# Definition and proof of Strict Stationarity

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.

• What does the definition say when $h=t-1$ and $n=1+h$?
– whuber
Oct 29, 2014 at 15:10
• @whuber and what if $t$ is smaller than -1? $h$ could be negative, but $n$ cannot... Oct 29, 2014 at 15:38
• 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 $\{1,2,3,\ldots,n,\ldots\}$.
– whuber
Oct 29, 2014 at 16:10
• @whuber ok, I'll edit the question. By the way, I forgot to say thanks for your interest in this question. :) Oct 29, 2014 at 16:20
• @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). Oct 30, 2014 at 2:06

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.
• 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 $j\ge i$.
• On your second point, and from what I understand, you prove that $(X_1,X_n)=^d(X_{1+h},X_{n+h})=^d(X_{i},X_{j})$? Oct 30, 2014 at 9:57