I'm having some trouble coming up with a counter-example for this problem:

Give an example of a stochastic process $\{X_n : n \in \mathbb{Z}^+\}$ on $(\Omega, \mathcal{F}, P)$ such that $P_{X_n} = P_{X_0}$ for all $n$ but $\{X_n\}$ is not stationary (I would assume a stationary process?). Here $P_{X_n}$ denotes the pushforward measure of $P$ under $X_n$.

I feel as if I am deeply misunderstanding either the question or the difference of two random variables having the same push-forward measure but bot being iid. Even a hint on the right direction would be much appreciated.

Many thanks in advance!

  • $\begingroup$ Pick some bits $Y_i \in \{0,1\}$. Throw a coin. If heads, $X_i = Y_i$ for all $i$. If tails, $X_i = 1-Y_i$ for all i? $\endgroup$ – Ville Salo Mar 9 at 10:20
  • $\begingroup$ Thanks! I guess I got stuck since I always though of $P_{X_n} = P_{X_0}$to be iid. In this case, if I am understanding correctly, we can choose to be shrinking our transition porbabilities? $\endgroup$ – Flustered_Potato Mar 9 at 18:25
  • $\begingroup$ Transition probabilities as in a Markov chain? In my suggestion you pick $X_0$ with uniform distribution, and then all transitions are deterministic, but the matrix depends on $i$. You can make a small modification to make all transitions positive, e.g. as a final step flip each bit independently with a small probability (you can see it as something like a non-stationary hidden Markov chain still). $\endgroup$ – Ville Salo Mar 9 at 19:01

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