2
$\begingroup$

This is, can we find an $m-$dependent ($1\leq m<\infty$) non-stationary, non-independent stochastic process $(X_{k})_{k}$ with the property that $\mathbb{P}[X_{k+1}\in A |X_{k}]=\nu(X_{k},A)$ for a fixed (independent on $k$) conditional probability distribution function $\nu:\mathbb{R}\times\mathcal{R}\to [0,1]$? ($\mathcal{R}$ is the Borel sigma algebra of $\mathbb{R}$).

If the answer is ''no'', can we do it if we require only that $E[X_{k+1}|X_{k}]=f\circ X_{k}$ for a fixed function $f:\mathbb{R}\to\mathbb{R}$ and every $k$?

Improve this question
$\endgroup$
2
  • $\begingroup$ What is the point emphasizing m-dependent and non-independent at the same time? Do "dependent"s have different meanings in this setting? And what do you mean by "conditionally stationary" since it seems that you are asking for a stationary process? $\endgroup$
    – Henry.L
    Commented Apr 2, 2017 at 13:19
  • $\begingroup$ According to some definitions, an independent process is $m-$dependent for every $m\geq 0$, but that is not important. An homogeneous Markov process is conditionally stationary even when it is not stationary. $\endgroup$
    – David
    Commented Apr 2, 2017 at 15:58

1 Answer 1

Reset to default
0
$\begingroup$

Let $\{X_i\}_{i=0}^{\infty}$ be an i.i.d. Bernoulli sequence with $P[X_i=1]=P[X_i=0]=1/2$. Consider:

$$\{X_1, X_0, X_2, X_1, X_3, X_2, X_4, X_3, X_5, X_4, X_6, X_5, ...\}$$

Improve this answer
$\endgroup$
4
  • $\begingroup$ Any reason for the $-1$? Not very mathematical or constructive to downvote with no explanation. My answer above seems to answer the question very simply. Is there something I am missing? $\endgroup$
    – Michael
    Commented May 11, 2017 at 22:56
  • $\begingroup$ Is it possible that someone did not notice the pattern for $\{Y_1, Y_2, Y_3, …\}$? $Y_i = X_{(i+1)/2}$ for $i$ odd; $Y_i = X_{(i-2)/2}$ for $i$ even. $\endgroup$
    – Michael
    Commented May 11, 2017 at 22:56
  • $\begingroup$ I think this is right. Thanks for answer! (which I had not seen for some reason) $\endgroup$
    – David
    Commented Sep 24, 2018 at 23:16
  • $\begingroup$ I wonder if we can modify this to produce an example which is not 1-stationary. $\endgroup$
    – David
    Commented Sep 24, 2018 at 23:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .