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I'm studying thistleton and sadigov ts analysis course, and the text says that

for a strict stationary stochastic process:

(A) The joint distribution of $X(t1),X(t2)$ is the same as the joint distribution of $X(t1+tau), X(t2+tau)$.

(B) Which means that the joint distribution depends only on the lag spacing, so the autocovariance function $l(t1, t2) = l(t2-t1) = l(tau)$

I'm at loss here. Is it the same t1, t2 in (A) and (B)? is it the same tau? Why is $t2-t1=tau$ in (B)? Why is (A) talking specifically about joint distribution of 2 random variables in the process and not about 3/4/7 random variables - is this the minimum needed for proving (B)?

thanks

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  • $\begingroup$ Hi A) if it holds for 2 rv's, it holds for 2 or more so it was just a decision to use 2 rv's. the author could have used $n$ rv's. B) the covariance of 2 rv's is a function of their joint dist so, if their joint dist only depends on tau, then the covariance only depends on tau. $\endgroup$
    – mlofton
    Aug 10, 2019 at 8:37
  • $\begingroup$ @mlofton I don't see how strict stationarity of all pairs of RVs implies strict stationarity of all triples. Indeed, it seems straightforward to find counterexamples for discrete stochastic processes. $\endgroup$
    – whuber
    Aug 12, 2019 at 20:47
  • $\begingroup$ ihadanny: Please be aware that (A) is not the usual definition of strict stationarity. See Wikipedia at en.wikipedia.org/wiki/Stationary_process#Definition for the standard meaning. Indeed, a video for the course you reference doesn't use definition (A): it uses the Wikipedia definition. $\endgroup$
    – whuber
    Aug 12, 2019 at 20:50
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    $\begingroup$ @Whuber: I don't know how to prove it but I think you're right with your point that the statement about 2 does not necessarily imply that it's also true for more than 2. What I should have said is that the same assumption about two might as well have been made about n because it's an assumption that is intended to be true for any number of (consecutive ) n. Atleast I think that's the intention. Good catch. $\endgroup$
    – mlofton
    Aug 12, 2019 at 23:28
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    $\begingroup$ @Whuber: Thanks. I think we're on the same page. I was wrong about 2 implying 2 or more. My point is that the author probably meant $n$ and didn't realize that writing two does not imply $n$. The author and I made the same mistake !!!!! $\endgroup$
    – mlofton
    Aug 14, 2019 at 1:38

2 Answers 2

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A) One of the infinitely many requirements for a process to be called strictly stationary is that the joint distribution of $X(t_1)$ and $X(t_2)$ is the same as the joint distribution of $X(t_1+\tau)$ and $X(t_2+\tau)$, that is, two random variables $X(t_1)$ and $X(t_2)$ that are separated by a given amount of time ($t_2-t_1$ seconds here) have the same joint distribution as any other pair of random variables $X(t_1+\tau)$ and $X(t_2+\tau)$ separated by the same amount of time. Note that $X(t_1+\tau)$ and $X(t_2+\tau)$ are separated by $(t_2+\tau) - (t_1 + \tau) = t_2-t_1$ seconds, just as $X(t_1)$ and $X(t_2)$ are separated by $t_2-t_1$ seconds.

B) is very confusingly written. The autocovariance function of a random process or time series is a function of two variables -- the time instants of the two random variables whose covariance is being calculated. That is, $l(t_1, t_2)$ is defined as $\operatorname{cov}\big(X(t_1), X(t_2)\big)$. Now, the covariance of $X(t_1)$ and $X(t_2)$ depends on the joint distribution of $X(t_1)$ and $X(t_2)$ which A) tells us is the same as the joint distribution of $X(t_1+\tau)$ and $X(t_2+\tau)$. Thus, for a strictly stationary process, the autocovariance function $l(t_1, t_2) \stackrel{\Delta}{=} \operatorname{cov}\big(X(t_1), X(t_2)\big)$ has the same value at the point $(t_1+\tau, t_2+\tau)$ and the same value at the point $(t_1+\tau^\prime, t_2+\tau^\prime)$ etc. So, $l(t_1, t_2)$ cannot possibly depend on the individual values of $t_1$ and $t_2$, only on their difference $t_2-t_1$ which we will denote by $\lambda$. It is a poor choice of notation to use the same letter $l$ to denote the autocovariance function when expressed as function of just the one variable $\lambda$ instead of $t_1$ and $t_2$ but the usage is far too entrenched in the literature to get rid of it here. What is generally written is $$l(\lambda) = E\left[(X(t)-\mu_X)(X(t+\lambda)-\mu_X)\right]\tag{1}$$ where we are relying on the fact that $X(t)$ and $X(t+\lambda)$ have the same mean $\mu_X$ (a consequence of stationarity) and the implicit assumption is that the RHS of $(1)$ does not depend on what value of $t$ is used; all that matters is that $X(t+\lambda)$ and $X(t)$ are separated by $\lambda$ seconds and thus the RHS depends on $\lambda$ alone.

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  • $\begingroup$ thanks! what did it for me is the distinction between tau and lambda $\endgroup$
    – ihadanny
    Aug 15, 2019 at 4:37
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I'm sure someone with a lot more knowledge will chime in, but in the mean time, I might have a decent answer.

I think you are correct in noticing that in both (A) and (B) the author has chosen to use $\tau$ to represent ANY difference in time, perhaps equivalent to $\Delta t$. But the $\Delta$s in (A) and (B) are not linked in any way.

If both (A) and (B) started out with "For any $\tau = t_2 - t_1$..." it would be clearer.

Note that, in general, the autocovariance is defined as:

$$ \begin{align} a(t_1, t_2) & = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}X(t_1)X(t_2)dt_1 dt_2 \\ & = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}x^2P[x,t_1]P[x,t_2]dt_1 dt_2 \end{align} $$

If we apply (A), then

$$a(t, \tau) = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} x^2P[x, \tau]dt dt$$

Which is independent of $t$. Consequently, the autocovariance is a function only of $\tau$

Gosh I hope someone comes along and makes that more rigorous.

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  • $\begingroup$ what do you mean by $\Delta(s)$ are not linked? you mean that the taus are different? $\endgroup$
    – ihadanny
    Aug 11, 2019 at 10:57
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    $\begingroup$ @ihadanny -- Yes. I believe that $\tau$ is simply a stand-in for ANY time difference. $\endgroup$
    – abalter
    Aug 11, 2019 at 18:24

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