Let ${\bf \tilde{x}}_1, {\bf \tilde{x}}_2, \ldots$ be a (possibly non-stationary) stochastic sequence of $d$-dimensional random vectors that converges in distribution. Does it immediately follow that the stochastic sequence $[{\bf \tilde{x}}_1, {\bf \tilde{x}}_2], [{\bf \tilde{x}}_3, {\bf \tilde{x}}_4], \ldots$ converges in distribution as well? Why or why not? More generally, if ${\bf \tilde{x}}_1, {\bf \tilde{x}}_2, \ldots$ converges in distribution, then does it follow that ${\bf \tilde{x}}_{t+1}, {\bf \tilde{x}}_{t+2}, \ldots, {\bf \tilde{x}}_{t+m}$ converges in distribution as $t \rightarrow \infty$ for fixed positive integer $m$ where $m > 2$? My interest in this problem is that if I have an algorithm that is generating a non-stationary stochastic sequence which converges in distribution to a random vector I want to assess convergence by examining the autocorrelation function which assumes weak stationarity. My interpretation the asymptotic behavior of the stochastic sequence would be stronger if I a priori knew that the tail end of the generated stochastic sequence was asymptotically stationary in the strong sense.

  • 1
    $\begingroup$ "...if ${\bf \tilde{x}}_1, {\bf \tilde{x}}_2, \ldots$ converges in distribution, then does it follow that ${\bf \tilde{x}}_1, {\bf \tilde{x}}_2, \ldots$ are independent and identically distributed?" - that would seem to preclude the existence of CLTs for non-iid sequences, which evidently is not the case. $\endgroup$ Commented Dec 29, 2018 at 15:33
  • $\begingroup$ I fixed the question... $\endgroup$
    – RMG
    Commented Dec 29, 2018 at 16:21

1 Answer 1



Here is a counterexample to help you develop some intuition. It concerns a sequence of identically distributed variables with a common Bernoulli$(1/2)$ distribution. It is defined by letting $(X_{4i+1}, X_{4i+2}, X_{4i+3}, X_{4i+4})$ equal $(0,0,0,1)$ with probability $1/2$ and otherwise letting it equal $(1,1,1,0)$ with probability $1/2$ for $i=1,2,3,\ldots.$

The variables $(X_{4i+1}, X_{4i+2})$ are always equal whereas the variables $(X_{4i+3}, X_{4i+4})$ always differ.

The sequence $(X_{2j+1}, X_{2j+2})$ thus has two convergent subsequences, but they converge to distinctly different bivariate distributions, demonstrating this sequence of bivariate random variables does not converge in distribution.

You should have no difficulty generalizing this counterexample to blocks of length $m$ within the original sequence of random variables.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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