When is the autocorrelation function of a stationary process strictly decreasing or nonincreasing? Can being Markovian make it true?

When is the autocorrelation function of a stationary process (strictly) increasing?


  • $\begingroup$ The answer is no--but the question becomes more interesting when you replace "decreasing" by "nonincreasing." $\endgroup$ – whuber Feb 27 '14 at 23:14
  • $\begingroup$ Thanks, @whuber, can you explain why no? What conditions can I add to make it yes? I will add "nonincreasing" to my questions. $\endgroup$ – Tim Feb 27 '14 at 23:17
  • $\begingroup$ It has to decrease otherwise it would have memory $\endgroup$ – Aksakal Feb 27 '14 at 23:18
  • $\begingroup$ @Aksakal: can you prove/explain it? $\endgroup$ – Tim Feb 27 '14 at 23:19
  • $\begingroup$ I'd have to think about the proof. intuitively if today is correlated with distant past it seems that should contradict the short memory requirement of Markov process. $\endgroup$ – Aksakal Feb 28 '14 at 0:19

The autocorrelation function $R_X(n)$ of a stationary process $\{X(t)\colon t \in \mathbb Z\}$, whether Markovian or not, is given by $$R_X(n) = E[X(m)X(m+n)]$$ where, because of stationarity, the choice of $m$ does not matter: $(X(m), X(m+n))$ has the same joint distribution as $(X(m^\prime), X(m^\prime + n))$ and so $E[X(m)X(m+n)]= E[X(m^\prime)X(m^\prime+n)]$. Hence, using an argument which is essentially a version of the argument that allows us to assert that the Pearson correlation coefficient $\rho \in [-1,1]$, we get that $$R_X(0) \geq |R_X(n)|.$$ It is also easy to verify that $R_X(n) = R_X(-n)$ and so, if $R_X(n)$ is an increasing function of $n$ for $n < 0$ (as $R_X(n)$ rises towards its peak at $0$, say), then $R_X(n)$ must be a decreasing function of $n$ for $n > 0$. In short, $R_X(n)$ cannot be a strictly increasing function of $n$ for all $n$. What goes up must come down.

The only other possibilities are

  1. $R_X(n)=R_X(0)$ for all $n$, that is, the autocorrelation function is a constant. An example of such a process is in @whuber's example in the comments on the main question: $X(n) = X$ for all $n$ (Aksakal's suggestion of $X$ being a constant is just a special (degenerate) case of this when the random variable $X$ equals a constant almost surely).

  2. $R_X(n) = (-1)^n R_X(0)$ for all $n$, that is, the autocorrelation function is periodic with period $2$ as in whuber's example in the comments where $X(n) = (-1)^nX$ with $X$ being a zero-mean random variable with symmetric distribution, that is, $X$ and $-X$ are identically distributed and so all the $X(n)$'s have the same distribution (important for stationarity).

  1. The autocorrelation function of the stationary process does not depend on the time $m$, i.e.

$$R_X(m,n)=E[X(m)X(m+n)] = R_X(n)$$

Then we can make a shift $m\rightarrow m-n$

$$R_X(n) = E[X(m-n)X(m-n+n)] = E[X(m-n)X(m)] = E[X(m)X(m-n)] = R_X(-n)$$

Thus we have the symmetry property

$$R_X(n) = R_X(-n)$$

  1. The proof of the non-increasing property is borrowed from Kun Il Park, Fundamentals of Probability and Stochastic Processes with Applications to Communications:

$$E[(X(m) - X(m+n))^2] = E[X^2(m) + X^2(m+n) - 2X(m)(m+n)] = 2 E[X^2(m)] + 2E[X(m)(m+n)] \geq 0$$

from which we get

$$E[X^2(m)] \geq E[X(m)(m+n)]$$


$$R(0) \geq R(n)$$


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