If a random process is first order stationary, its mean is constant. However, if a random process has a constant mean say $3$ and an autocorrelation equal to $9 + 15e^{|-\tau|}$. The process is clearly wide sense stationary. Is the process first order stationary?

  • $\begingroup$ What is your understanding of the term "first-order stationary"? Be as specific as you can. $\endgroup$ Commented Mar 28, 2017 at 23:07
  • $\begingroup$ @Dilip By first-order stationarity, I mean the first order density function is independent of absolute time. $\endgroup$
    – Ehsa
    Commented Mar 29, 2017 at 2:42

1 Answer 1


In the usual sense of the term, first-order stationarity means that the first-order distribution of all the random variables is the same: each $X_t$ has the same CDF, and so the same pdf (or pmf) too if the random variables are continuous (or discrete). If the random variables have a mean, then they all have the same mean. But, a sequence of independent identically distributed Cauchy random variables is a first-order stationary process (as well as a strictly stationary process for that matter) but it cannot be said to have constant mean because none of the variables have a mean at all.

Since your response to my query on the problem statement is that your definition of first-order stationarity is the standard one, then the answer to the question in the title of your question, viz.

If the mean of a random process is constant, does it imply that the process is first-order stationary?

is NO, and an example of a process that has constant mean but is not first-order stationary is a sequence of constant-mean random variables whose variances depend on time.

The answer to the question in the text, viz.

Is a WSS process necessarily first-order stationary?

ia also NO. Consider the random process $\{X(t)\colon X(t)= \cos (t + \Theta), -\infty < t < \infty\}$ where $\Theta$ takes on four equally likely values $0, \pi/2, \pi$ and $3\pi/2$ and so $X(t)$ takes on four equally likely values $\cos(t)$, $\cos(t+\frac{\pi}{2}) = -\sin(t)$, $\cos(t+\pi) = -\cos(t)$, and $\cos(t+\frac{3\pi}{2}) = \sin(t)$. It is easy to see that in general $X(t)$ and $X(s)$ have different distributions and so the process is not first-order stationary. On the other hand, $E[X(t)] = 0$ for all $t$ while \begin{align} E[X(t)X(s)]&= \left.\left.\frac 14\right[2\cos(t)\cos(s) + 2\cos(t+\pi/2)\cos(s+\pi/2)\right]\\ &= \left.\left.\frac 12\right[\cos(t)\cos(s) + \sin(t)\sin(s)\right]\\ &= \frac 12 \cos(t-s) \end{align} and so the process is wide sense stationary, but its first-order distribution does depend the choice of $t$ and so the process is not first-order stationary.

Note: the above proof uses a very commonly accepted definition of wide sense stationarity as found in numerous books, online resources, Wikipedia etc, viz. a random process $\{X(t)\}$ is called a wide sense stationary (WSS) process if

  1. $E[X(t)]$ is a constant, and
  2. $R_{X}(t,s) = E[X(t)X(s)]$ depends only on the difference $t-s$ of the two arguments and not on the individual arguments $t$ and $s$.
  • $\begingroup$ Ok I've checked all of the (engineering) books in my library on stochastic processes and it seems that the definition of WSS random process agrees with your statement; a process with a constant mean and lag dependent ACF... I've deleted my answer. $\endgroup$
    – Fat32
    Commented Apr 27, 2017 at 22:38
  • $\begingroup$ @Dilip Sarwate "but its first-order distribution does depend the choice of t and so the process is not first-order stationary" Is that a typo? Doesn't it depend only on the difference t−s ? $\endgroup$
    – VMMF
    Commented Feb 7, 2018 at 21:26
  • $\begingroup$ @VMMF There is no typo. The first-order distribution tells us the values that the random variable $X(t)$ takes on, and these are $\pm \cos t$ and $\pm \sin t$, very obviously different for different choices of $t$. On the other hand, the autocorrelation function $R_X(t-s) = E[X(t)X(s)]$ is a function of $t-s$ which is why the process that I give as an example is wide-sense stationary but not first-order stationary. $\endgroup$ Commented Feb 7, 2018 at 22:04
  • $\begingroup$ @Dilip Sarwate So as the phase takes 4 different distribution values the first-order density function is not independent of time origin, hence the process is not first-order stationary even though the conditions for wide sense stationarity are present? $\endgroup$
    – VMMF
    Commented Feb 7, 2018 at 22:55
  • $\begingroup$ @VMMF You got it! $\endgroup$ Commented Feb 7, 2018 at 23:04

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