Cross-Correlation

The cross-correlation function (ccf) $\rho_{xy}(h)$ of two time series $x_t$ and $y_t$ is the correlation between $x_{t+h}$ and $y_t$ , one series set against the other shifted by $h$ periods. Read across the lags, it shows whether one series leads the other, which makes it a forecasting tool: a variable that moves ahead of the one to be forecast is a ready-made predictor.

$\gamma_{xy}(h) = \mathrm{Cov}(x_{t+h},\, y_t) = E\big[(x_{t+h} - \mu_x)(y_t - \mu_y)\big]$

$\rho_{xy}(h) = \frac{\gamma_{xy}(h)}{\sqrt{\gamma_x(0)\,\gamma_y(0)}}$

$\hat{\gamma}_{xy}(h) = \frac{1}{n}\sum_{t=1}^{n-h}(x_{t+h} - \bar{x})(y_t - \bar{y})$

$\hat{\rho}_{xy}(h) = \frac{\hat{\gamma}_{xy}(h)}{\sqrt{\hat{\gamma}_x(0)\,\hat{\gamma}_y(0)}}$

$\gamma_{xy}(h)$ is the cross-covariance function (ccvf), the covariance of the pairs $(x_{t+h}, y_t)$ . It depends on the lag $h$ alone when the two series are jointly stationary: each is stationary, and the cross-covariance does not change with $t$ . $\gamma_x(0)$ and $\gamma_y(0)$ are the two variances, so $-1 \le \rho_{xy}(h) \le 1$ . With $y = x$ the ccf is the ACF.
The sample ccf divides by the lag-0 autocovariances. The lag- $h$ cross-covariance is scaled by $\hat{\gamma}_x(0)$ and $\hat{\gamma}_y(0)$ , each series' sum of squared deviations divided by $n$ . The autocovariances at lag $h$ do not enter. As in the sample ACF, the sum has $n - h$ terms but is divided by $n$ . Negative lags come from $\hat{\gamma}_{xy}(-h) = \hat{\gamma}_{yx}(h)$ .
Not symmetric. Unlike the ACF, $\rho_{xy}(h) \ne \rho_{xy}(-h)$ in general. Instead $\rho_{xy}(h) = \rho_{yx}(-h)$ , so the plot runs over negative and positive lags, and swapping the two series mirrors it.
Reading lead and lag. In R, ccf(x, y) at lag $k$ estimates the correlation between $x_{t+k}$ and $y_t$ (type = "covariance" gives the ccvf). A peak at a negative lag $h = -\ell$ means $x_{t-\ell}$ moves with $y_t$ , so $x$ leads $y$ by $\ell$ periods. A peak at a positive lag means $x$ lags $y$ . R returns and plots the lag in units of time, not observations, so on a quarterly ts object a one-quarter lead peaks at $-0.25$ . A negative peak means the two move in opposite directions. Minitab shifts $y$ instead, so its plot reads the other way round.
Why the peak marks the lead. If $y_t = A x_{t-\ell} + w_t$ , with noise $w_t$ uncorrelated with $x_t$ , then $\gamma_{yx}(h) = A\,\gamma_x(h - \ell)$ . The cross-covariance is the input's own autocovariance moved to lag $\ell$ , and it is largest in absolute value there.
Significance bounds. If $x_t$ and $y_t$ are independent and at least one of them is iid white noise (independent, not merely uncorrelated), $\hat{\rho}_{xy}(h)$ is approximately normal with mean $0$ and standard deviation $1/\sqrt{n}$ , so values outside $\pm 2/\sqrt{n}$ point to a real relationship. When both series carry their own structure (a trend or a seasonal cycle), independent series can show sizeable cross-correlations. Remove the trend, or prewhiten one series (strip its own structure so that what is left is white noise), before reading the plot.
Use in forecasting. A related variable that leads the one to be forecast by one or more periods is an efficient basis for a forecast. The closer the relationship and the longer the lead, the better. Cowpertwait's example is Australian building activity, which tends to lag building approvals by about a quarter. The lags the ccf picks out are candidate predictors in a lagged regression of $y_t$ on past values of $x_t$ .

Do New Policies Lead Claims? Over six quarters an insurer wrote $x_t = 3, 6, 4, 7, 5, 5$ (hundreds of new policies) and received $y_t = 5, 3, 6, 4, 6, 6$ (tens of claims). Compute the sample cross-correlations at lags $-1$ , $0$ and $1$ , and say which series leads.
Answer Both means are $5$ , so the deviations are $(-2, 1, -1, 2, 0, 0)$ for $x$ and $(0, -2, 1, -1, 1, 1)$ for $y$ . Their sums of squares are $10$ and $8$ . The $1/n$ factors cancel, so each $\hat{\rho}_{xy}(h)$ is a sum of deviation products divided by $\sqrt{10 \times 8} = 8.944$ .
 $\begin{align*} \hat{\rho}_{xy}(-1) &= \frac{(-2)(-2) + (1)(1) + (-1)(-1) + (2)(1) + (0)(1)}{8.944} \\ &= \frac{8}{8.944} \\ &= 0.894 \end{align*}$ 
 $\begin{align*} \hat{\rho}_{xy}(0) &= \frac{(-2)(0) + (1)(-2) + (-1)(1) + (2)(-1) + (0)(1) + (0)(1)}{8.944} \\ &= \frac{-5}{8.944} \\ &= -0.559 \end{align*}$ 
 $\begin{align*} \hat{\rho}_{xy}(1) &= \frac{(1)(0) + (-1)(-2) + (2)(1) + (0)(-1) + (0)(1)}{8.944} \\ &= \frac{4}{8.944} \\ &= 0.447 \end{align*}$ 
Lag $-1$ pairs $x_{t-1}$ with $y_t$ , and it is the peak. New policies lead claims by one quarter. The lag-0 correlation is negative, so correlating the two columns side by side would have missed the relationship.

Is the Lead Real? Over $n = 100$ months, $x_t$ is the monthly change in a construction-cost index, already prewhitened to iid white noise, and $y_t$ is the monthly change in average property claim severity. The sample variances are $\hat{\gamma}_x(0) = 0.36$ and $\hat{\gamma}_y(0) = 2.25$ , and the sample cross-covariances are:

$h$ $-3$ $-2$ $-1$ $0$ $1$ $2$

$\hat{\gamma}_{xy}(h)$ $0.063$ $0.405$ $0.108$ $0.045$ $-0.081$ $0.027$

Which cross-correlations are significant, and what do they say?
Answer
 $\begin{align*} \sqrt{\hat{\gamma}_x(0)\,\hat{\gamma}_y(0)} &= \sqrt{0.36 \times 2.25} \\ &= 0.9 \end{align*}$ 
Dividing each cross-covariance by $0.9$ gives $\hat{\rho}_{xy}(h) = 0.07,\ 0.45,\ 0.12,\ 0.05,\ -0.09,\ 0.03$ for $h = -3, \ldots, 2$ .

$x_t$ is iid white noise, so the bounds apply:
 $\begin{align*} \pm\frac{2}{\sqrt{n}} &= \pm\frac{2}{\sqrt{100}} \\ &= \pm 0.2 \end{align*}$ 
Only lag $-2$ is outside them. Cost changes lead severity changes by two months, and the positive sign says they move in the same direction. That makes $x_{t-2}$ a candidate predictor of $y_t$ .