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After an ARMA model is fit to a time series, it is common to check the residuals via the Ljung-Box portmanteau test (among other tests). The Ljung-Box test returns a p value. It has a parameter, h, which is the number of lags to be tested. Some texts recommend using h=20; others recommend using h=ln(n); most do not say what h to use.

Rather than using a single value for h, suppose that I do the Ljung-Box test for all h<50, and then pick the h which gives the minimum p value. Is that approach reasonable? What are the advantages and disadvantages? (One obvious disadvantage is increased computation time, but that is not a problem here.) Is there literature on this?

To elaborate slightly.... If the test gives p>0.05 for all h, then obviously the time series (residuals) pass the test. My question concerns how to interpret the test if p<0.05 for some values of h and not for other values.

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    $\begingroup$ @user2875, I've deleted my answer. Fact is that for large $h$ the test is not reliable. So the answer really depends for which $h$, $p<0.05$. Furthermore what is exact value of $p$? If we decrease the threshold to $0.01$, does the result of the test changes? Personally in case of conflicting hypotheses I look for other indicators whether model is good or not. How well model fits? How does the model compare to alternative models? Do the alternative model have the same problems? For what other violations the test rejects the null? $\endgroup$ – mpiktas Jan 23 '11 at 19:52
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    $\begingroup$ @mpiktas, The Ljung-Box test is based on a statistic whose distribution is asymptotically (as h becomes large) chi-squared. As h gets large relative to n, though, the power of the test decreases to 0. Hence the desire to choose h large enough that the distribution is close to chi-squared but small enough to have useful power. (I do not know what the risk of a false negative is, when h is small.) $\endgroup$ – user2875 Jan 24 '11 at 0:05
  • $\begingroup$ @user2875, this the third time you changed the question. First you ask about the strategy of picking $h$ with smallest value, then how to interpret the test if $p<0.05$ for some values of $h$, and now what is the optimal $h$ to choose. All three questions have different answers and may even have different answers depending on the context of particular problem. $\endgroup$ – mpiktas Jan 24 '11 at 12:12
  • $\begingroup$ @mpiktas, the questions are all the same, just different ways of looking at it. (As pointed out, if p>0.05 for all h, then we know how to interpret the smallest p; if we knew the optimal h--we don't--then we would not be concerned with choosing the smallest p.) $\endgroup$ – user2875 Jan 24 '11 at 18:43
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The answer definitely depends on: What are actually trying to use the $Q$ test for?

The common reason is: to be more or less confident about joint statistical significance of the null hypothesis of no autocorrelation up to lag $h$ (alternatively assuming that you have something close to a weak white noise) and to build a parsimonious model, having as little number of parameters as possible.

Usually time series data has natural seasonal pattern, so the practical rule-of-thumb would be to set $h$ to twice this value. Another one is the forecasting horizon, if you use the model for forecasting needs. Finally if you find some significant departures at latter lags try to think about the corrections (could this be due to some seasonal effects, or the data was not corrected for outliers).

Rather than using a single value for h, suppose that I do the Ljung-Box test for all h<50, and then pick the h which gives the minimum p value.

It's a joint significance test, so if the choice of $h$ is data-driven, then why should I care about some small (occasional?) departures at any lag less than $h$, supposing that it is much less than $n$ of course (the power of the test you mentioned). Seeking to find a simple yet relevant model I suggest the information criteria as described below.

My question concerns how to interpret the test if $p<0.05$ for some values of $h$ and not for other values.

So it will depend on how far from the present it happens. Disadvantages of far departures: more parameters to estimate, less degrees of freedom, worse predictive power of the model.

Try to estimate the model including the MA and\or AR parts at the lag where the departure occurs AND additionally look at one of information criteria (either AIC or BIC depending on the sample size) this would bring you more insights on what model is more parsimonious. Any out-of-sample prediction exercises are also welcome here.

