# A bjection between two stochastic processes

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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What assumptions are you putting on the index set T and the function f? –  Yemon Choi Jun 1 '11 at 17:30
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. –  Mikola Jun 1 '11 at 17:36
@Mikola: if the function f is a strictly increasing function and not affine ... –  Ghassen Hamrouni Jun 1 '11 at 17:46
–  Steve Huntsman Jun 1 '11 at 18:09