Autocorrelations of series ‘c(0, 1, 2, 3, 4, 5)’, by lag

     0      1      2      3      4      5 
 1.000  0.500  0.057 -0.271 -0.429 -0.357 

Why does the ACF output becomes negative as lag increases? My understanding is that no matter what the lag is, the series is in general increasing. Therefore the auto-correlation should be positive. For example, at lag 2, we are calculating the correlation function of the two series [0,1,2,3] and [2,3,4,5], where the positive correlation still holds. Where do I get this wrong?


Here is my intuitive understanding of the acf of an monotonically increasing sequence:

ACF of a sequence is a function $\gamma(k)$ of the lag, k. By definition, this function indeed measures the correlation between $y_t$ and $y_{t-k}$. The misunderstanding comes from the understanding of correlation. A monotonically increasing sequence is not stationary, so the mean is not stable. In another word, the sequence does not exhibit mean reverting behavior. This distorts my usual understanding of correlation (when we think about the mean level at 0). Since the mean increases over time, those observations come earlier are more likely to be lower than the sample mean, thus inducing a negative sample acf when lag is larger.


2 Answers 2


Let $x = (x_1, x_2, \ldots, x_n)$ be the series. Set

$$y_t = x_t - \bar{x}.$$

These are the residuals with respect to the estimated mean $\bar{x} = \frac{1}{n}\sum_{t=1}^n x_t$ of the series.

For $k=0, 1, 2, \ldots, n-1$ the acf function is computing

$$\text{acf}(x)_k = \frac{\sum_{t=1}^{n-k} y_t y_{t+k}}{\sum_{t=1}^n y_t^2}.$$

Notice that as the lag $k$ grows, there are fewer and fewer terms in the numerator as well as a shift of the indexes in the product. The reduction in number of terms in the numerator essentially forces a decrease in the value as $k$ increases. Most time series analyses consider only lags $k$ much smaller than $n$ for which this effect is negligible.

In your example where $x = (0, 1, 2, 3, 4, 5)$, $y = (-5/2, -3/5, -1/2, 1/2, 3/2, 5/2)$ initially has negative values and then moves into positive territory. For lags $k \ge 3$, the products $y_ty_{t+k}$ are pairing the early negative values with the later positive values, producing negative numbers.

Edit: Intuitive Explanation

Intuitively, $\text{acf}(x)_k$ is supposed to be telling us the correlation between a series and its lag-$k$ version. The motivation for the question is that a series like $(0, 1, \ldots, n-1)$ is perfectly correlated with all its lags for $k=0$ right through $k=n-2$. How, then, can the ACF plot produce near zero and even negative values?

There are two factors in play here. They can be seen by comparing the ACF formula to that of the usual correlation coefficient. For two series $(u_t)$ and $(w_t)$ of the same length $n-k$, let $\upsilon_t = u_t - \bar{u}$ and $\omega_t = w_t - \bar{w}$ be their residuals. (In the ensuing discussion, $(u_t)$ will be the prefix $(x_1, x_2, \ldots, x_{n-k}$ and $(w_t)$ will be the suffix $(x_{k+1}, x_{k+2}, \ldots, x_n)$.) By definition, their correlation coefficient is the average standardized residual,

$$\rho(u, w) = \frac{\sum_{t=1}^{n-k} \upsilon_t \omega_t}{\sqrt{\sum_{t=1}^{n-k} \upsilon_t^2 \sum_{t=1}^{n-k} \omega_t^2}}.$$

(The constants $\frac{1}{n-k}$ that usually appear in formulas for averages cancel in this ratio, so I have omitted them.)

