 | Y(n)=1, Y(n-1)=0)*P(Y(n)=1, Y(n-1)=0)

= 1/2*1/2 + 0*1/2 = 1/4.

Similarly, P(1|1) is found to be 1/2 and P(2|1) is 1/4.

What is confusing is that, I thought, when X(n) is given, there must be a certain realization for Y(n) and Y(n-1), so we couldn't do the above...

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# Homework Help: Is X(n) a markov sequence?

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