ARCH Effects and Engle's LM Test

Why Examine Squared Residuals?

A conditional-mean model describes how the expected value of a time series depends on its information set. Even when that mean is adequately modeled, the size of its innovations may vary predictably over time. Large innovations can be followed by large innovations, and small ones by small ones. This pattern is called volatility clustering. It may be visible in squared innovations even when the innovations themselves have little or no serial correlation.

The ARCH test is a diagnostic for this kind of dependence. It asks whether past squared innovations help explain the current conditional variance. In practice, the innovations are unobserved, so the test uses residuals from an estimated conditional-mean model. A rejection is evidence against constant conditional variance with respect to the lags tested; it does not by itself identify why variance changes or establish that a specific volatility model is correct.

Definition: Let $u_t$ be the innovation in a conditional-mean model, with information available before time $t$ denoted by $\mathcal{F}_{t-1}$. An ARCH($q$) specification has $$ E(u_t\mid\mathcal{F}_{t-1})=0,\qquad \operatorname{Var}(u_t\mid\mathcal{F}_{t-1})=h_t =\omega+\sum_{j=1}^{q}\alpha_j u_{t-j}^{2}, $$ where $\omega>0$ and $\alpha_j\geq 0$. The conditional variance $h_t$ depends on past squared innovations. The no-ARCH null sets all $q$ lag coefficients to zero.

This specification describes dependence in the conditional second moment, not necessarily in the conditional mean. Under the usual ARCH interpretation, an innovation’s sign does not determine its contribution to future variance: its square does. Consequently, a series can have uncorrelated innovations but autocorrelated squared innovations. The test targets the latter pattern.

Engle’s LM Test

Suppose a conditional-mean model has been fitted and gives residuals $e_t$. Choose a lag order $q$ before examining the test result. For each usable time point, regress the squared residual on a constant and $q$ lagged squared residuals:

e_t^2=\beta_0+\beta_1e_{t-1}^2+\cdots+\beta_qe_{t-q}^2+v_t.

Here $v_t$ is the auxiliary-regression error, not the innovation in the original conditional-mean model. If the original series has $T$ observations and the auxiliary regression uses lags $1,\ldots,q$, its effective sample size is $N=T-q$, assuming no additional observations are lost. Let $R^2$ be the ordinary coefficient of determination from this auxiliary regression.

Test result: Under the null of no ARCH dependence through lag $q$, and under the regularity conditions described below, Engle’s LM statistic is $$ LM=N R^2 \ \overset{a}{\sim}\ \chi_q^2. $$ The null hypothesis is $H_0:\beta_1=\cdots=\beta_q=0$. The alternative is that at least one of these lag coefficients is nonzero. The reference distribution has $df=q$. Reject $H_0$ at significance level $\alpha$ when the statistic exceeds the $1-\alpha$ quantile of $\chi_q^2$, or equivalently when its p-value is below $\alpha$.

The null concerns the lagged squared residuals in the specified auxiliary regression. A conventional justification assumes the conditional mean is correctly specified, the innovations have conditional mean zero and constant conditional variance under $H_0$, and the process satisfies suitable stationarity, dependence, and moment conditions. Finite fourth moments are important because the test uses squared innovations. Independent Gaussian innovations provide a familiar sufficient setting for the likelihood-based derivation. With estimated residuals, the asymptotic result also relies on regular estimation of the conditional-mean model. In small samples, the chi-square approximation may be imperfect.

Why the Auxiliary Regression Gives an LM Statistic

The connection can be seen from the Gaussian conditional log likelihood for an ARCH variance. For this derivation, hold the conditional-mean parameters fixed and write the conditional variance as $h_t=\omega+\sum_{j=1}^{q}\alpha_j u_{t-j}^2$. Ignoring terms that do not depend on the parameters, the log likelihood is

\ell=-\frac{1}{2}\sum_t\left(\log h_t+\frac{u_t^2}{h_t}\right).

At the no-ARCH null, $\alpha_1=\cdots=\alpha_q=0$, so $h_t=\omega$. Differentiating with respect to $\alpha_j$ and evaluating at that null gives the score component

\left.\frac{\partial\ell}{\partial\alpha_j}\right|_{H_0} =\frac{1}{2\omega^2}\sum_t (u_t^2-\omega)u_{t-j}^2.

The nuisance parameter $\omega$ is estimated under the null. Accounting for it removes the constant component, just as including an intercept in the auxiliary regression centers the squared innovations and their lagged values. The remaining score tests whether lagged squares explain variation in current squares. In the auxiliary regression, the explained sum of squares divided by the total sum of squares is $R^2$; the corresponding LM quadratic form is asymptotically $N R^2$. Under $H_0$, this quadratic form has a chi-square limit with one degree of freedom for each tested lag, hence $\chi_q^2$. Replacing the unobserved innovations with fitted residuals gives the operational Engle test, subject to the stated regularity conditions.

