3.3 Properties of the autocorrelation function

Suppose \(\{X_t\}\) is stationary with mean \(\mu \), variance \(\sigma ^2\), autocovariance \(\gamma (k)\) and ac.f. \(\rho (k)\). Then \(\rho (k) = \gamma (k)/\gamma (0) = \gamma (k)/\sigma ^2\), and the following properties hold.

Theorem 3.3.1.

(i): \(\rho (0) = 1\).
(ii): The ac.f. is an even function: \(\rho (-k) = \rho (k)\), and likewise \(\gamma (-k) = \gamma (k)\).
(iii): \(|\rho (k)| \le 1\) for every \(k\).
(iv): \(\gamma \) is non-negative definite: for any times \(t_1, \dots , t_n\) and real numbers \(a_1, \dots , a_n\), \[ \sum _{i=1}^n\sum _{j=1}^n a_ia_j\gamma (t_i - t_j) \ge 0 . \] Equivalently, every matrix \(\boldsymbol \Gamma _n\) with \((i, j)\) entry \(\gamma (i - j)\) is non-negative definite.

Proof. (i) \(\rho (0) = \gamma (0)/\gamma (0) = 1\).

(ii) By stationarity, \(\gamma (-k) = \Cov (X_t, X_{t-k}) = \Cov (X_{t-k}, X_t)\), and shifting time by \(k\) gives \(\Cov (X_t, X_{t+k}) = \gamma (k)\).

(iii) For either sign, \(0 \le \Var (X_t \pm X_{t+k}) = \sigma ^2 + \sigma ^2 \pm 2\gamma (k) = 2\sigma ^2(1 \pm \rho (k))\), so \(1 + \rho (k) \ge 0\) and \(1 - \rho (k) \ge 0\).

(iv) The left-hand side is \(\Var \bigl (\sum _ia_iX_{t_i}\bigr )\), since \(\Var (\sum _ia_iX_{t_i}) = \sum _i\sum _ja_ia_j\Cov (X_{t_i}, X_{t_j}) = \sum _i\sum _ja_ia_j\gamma (t_j - t_i)\), and \(\gamma (t_j - t_i) = \gamma (t_i - t_j)\) by (ii). A variance is never negative. □

Property (iv) is a real restriction. Not every even function with \(\rho (0) = 1\) and \(|\rho (k)| \le 1\) is the ac.f. of some stationary process, and (iv) is the test. Its most usable form is that all the matrices \(\mathbf {R}_n = \boldsymbol \Gamma _n/\sigma ^2\), with entries \(\rho (i - j)\), must have non-negative determinants.

Note also that the ac.f. does not determine the process. Different processes can have the same ac.f.; an example appears in Section 7, on invertibility. And a process with a trend or with a changing variance has no ac.f. in this sense at all, which is why trend and seasonality are removed before the correlogram is used to choose a model.

Example 3.3.2. Can \(\rho (0) = 1\), \(\rho (\pm 1) = 0.8\) and \(\rho (k) = 0\) for \(|k| \ge 2\) be the autocorrelation function of a stationary process?

Solution. The function is even, with \(\rho (0) = 1\) and \(|\rho (k)| \le 1\), so (i)–(iii) of Theorem 3.3.1 hold. Check (iv) with \(n = 3\): \begin {align*} \det \mathbf {R}_3 &= \det \begin {pmatrix} 1 & 0.8 & 0 \\ 0.8 & 1 & 0.8 \\ 0 & 0.8 & 1 \end {pmatrix} \\ &= 1(1 - 0.64) - 0.8(0.8 - 0) = 0.36 - 0.64 = -0.28 < 0 . \end {align*}

So \(\mathbf {R}_3\) has a negative eigenvalue, and there are weights \(a_1, a_2, a_3\) for which \(\Var (a_1X_1 + a_2X_2 + a_3X_3)\) would be negative. For instance \(\mathbf {a} = (1, -1, 1)\) gives \(\mathbf {a}^\top \mathbf {R}_3\mathbf {a} = 3 - 4(0.8) = -0.2\). The function is not an ac.f. Intuitively, if \(X_1\) and \(X_2\) are strongly correlated, and \(X_2\) and \(X_3\) are too, then \(X_1\) and \(X_3\) cannot be uncorrelated. (Section 6 shows that a function of this shape is an ac.f. exactly when \(|\rho (1)| \le \frac 12\).)

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