Spurious Regression Between Independent Random Walks

Why Independent Series Can Look Related

A regression can look persuasive even when its two variables have no economic or statistical relationship. This is especially striking when both variables are persistent: each series wanders over time, and a level regression can mistake their separate stochastic trends for a shared one. The resulting fit may have a high \(R^2\), and the usual regression t-statistic may appear highly significant, even though the series were generated independently.

The issue is not simply that random walks have large variances. The ordinary regression calculations are dominated by the low-frequency movement in the levels, while the usual t-test relies on assumptions appropriate to stationary regression errors. When the fitted residuals retain stochastic-trend behavior, those assumptions fail.

Model and Regression Quantities

Consider two series with no drift:

Assume the initial values are fixed, the innovations have mean zero and finite positive variances \(\sigma_x^2\) and \(\sigma_y^2\), and the innovation sequences are independent of one another and serially independent over time. For the standard Brownian-limit expressions below, assume the usual functional central limit theorem conditions, such as iid innovations with finite second moments. In particular, the two random walks are independent and neither has a deterministic trend.

Fit an OLS regression with an intercept, \(y_t=a+b x_t+u_t\). Define the centered sums \(S_{xx}=\sum_{t=1}^n(x_t-\bar{x})^2\), \(S_{yy}=\sum_{t=1}^n(y_t-\bar{y})^2\), and \(S_{xy}=\sum_{t=1}^n(x_t-\bar{x})(y_t-\bar{y})\). Then \(\hat{b}=S_{xy}/S_{xx}\), \(R^2=S_{xy}^2/(S_{xx}S_{yy})\), and the conventional t-statistic for testing \(H_0:b=0\) is \(\hat{b}/SE(\hat{b})\), where \(SE(\hat{b})=\sqrt{\hat{s}^2/S_{xx}}\) and \(\hat{s}^2=\mathrm{SSR}/(n-2)\). These are the usual OLS formulas; their algebraic validity does not guarantee that their usual inferential interpretation applies.

What the Brownian Limits Imply

For an integrated series, its level after \(n\) steps is typically of order \(\sqrt{n}\). Consequently, centered sums of squares and cross-products of two random walks are of order \(n^2\), not order \(n\). The earlier material on Brownian limits for integrated processes gives the relevant scaling: after rescaling time and dividing levels by \(\sqrt n\), each random walk converges to a Brownian path.

Let \(W_x(r)\) and \(W_y(r)\), for \(0\leq r\leq1\), be independent standard Brownian motions. Write \(\bar W_j=\int_0^1W_j(r)\,dr\) for \(j=x,y\), and define the path functionals

The key point is that these are random quantities from two independent Brownian paths. In general, their cross-product \(A_{xy}\) is not zero for a particular pair of paths, even though the paths are independent in the probabilistic sense.

A short scaling argument explains why the t-statistic can grow. The functional limit gives \(S_{xx}/n^2\to \sigma_x^2 A_{xx}\), \(S_{yy}/n^2\to \sigma_y^2 A_{yy}\), and \(S_{xy}/n^2\to \sigma_x\sigma_y A_{xy}\), in the corresponding weak-convergence sense. Thus the OLS slope remains of order one, with limiting value \((\sigma_y/\sigma_x)A_{xy}/A_{xx}\). The residual sum of squares is \(S_{yy}-S_{xy}^2/S_{xx}\), so it is of order \(n^2\). Dividing by \(n-2\) makes \(\hat{s}^2\) of order \(n\), while \(S_{xx}\) is of order \(n^2\). Therefore \(SE(\hat b)\) is of order \(n^{-1/2}\), and a slope that is typically nonzero produces a t-statistic of order \(\sqrt n\).

More explicitly, substituting the limits into the t-statistic gives

The expression in the denominator is positive when the two centered Brownian paths are not proportional in squared-integral distance, which occurs with probability one for independent Brownian paths. The numerator is generally nonzero as well. Thus the conventional t-statistic does not remain bounded in the way a standard t reference distribution would suggest. Under the independent-random-walk null, a usual two-sided test can reject with probability approaching one as the sample grows; its nominal significance level is not its actual rejection probability.

How \(R^2\) and the t-Statistic Behave

The Brownian limits describe two different features of the apparent fit. First, \(R^2\) converges to a random path-dependent quantity between zero and one. It does not generally converge to zero just because the random walks are independent. Some independent path pairs happen to move together over the observed interval, and a large \(R^2\) is therefore possible. It is not correct to conclude that every such regression has a high \(R^2\): the limiting value is random, and its size varies across realizations.

Second, the conventional t-statistic grows because the coefficient estimate and its conventional standard error scale differently. The slope is typically order one, while the standard error is typically order \(n^{-1/2}\). A nonzero slope estimate is therefore divided by an increasingly small estimated standard error. The resulting apparent precision is an artifact of applying a stationary-error formula to a regression dominated by stochastic trends.

Why the Residuals Matter

A stationary-error regression requires residuals that fluctuate around a stable level rather than retaining a stochastic trend. In a regression between independent random walks, OLS chooses a slope to fit the two paths over the sample, but that choice generally does not remove the random wandering in either path. The residuals therefore remain large-scale, persistent level movements rather than a sequence of stationary deviations around zero.

The distinction can be seen by considering a fixed linear combination of the two walks. For a fixed coefficient \(b\), define \(u_t=y_t-bx_t\). Since the innovations are independent, the variance of its change is

This is positive for any fixed \(b\) when both innovation variances are positive, so the level \(u_t\) accumulates innovations rather than reverting to a fixed mean. In the fitted regression, the estimated slope is random and depends on the paths, but fitting it does not create a genuine common stochastic trend that cancels the independent walks. The residual levels still have the nonstationary behavior that invalidates the usual regression inference. Their first differences may be much less persistent; that fact does not make the residual levels stationary.

Common Mistake and Practical Implications

A common mistake is to read a high \(R^2\) and a tiny conventional t-test p-value as evidence that one random walk predicts or causes the other. The mistaken reasoning is: “The fitted line explains nearly all of the variation, and the coefficient is many standard errors from zero, so the relationship must be real.” The correction is that those familiar interpretations rely on a suitable regression error process. With independent nonstationary levels, both the apparent fit and the conventional t-statistic can be generated by unrelated stochastic trends.

A high \(R^2\) alone does not diagnose a spurious regression, and a low \(R^2\) does not establish that a regression is valid. The relevant question is whether the modeled relationship leaves stationary residuals, under assumptions appropriate to the time-series setting. If residuals retain a unit root, the usual OLS t and F reference distributions cannot be used as if the errors were stationary. Unit-root and cointegration methods address this issue using their own hypotheses and nonstandard critical values; a routine stationary-regression p-value is not a replacement.