I was reading the following question: A stochastic process that is 1st and 2nd order (strictly) stationary, but not 3rd order stationary

The following matrix is given representing the correlation structure of a stochastic process $X[t], X[t+1], X[t+2]$:

$$\left[\begin{array}{cc} \sigma^2 & a & b \newline a & \sigma^2 & c \newline b & c& \sigma^2 \end{array}\right]$$

I noticed that the correlation structure matrix is not a Toeplitz matrix. However, this seems to result in a contradiction: if you take $t=k$ then you find that the correlation between $X(k+1)$ and $X(k+2)$ is $c$, whereas if you take $t=k+1$, then you find that the correlation between $X(k+1)$ and $X(k+2)$ is $a$. This seems to mean that the matrix is not a valid correlation structure unless it is a Toeplitz matrix.

Is my reasoning valid?


1 Answer 1


This is not a stationary process, so there is no requirement for $X(t)$ and $X(t+1)$ to have the same correlation as $X(t+1)$ and $X(t+2)$.

  • $\begingroup$ But isn’t that the very definition given - that it is a first order stationary process that is not also second order? $\endgroup$
    – Resquiens
    Feb 11, 2019 at 19:52
  • $\begingroup$ And if the above matrix describes the general correlations between the t, t+1’th and t+2th Random Variables for any t then surely it is contradictory as I have shown? $\endgroup$
    – Resquiens
    Feb 11, 2019 at 20:00
  • 1
    $\begingroup$ "First order stationary" just says the distributions of the individual $X(t)$ are all the same. For example, they might be normal with mean $0$ and variance $\sigma^2$. Robby didn't mean to have the covariance of $X(t)$ and $X(t+1)$ be $a$ for all $t$, just for some particular $t$. You might, for example, have this alternate between $a$ and $c$ depending on whether $t$ is odd or even. $\endgroup$ Feb 12, 2019 at 17:05
  • $\begingroup$ Ah ok. I didn't realise that the variable $t$ was meant to refer to only one time. In that case, it doesn't seem like a proof that first order stationary processes need not be secondary in Robby's question when he says that the matrix is not a Toeplitz matrix. It doesn't show how the first order statistics can be the same but the second order ones can differ... $\endgroup$
    – Resquiens
    Feb 13, 2019 at 2:23
  • $\begingroup$ I also see (I think) that you were referring to strict sense stationary in your answer (which would imply second order stationarity too)? $\endgroup$
    – Resquiens
    Feb 13, 2019 at 2:25

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