The question is related to this one.

In this question @mpiktas gives an answer on why checking correlation is not enough but the answer doesn't seem completely correct to me for the following reason:

If 2 time-series are cointegrated, i.e. there is a linear relation between them $$y_t = a + b x_t + \varepsilon_t$$ with stationary $\varepsilon_t$ it implies a linear relation between their differences $$\Delta y_t = b \Delta x_t + \Delta \varepsilon_t.$$ So if series are cointegrated, their differences should be correlated. And this means that if we don't see significant correlation between differences in our data, then there is no cointegration either. Is this correct or am I missing something?

The question arises because I look for relations between hundreds of time-series (mainly non-stationary) and the way I do this is by considering correlations between their differenced counterparties. And I assume that if I don't see correlation between differenced series there is no cointegration either - it suffices to check only correlations.


3 Answers 3


The existence or not of a linear relationship does not necessarily go hand-in-hand with co-integration. Variables co-integrated in levels won't necessarily exhibit correlation in first-differences.

Assume that the following relation holds: $$y_t = a + b x_t + \varepsilon_t, \; \varepsilon_t=\text {i.i.d} $$

i.e. the variables are co-integrated. Then the relation

$$\Delta y_t = b \Delta x_t + \Delta \varepsilon_t$$
also holds. Calculating the sample correlation of first-differences we will estimate the Covariance as

$$ \begin{align}\operatorname{\hat Cov}(\Delta y_t,\Delta x_t)=& \frac 1{T-1} \sum_{t=2}^{T}\left(b \Delta x_t + \Delta \varepsilon_t\right)\Delta x_t \\-&\left(\frac 1{T-1} \sum_{t=2}^{T}\left(b \Delta x_t + \Delta \varepsilon_t\right)\right)\left(\frac 1{T-1} \sum_{t=2}^{T}\Delta x_t\right)\end{align} $$

$$ \begin{align}=b\frac 1{T-1}& \sum_{t=2}^{T}\left(\Delta x_t \right)^2 + \frac 1{T-1} \sum_{t=2}^{T}\left(\Delta x_t \Delta \varepsilon_t\right) \\ -& b\left(\frac 1{T-1}\sum_{t=2}^{T} \Delta x_t\right)^2 -\left(\frac 1{T-1} \sum_{t=2}^{T}\Delta \varepsilon_t\right)\left(\frac 1{T-1} \sum_{t=2}^{T}\Delta x_t\right) \end{align}$$

To the degree that $x_t$ and $\varepsilon_t$ are independent, the terms involving the error will tend to vanish and so

$$ \operatorname{\hat Cov}(\Delta y_t,\Delta x_t)\rightarrow bs^2_{\Delta x_t} $$

where $s^2$ is the sample variance (irrespective of whether the variance of $x_t$, or $\Delta x_t$ is constant or not).

The sample variance of $\Delta y_t$ will be

$$s^2_{\Delta s_t} \approx b^2s^2_{\Delta x_t} + s^2_{\Delta \varepsilon_t}$$

again, irrespective of whether these sample moments estimate anything meaningfull.

So $$\operatorname {\hat Corr}(\Delta y_t,\Delta x_t) \approx \frac {bs^2_{\Delta x_t}}{\sqrt {\left(b^2s^2_{\Delta x_t} + s^2_{\Delta \varepsilon_t}\right)}\sqrt {s^2_{\Delta x_t} }} = \frac {bs_{\Delta x_t}}{\sqrt {\left(b^2s^2_{\Delta x_t} + s^2_{\Delta \varepsilon_t}\right)}}$$

So the magnitude of the empirically estimated correlation of first differences, will depend on the magnitude of the variance of the error term (which moreover enters the expression doubled since we consider first differences). If this (constant) variance is large compared to the variance of $x_t$, then the estimated correlation of first-differences may be small to non-existent, even though the variables are co-integrated in levels.

  • $\begingroup$ Thanks a lot to Alecos, your analysis is very impressive and quite insightful! I have couple comments though, but they are long so I post them as a separate post below. $\endgroup$
    – Kochede
    Commented Oct 30, 2013 at 9:52

Since correlation is a measure of the degree of linear dependence, first differences should tease this out. Now, I am assuming you check cointegration across multiple lags, not just contemporaneous values, since there could be something like $y_t = a + b x_{t-1} + \varepsilon_t$ going on, which may complicate matters. Alecos' observation that there may be not detectible cointegration is also important.


So if my understanding of @Alecos very insightful analysis is correct, he has 2 points: even if returns are linearly related $\Delta y_t = b\cdot\Delta x_t + \Delta\varepsilon_t$ then

  1. true correlation between $\Delta y_t$ and $\Delta x_t$ can be anything between 0 and 1 depending on noise/signal ratio $var(\Delta \varepsilon_t)/ var(\Delta x_t)$

  2. because we estimate above true correlation from a finite sample, our estimate can be something different from the true one.

Now, for point 1 I can object that if noise/signal ratio is big then both correlation and original cointegration will be "weak" (not sure if there is a measure of cointegration strength - probably ADF-test p-value?).

So if we know that true correlation between differenced series is ~0 due to high $var(\Delta \varepsilon)$, we probably still can conclude that cointegration between original series is very weak due to same high $var(\varepsilon)$.

Now, the second point probably makes this conclusion less certain - if finite sample estimate of correlation is ~0 this doesn't mean true one is ~0 and thus cointegration may be in place. So the question is how far can sample correlation be from the true one given a sample size :)

  • $\begingroup$ You raise an interesting question - I don't remember anything related to "co-integration strength", perhaps because it is a "yes/no" property: "the error term is/is not stationary". It could be an fruitful area of research. In practice if the relative magnitude of the error is very high, then while co-integration will exist as defined, still inference on $y$ that will depend on $x$ will be most likely poor, exactly because most of the variability in $y$ is generated by the error term. So in a sense, the existence of co-integration won't be very useful. (CONT'D) $\endgroup$ Commented Oct 30, 2013 at 10:25
  • $\begingroup$ (CONT'D). The correlation coefficient is a complicated and sensitive statistic, see for example en.wikipedia.org/wiki/… $\endgroup$ Commented Oct 30, 2013 at 10:41

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