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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Sign up to join this communityLet 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 ?
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.