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This question already has an answer here:

$ (X_n : n = 1, 2, ...) $ is a Markov chain with state space $(-1, 0, 1)$. Then which of the following is the correct answer?

  1. $(sin(X_n) : n = 1, 2, ...$) is a Markov chain.
  2. $(cos(X_n) : n = 1, 2, ...$) is a Markov chain.
  3. $(|X_n| : n = 1, 2, ...$) is a Markov chain.
  4. $(X_n^2 : n = 1, 2, ...$) is a Markov chain.
  5. None of the above is correct.

Is it option 5 is the correct answer since option 1&2 are continuous, option 3&4 cannot be negative value, so that the state space cannot be -1?

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marked as duplicate by user940, user296602, JMP, Pragabhava, Em. Feb 19 '16 at 9:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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$\sin(X_n)$ and $\cos(X_n)$ are not continuous processes. These are discrete time and discrete state space stochastic processes. The state spaces are respectively $\{\sin(-1),0,\sin(1)\}$ and $\{\cos(-1),1,\cos(1)\}$.

The problem with 2, 3 and 4 is actually the same, the function is not one-to-one: $\cos(-1)=\cos(1)$, $|-1|=|1|$ and $(-1)^2=1^2$. So these processes are not Markov chains because when you observe e.g. $|X_n|=1$, you don't know if $X_n=1$ or $X_n=-1$, so you don't know the transition probability from $1$ to $0$ (it may depend on the actual value of $X_n$), you need some more information.

You don't have this problem with 1, and you can easily show that it's a Markov chain. It's actually the same as $(X_n)$ except you renamed the state space.

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The question as posed is not a great one because even if the mapping is not one-to-one, the resulting sequence can still be a Markov chain. For example, if the transition matrix for the original sequence is $$ \left( \begin{matrix} 0&1&0\\ \frac12&0&\frac12\\ 0&1&0 \end{matrix} \right) $$ (labeling the rows and columns with states $-1,0,1$ respectively) then the transformed sequences for options 2, 3, 4, and 5 are all Markov chains, with transition matrix $$ \left(\begin{matrix}0&1\\1&0\end{matrix}\right)\;. $$ The reason is that in the original sequence the states $-1$ and $1$ are essentially indistinguishable, so transforming the sequence via an even function (such as cosine, absolute value, or square) creates a new sequence that deterministically bounces back and forth between the two transformed states.

OTOH if the question were "which of the following is guaranteed to be a Markov chain" then option 1 is the correct choice.

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