Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 ?

share|improve this question
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
add comment

1 Answer

up vote 1 down vote accepted

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.

share|improve this answer
Funny: neither continuous, nor differentiable nor strictly are necessary. –  Did Jun 23 '11 at 16:06
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.