Let x(t) be a Markov process. We define the stochastic process y(t) such that :
y(t) = x(f(t))
f : T -> T
T is the parameter set of the process x(t).
If we know that f is bijective, is y(t) a Markov process ?
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$\begingroup$ What assumptions are you putting on the index set T and the function f? $\endgroup$– Yemon ChoiCommented Jun 1, 2011 at 17:30
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$\begingroup$ Only if f is affine. To see why, try applying the change of variables theorem. In order for the coefficients for the process to be constant, the function f must be at most first order. $\endgroup$– MikolaCommented Jun 1, 2011 at 17:36
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$\begingroup$ @Mikola: if the function f is a strictly increasing function and not affine ... $\endgroup$– Ghassen HamrouniCommented Jun 1, 2011 at 17:46
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$\begingroup$ See books.google.com/books?id=_gOMJL5AEdQC&pg=PA99 $\endgroup$– Steve HuntsmanCommented Jun 1, 2011 at 18:09
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1 Answer
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If f is continuous, differentiable and strictly increasing, the answer is yes. It is easy to see that y(t_n) is conditionally independent of y(t_1), y(t_2), y(t_{n-2}) given y(t_{n-1}), as this is the equivalent property enjoyed by x(t) when it is Markov. x(\tilde{t}_n}) is independent of x(\tilde{t}1),...,x(\tilde{t}{n-2}) conditional on x(\tilde{t}_{n-1}) as long as the \tilde{t}_k form a strictly increasing sequence.
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$\begingroup$ Funny: neither continuous, nor differentiable nor strictly are necessary. $\endgroup$– DidCommented Jun 23, 2011 at 16:06