Suppose I have a time series process $\{X_t\}$ that is strictly stationary in the sense that the joint distribution of $[X_{t_1},...,X_{t_k}]$ and $[X_{t_1+a},...,X_{t_k+a}]$ are the same for any set of integers $t_1,...,t_k$ and any integer $a$. If in addition, I know that this process have mutually independent observations. Does this imply that $\{X_t\}$ is an i.i.d. process? My guess is that it is. Because strict stationarity implies that every term in this time series have the same distribution as $X_1$, and thus when independence also hold, it is an i.i.d. process. Does this look correct?

  • $\begingroup$ "Because strict stationarity implies that every term in this time series have the same distribution as $X_1$" This is not what strict stationarity means. This contradicts the definition that you gave, which is about the joint distribution, and not the marginals. $\endgroup$
    – mhdadk
    Apr 23 at 23:05
  • 2
    $\begingroup$ @mhdadk Yeah, it's about joint distribution, but I meant it IMPLIES marginals are have identical distribution. Just take $t_1,...,t_k$ as 1, the definition says $X_1$ has the same distribution as $X_{1+a}$ for any $a$. What's wrong with this? $\endgroup$ Apr 23 at 23:16

1 Answer 1


Yes, that is correct. Strict stationarity implies a common marginal distribution for the variables in the series, which is the ID part in IID. If you combine this with an assumption of independence you then get IID.

  • $\begingroup$ Thanks a lot! This is very helpful. $\endgroup$ Apr 24 at 4:02
  • $\begingroup$ But isn't the assumption of "mutual independence" in the question weaker than independence? That distinction appears to be the crux of the matter. $\endgroup$
    – whuber
    May 6 at 15:38
  • $\begingroup$ @whuber: Isn't mutual independence the stronger one (i.e., independent conditional on any of the observations in the series)? I was under the impression that pairwise, triplewise, etc., were the weaker and mutual independence is full independence. $\endgroup$
    – Ben
    Oct 3 at 5:02
  • $\begingroup$ You could be right: I had understood "mutual" as meaning "pairwise," for otherwise why modify the word "independence" at all? $\endgroup$
    – whuber
    Oct 3 at 13:10
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    $\begingroup$ @whuber: For now I'll leave the answer unchanged, since I think that mutual independence is the strong form. Like you, I always take the unqualified term "independence" as meaning the strong (mutual) form, so don't need the qualifier, but perhaps OP wanted to specify to be sure. $\endgroup$
    – Ben
    Oct 3 at 15:02

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