When we are dealing with a single series $(x_t)$ of length $n$ and its (short) lags $k$, both $\upsilon_t$ and $\omega_t$ are essentially the same, apart from the shift of $k$ in their indexes: the first consists of the $(y_t)$ for $t$ from $1$ through $n-k$ (the high-$t$ end has been trimmed off) while the second consists of the same $(y_t)$ for $t$ from $k$ through $n$ (the low-$t$ end has been removed). If we ignore these slight differences, the denominator of $\rho(u, w)$ simplifies to

$$\sqrt{\sum_{t=1}^{n-k} \upsilon_t^2 \sum_{t=1}^{n-k} \omega_t^2} = \sqrt{\sum_{t=1}^{n-k} y_t^2 \sum_{t=1}^{n-k} y_{t+k}^2} \approx \sqrt{\sum_{t=1}^{n} y_t^2 \sum_{t=1}^n y_{t}^2} = \sqrt{\left(\sum_{t=1}^{n} y_t^2\right)^2 } = \sum_{t=1}^{n} y_t^2.$$

In making this approximation I have inserted the first $k$ terms $y_1^2 + \cdots + y_k^2$ into the sum for the suffix ($\omega_t$) and the last $k$ terms $y_{n-k+1}^2 + \cdots + y_{n}^2$ into the sum for the prefix ($\upsilon_t$). Because these are both sums of squares, they cannot decrease the denominator, and usually increase it a little bit. Accordingly, we see that using $\sum_{t=1}^n y_t^2$ in the denominator decreases the apparent correlation $\rho(u, w)$. The greater the lag $k$, the more the denominator will tend to increase, so this factor tends to reduce the high-lag values of the ACF no matter what.

The second factor has to do with the difference between the mean of the entire series $\bar{x}$ and the means of the prefix $\bar{\upsilon} = \frac{1}{n-k}\sum_{t=1}^{n-k} y_t$ and suffix $\bar{\omega} = \frac{1}{n-k}\sum_{t=k+1}^n y_t$. The ACF formula uses the former whereas the correlation coefficient formula uses the latter. We can work out the change in the numerator by comparing the ACF and correlation coefficient formulas, working algebraically to make the ACF numerator look like the $\rho$ numerator:

$$\eqalign{ \sum_{t=1}^{n-k} y_t y_{t+k} &= &\sum_{t=1}^{n-k} (x_t-\bar{x})(x_{t+k}-\bar{x}) \\ &= &\sum_{t=1}^{n-k} (x_t-\bar u + \bar u - \bar{x})(x_{t+k}-\bar w + \bar w - \bar{x}) \\ &= &\sum_{t=1}^{n-k} \left((x_t-\bar u)(x_{t+k}-\bar w) + (\bar u - \bar{x})(\bar w - \bar{x})\right) \\ &= &\left(\sum_{t=1}^{n-k} \upsilon_t \omega_t\right) + (n-k)(\bar u - \bar{x})(\bar w - \bar{x}). }$$

(The cross terms disappeared after the second line for the usual reason: they sum to zero.)

Comparing to the formula for $\rho$, we see that the discrepancy in numerators depends on the lag (in terms of $n-k$) and the products of the changes in the means, $\bar u - \bar{x}$ and $\bar w - \bar{x}$. For a stationary series and large $k$ those changes ought to be small; for small $k$ we hope they will be small but perhaps not. In the example, for instance, at lag $k=1$ the mean after dropping off the last term decreases by $1/2$ and the mean after dropping off the first term similarly increases by $1/2$. The product

$$(n-k)(\bar u - \bar{x})(\bar w - \bar{x}) = (6-1)(-1/2)(1/2) = -5/4$$

decreases the numerator in the ACF compared to the numerator in $\rho$.

The net effect of these two factors in the example is that both conspire to decrease the apparent correlation: the denominator goes up, because it includes a few more positive terms overall, and the numerator goes down, because one end of the series tends to be less than the average and the other end tends to be greater than the average. (That's more or less what a "long term trend" means, suggesting there is some evidence of non stationarity in this series.)

To illustrate the formula for the ACF, here is direct (but less efficient) R code to compute acf:

acf.0 <- function(x) {
  n <- length(x)
  y <- x - mean(x)
  sapply(1:n - 1, function(k) sum( y[1:(n-k)] * y[1:(n-k) + k] )) / sum(y * y)

As a test, compare the two results:

> sum((acf.0(0:5) - acf(0:5, plot=FALSE)$acf)^2)
> 6.162976e-33

The answers agree to within double precision floating point roundoff error.