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  • $\begingroup$ +1, this is what I was trying to express but was not able to :) $\endgroup$ – mpiktas Jan 25 '11 at 14:20
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Assume that we specify a simple AR(1) model, with all the usual properties,

$$y_t = \beta y_{t-1} + u_t$$

Denote the theoretical covariance of the error term as

$$\gamma_j \equiv E(u_tu_{t-j})$$

If we could observe the error term, then the sample autocorrelation of the error term is defined as

$$\tilde \rho_j \equiv \frac {\tilde \gamma_j}{\tilde \gamma_0}$$

where

$$\tilde\gamma_j \equiv \frac 1n \sum_{t=j+1}^nu_tu_{t-j},\;\;\; j=0,1,2...$$

But in practice, we do not observe the error term. So the sample autocorrelation related to the error term will be estimated using the residuals from estimation, as

$$\hat\gamma_j \equiv \frac 1n \sum_{t=j+1}^n\hat u_t\hat u_{t-j},\;\;\; j=0,1,2...$$

The Box-Pierce Q-statistic (the Ljung-Box Q is just an asymptotically neutral scaled version of it) is

$$Q_{BP} = n \sum_{j=1}^p\hat\rho^2_j = \sum_{j=1}^p[\sqrt n\hat\rho_j]^2\xrightarrow{d} \;???\;\chi^2(p) $$

Our issue is exactly whether $Q_{BP}$ can be said to have asymptotically a chi-square distribution (under the null of no-autocorellation in the error term) in this model.
For this to happen, each and everyone of $\sqrt n \hat\rho_j$ must be asymptotically standard Normal. A way to check this is to examine whether $\sqrt n \hat\rho$ has the same asymptotic distribution as $\sqrt n \tilde\rho$ (which is constructed using the true errors, and so has the desired asymptotic behavior under the null).

We have that

$$\hat u_t = y_t - \hat \beta y_{t-1} = u_t - (\hat \beta - \beta)y_{t-1}$$

where $\hat \beta$ is a consistent estimator. So

$$\hat\gamma_j \equiv \frac 1n \sum_{t=j+1}^n[u_t - (\hat \beta - \beta)y_{t-1}][u_{t-j} - (\hat \beta - \beta)y_{t-j-1}]$$

$$=\tilde \gamma _j -\frac 1n \sum_{t=j+1}^n (\hat \beta - \beta)\big[u_ty_{t-j-1} +u_{t-j}y_{t-1}\big] + \frac 1n \sum_{t=j+1}^n(\hat \beta - \beta)^2y_{t-1}y_{t-j-1}$$

The sample is assumed to be stationary and ergodic, and moments are assumed to exist up until the desired order. Since the estimator $\hat \beta$ is consistent, this is enough for the two sums to go to zero. So we conclude

$$\hat \gamma_j \xrightarrow{p} \tilde \gamma_j$$

This implies that

$$\hat \rho_j \xrightarrow{p} \tilde \rho_j \xrightarrow{p} \rho_j$$

But this does not automatically guarantee that $\sqrt n \hat \rho_j$ converges to $\sqrt n\tilde \rho_j$ (in distribution) (think that the continuous mapping theorem does not apply here because the transformation applied to the random variables depends on $n$). In order for this to happen, we need

$$\sqrt n \hat \gamma_j \xrightarrow{d} \sqrt n \tilde \gamma_j$$

(the denominator $\gamma_0$ -tilde or hat- will converge to the variance of the error term in both cases, so it is neutral to our issue).

We have

$$\sqrt n \hat \gamma_j =\sqrt n\tilde \gamma _j -\frac 1n \sum_{t=j+1}^n \sqrt n(\hat \beta - \beta)\big[u_ty_{t-j-1} +u_{t-j}y_{t-1}\big] \\+ \frac 1n \sum_{t=j+1}^n\sqrt n(\hat \beta - \beta)^2y_{t-1}y_{t-j-1}$$

So the question is : do these two sums, multiplied now by $\sqrt n$, go to zero in probability so that we will be left with $\sqrt n \hat \gamma_j =\sqrt n\tilde \gamma _j$ asymptotically?

