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 ?



A bjection between two stochastic processes
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. 