  • $\begingroup$ Thanks for your answer. Intuitively, what does acf measure exactly? $\endgroup$
    – zsljulius
    Jan 10, 2014 at 0:15
  • 1
    $\begingroup$ I inserted a new section to answer that question, zsljulius. It concludes that the ACF approximates the correlation coefficient between the beginning and end of the series, as lagged by $k$, but that it tends to be a little smaller in absolute value and in your example grows more negative, too, because the series starts off with less-than-average values and ends up with greater-than-average values. $\endgroup$
    – whuber
    Jan 10, 2014 at 22:14
  • $\begingroup$ Regarding the second factor: Is it because we assume a stationary series that we use the mean of the entire series? Does it become incorrect if we use their corresponding means (and calculate correlation)? $\endgroup$
    – ayhan
    Jan 3, 2019 at 11:28
  • 1
    $\begingroup$ @whuber I was referring to the means of the series you called prefix and suffix. If I have 50 observations and want to calculate the autocorrelation at lag 5 that would give me the first 45 observations as the prefix and the last 45 observations as the suffix. They would have different means. In the Wikipedia autocorrelation article the difference between Eq. 2 and 4 is basically what I was asking. Would it be correct to calculate autocorrelation like in Eq 2? $\endgroup$
    – ayhan
    Jan 3, 2019 at 15:43
  • 1
    $\begingroup$ @ayhan Yes, that makes sense and is a method frequently used when it is known or suspected that the process will not be quite stationary but its deviations from stationarity are locally small and when the analysis (such as forecasting, interpolation, or whatever) also is going to be a local one. $\endgroup$
    – whuber
    Jan 3, 2019 at 16:17

I am gonna show it for the case when the time series is a special form of $Y_t=0,1,2, ...,n.$ First assume $n$ is odd. So the $\bar{Y}=n/2$ that will happen at the time $(n+1)/2$. The formula for the sample acf at lag $k$ is $r_k=\dfrac{\sum_{t=k+1}^n(Y_{t}-\bar{Y})(Y_{t-k}-\bar{Y})}{\sum_{t=1}^n(Y_{t}-\bar{Y})^2}$.
Now for $k\geq \bar{Y}$, the time index for $(Y_{t}-\bar{Y})$ in the numerator of $r_k$ goes from $\{\dfrac{n+1}{2}+1,...,n\}.$ Therefore, $(Y_{t}-\bar{Y})\geq 0$ for $t\in \{\dfrac{n+1}{2}+1,...,n\}$. On the other hand the time index for the 2nd term in the numerator of $r_k$ i.e. $(Y_{t-k}-\bar{Y})$ goes from $\{1,2,...,n-k\}$. Note that since we assumed $k\geq \bar{Y}$ then $k\geq (n+1)/2$. So $n-k\leq (n-1)/2\leq (n+1)/2.$ Therefore, $(n-k)\leq \bar{Y}$. So $(Y_{t-k}-\bar{Y})\leq 0$ for $t \in \{1,2,...,n-k\}$. Hence the product of $(Y_{t}-\bar{Y})(Y_{t-k}-\bar{Y})$ in the numerator of $r_k$ would be all negative for $t\in \{k+1,...,n\}$. And the summation of all negative numbers is again negative. So $r_k\leq 0$. (We don't care about the denominator coz its always positive). Same argument can be applied for the case when $n$ is even.

In your example, we have: $Y=0,1,2...,5.$ So $\bar{y}=2.5.$ Based on above argument when $k\geq 2.5$ , we will have $r_k\leq 0$. Here is another example:

> y=0:10
> mean(y)
[1] 5
> acf(y,plot=FALSE)

Autocorrelations of series ‘y’, by lag

     0      1      2      3      4      5      6      7      8      9     10 
 1.000  0.727  0.464  0.218  0.000 -0.182 -0.318 -0.400 -0.418 -0.364 -0.227 

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