# autocovariance for a strict stationary stochastic process

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

• 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. Aug 10, 2019 at 8:37
• @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.
– whuber
Aug 12, 2019 at 20:47
• 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.
– whuber
Aug 12, 2019 at 20:50
• @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. Aug 12, 2019 at 23:28
• @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 !!!!! Aug 14, 2019 at 1:38

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.

• thanks! what did it for me is the distinction between tau and lambda Aug 15, 2019 at 4:37

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.

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