Worked Examples

Worked Example: Evidence of ARCH at Two Lags

Hypothetical data. A researcher fits a conditional-mean model to 102 monthly observations of an economic index. The residuals show no obvious remaining pattern in their signs. To check for ARCH dependence through two months, the researcher regresses $e_t^2$ on a constant, $e_{t-1}^2$, and $e_{t-2}^2$. The 100 usable rows give $R^2=0.164$. Assume the mean model is correctly specified and the usual large-sample conditions are reasonable.

The hypotheses are $H_0:\beta_1=\beta_2=0$ and $H_a:$ at least one of these coefficients is nonzero. There are $N=102-2=100$ usable observations, so

$$ LM=N R^2=100(0.164)=16.4. $$

Under $H_0$, the reference distribution is $\chi_2^2$. The upper-tail p-value is approximately $0.000275$. Since this is below $\alpha=0.05$, reject $H_0$. In this hypothetical example, the residual squares contain evidence of dependence on their two preceding values, consistent with volatility clustering at the tested lags. This result does not show that the conditional-mean model is wrong, nor does it prove that ARCH(2) is the uniquely appropriate variance specification.

Worked Example: A One-Lag Test Does Not Reject

Hypothetical data. For 81 quarterly observations, an analyst fits a conditional-mean model and uses the 80 available rows in a one-lag auxiliary regression. Its $R^2$ is $0.021$. Suppose residual diagnostics give no reason to doubt the mean specification, and the standard asymptotic conditions are treated as adequate.

The null is $H_0:\beta_1=0$, so $df=1$. The statistic and its approximate p-value are

LM=80(0.021)=1.68,\qquad P(\chi_1^2\geq 1.68)\approx 0.195.

Because $0.195>0.05$, do not reject $H_0$ at the 5% level. The data do not provide strong evidence of ARCH dependence at lag one under this test. This is not proof that the conditional variance is constant: the test may have limited power, and this one-lag specification does not test dependence at every possible lag.

Worked Example: The Lag Order Determines the Reference Distribution

Hypothetical data. A fitted model for 123 daily observations is examined for ARCH effects through three days. The auxiliary regression uses $N=120$ rows and has $R^2=0.041$. Assume the conditional-mean specification and the test’s regularity conditions are adequate.

For $q=3$, the null sets three slopes to zero and the reference distribution has $df=3$:

LM=120(0.041)=4.92,\qquad P(\chi_3^2\geq 4.92)\approx 0.178.

At $\alpha=0.05$, the result does not reject the no-ARCH null for the first three lags. The degrees of freedom are three, not one: the test evaluates the joint restriction on all three lag coefficients. A different choice of $q$ asks a different question and changes both the effective sample and the reference distribution.

Practical Interpretation and Common Pitfalls

An ARCH LM test is useful after modeling the conditional mean because predictable variance changes matter for uncertainty estimates and forecasts, even when the mean equation appears adequate. The test is a diagnostic, not a replacement for examining the original residuals and their squares. If it rejects, the result motivates further investigation of the variance specification and the conditional-mean model.

Confusing raw and squared residual dependence. No visible autocorrelation in $e_t$ does not rule out autocorrelation in $e_t^2$. ARCH concerns the conditional variance.
Using the wrong sample size. With $T$ original observations and $q$ lags, use the number $N$ of rows actually in the auxiliary regression. Do not automatically multiply $R^2$ by $T$ if rows have been lost for other reasons.
Using the wrong degrees of freedom. The chi-square degrees of freedom equal the number of lagged-square slopes tested, $q$, not the number of observations.
Treating rejection as a causal conclusion. Rejection indicates residual-square predictability at the tested lags. It does not establish what caused the pattern or identify a unique model.
Ignoring mean-model misspecification. Omitted dynamics in the conditional mean can leave structure in residuals that complicates variance diagnosis. Check the mean specification before attributing squared-residual dependence to changing variance.
Trying many lag orders and reporting only a favorable result. Choosing $q$ after inspecting many tests changes the interpretation of the reported p-value. Specify a substantively reasonable lag order, or report the set of diagnostics considered.

Key takeaway: Regress squared fitted residuals on an intercept and their chosen lags. The statistic $N R^2$ is compared with $\chi_q^2$ to assess whether those lagged squares jointly predict current squared residuals under the no-ARCH null.

Check Your Understanding

Answer each question using the definitions and test logic developed above.

In an ARCH($q$) specification, what aspect of the current innovation depends on past squared innovations?
An auxiliary regression uses 150 rows and has $R^2=0.032$ with four lagged-square predictors. Calculate the LM statistic and state its null reference distribution.
Why can innovations have little serial correlation while their squares display serial dependence?
A one-lag test has p-value $0.08$. What is the decision at $\alpha=0.05$, and what does that decision not establish?
Give one reason a rejection from residuals of a conditional-mean model should not automatically be interpreted as proof of an ARCH variance specification.