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 ?

  • $\begingroup$ What assumptions are you putting on the index set T and the function f? $\endgroup$
    – Yemon Choi
    Jun 1, 2011 at 17:30
  • $\begingroup$ 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. $\endgroup$
    – Mikola
    Jun 1, 2011 at 17:36
  • $\begingroup$ @Mikola: if the function f is a strictly increasing function and not affine ... $\endgroup$ Jun 1, 2011 at 17:46
  • $\begingroup$ See books.google.com/books?id=_gOMJL5AEdQC&pg=PA99 $\endgroup$ Jun 1, 2011 at 18:09

1 Answer 1


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.

  • $\begingroup$ Funny: neither continuous, nor differentiable nor strictly are necessary. $\endgroup$
    – Did
    Jun 23, 2011 at 16:06

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