The definition of strict stationarity I'm using is the following:

$(X_1,...,X_n)=^d(X_{1+h},...,X_{n+h})$, for any integer h, and positive integer n.

I'm trying to prove that $(X_1,X_{1+h})=^d(X_{t},X_{t+h})$ for any integer t, but the only close thing I managed to prove until now is that $(X_1,X_{t})=^d(X_{1+h},X_{t+h})$.

Any help would be appreciated.

  • $\begingroup$ What does the definition say when $h=t-1$ and $n=1+h$? $\endgroup$
    – whuber
    Oct 29, 2014 at 15:10
  • $\begingroup$ @whuber and what if $t$ is smaller than -1? $h$ could be negative, but $n$ cannot... $\endgroup$ Oct 29, 2014 at 15:38
  • $\begingroup$ Where in your question is it stipulated that $t$ could be anything less than $1$? Your notation strongly suggests the indexes of this process are $\{1,2,3,\ldots,n,\ldots\}$. $\endgroup$
    – whuber
    Oct 29, 2014 at 16:10
  • $\begingroup$ @whuber ok, I'll edit the question. By the way, I forgot to say thanks for your interest in this question. :) $\endgroup$ Oct 29, 2014 at 16:20
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    $\begingroup$ @AlecosPapadopoulos Why does strict stationarity not imply weak stationarity in general? A strictly stationary process for which $E[X_t^2]$ is finite is weakly stationary. It is only those processes for which $E[X_t^2]$ is not finite that fail to be weakly stationary (because we cannot define the autocovariance). $\endgroup$ Oct 30, 2014 at 2:06

1 Answer 1


Setting $i=1+h, j=n+h$, the definition of stationarity implies that the distribution of $(X_i,X_j)$ depends only on $j-i$ for all integers $i$ and $j$. The result follows immediately.

  • $\begingroup$ Thanks. Could you elaborate more on yourfirst sentence? I'm having a bit of difficulty. Thank you for your patience. ;) $\endgroup$ Oct 29, 2014 at 22:09
  • $\begingroup$ It is a consequence of the facts that (1) $n-1 = (n+h)-(1+h)$; (2) when $h$ can be any integer and $n$ can be any positive integer, then $(i,j)=(1+h, n+h)$ can be any pair of integers provided $j\ge i$, because given $(i,j)$ you can solve for $(h,n)=(i-1,j-i+1)$; and (3) since the distribution of $(X_i,X_j)$ determines the distribution of $(X_j,X_i)$, we may always choose $j\ge i$. $\endgroup$
    – whuber
    Oct 29, 2014 at 22:16
  • $\begingroup$ On your second point, and from what I understand, you prove that $(X_1,X_n)=^d(X_{1+h},X_{n+h})=^d(X_{i},X_{j})$? $\endgroup$ Oct 30, 2014 at 9:57
  • $\begingroup$ I don't really view this as a "proof" of anything: it is merely pointing out that what you want to conclude is just another way of interpreting the statement you started with. $\endgroup$
    – whuber
    Oct 30, 2014 at 13:41
  • $\begingroup$ But then may I conclude what I wrote in my comment above? I just want to be sure I understood your reasoning. $\endgroup$ Oct 31, 2014 at 0:02

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