For the second sum we have

$$\frac 1n \sum_{t=j+1}^n\sqrt n(\hat \beta - \beta)^2y_{t-1}y_{t-j-1} = \frac 1n \sum_{t=j+1}^n\big[\sqrt n(\hat \beta - \beta)][(\hat \beta - \beta)y_{t-1}y_{t-j-1}]$$

Since $[\sqrt n(\hat \beta - \beta)]$ converges to a random variable, and $\hat \beta$ is consistent, this will go to zero.

For the first sum, here too we have that $[\sqrt n(\hat \beta - \beta)]$ converges to a random variable, and so we have that

$$\frac 1n \sum_{t=j+1}^n \big[u_ty_{t-j-1} +u_{t-j}y_{t-1}\big] \xrightarrow{p} E[u_ty_{t-j-1}] + E[u_{t-j}y_{t-1}]$$

The first expected value, $E[u_ty_{t-j-1}]$ is zero by the assumptions of the standard AR(1) model. But the second expected value is not, since the dependent variable depends on past errors.

So $\sqrt n\hat \rho_j$ won't have the same asymptotic distribution as $\sqrt n\tilde \rho_j$. But the asymptotic distribution of the latter is standard Normal, which is the one leading to a chi-squared distribution when squaring the r.v.

Therefore we conclude, that in a pure time series model, the Box-Pierce Q and the Ljung-Box Q statistic cannot be said to have an asymptotic chi-square distribution, so the test loses its asymptotic justification.

This happens because the right-hand side variable (here the lag of the dependent variable) by design is not strictly exogenous to the error term, and we have found that such strict exogeneity is required for the BP/LB Q-statistic to have the postulated asymptotic distribution.

Here the right-hand-side variable is only "predetermined", and the Breusch-Pagan test is then valid. (for the full set of conditions required for an asymptotically valid test, see Hayashi 2000, p. 146-149).

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    $\begingroup$ You wrote "But the second expected value is not, since the dependent variable depends on past errors." That's called strict exogeneity. I agree that it's a strong assumption, and you can build AR(p) framework without it, just by using weak exogeneity. This the reason why Breusch-Godfrey test is better in some sense: if the null is not true, then B-L loses power. B-G is based on weak exogeneity. Both tests are not good for some common econometric, applications, see e.g. this Stata's presentation, p. 4/44. $\endgroup$ – Aksakal Apr 3 '16 at 17:14
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    $\begingroup$ @Aksakal Thanks for the reference. The point exactly is that without strict exogeneity, the Box-Pierce/Ljung-Box do not have an asymptotic chi-square distribution, this is what the mathematics above show. Weak exogeneity (which holds in the above model) is not enough for them. This is exactly what the presentation you link to says in p. 3/44. $\endgroup$ – Alecos Papadopoulos Apr 3 '16 at 17:39
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    $\begingroup$ @AlecosPapadopoulos, an amazing post!!! Among the few best ones I have encountered here at Cross Validated. I just wish it would not disappear in this long thread and many users would find and benefit from it in the future. $\endgroup$ – Richard Hardy Apr 3 '16 at 19:08
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Before you zero-in on the "right" h (which appears to be more of an opinion than a hard rule), make sure the "lag" is correctly defined.

http://www.stat.pitt.edu/stoffer/tsa2/Rissues.htm

Quoting the section below Issue 4 in the above link:

"....The p-values shown for the Ljung-Box statistic plot are incorrect because the degrees of freedom used to calculate the p-values are lag instead of lag - (p+q). That is, the procedure being used does NOT take into account the fact that the residuals are from a fitted model. And YES, at least one R core developer knows this...."

Edit (01/23/2011): Here's an article by Burns that might help:

http://lib.stat.cmu.edu/S/Spoetry/Working/ljungbox.pdf

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  • $\begingroup$ @bil_080, the OP does not mention R, and help page for Box.test in R mentions the correction and has an argument to allow for the correction, although you need to supply it manualy. $\endgroup$ – mpiktas Jan 22 '11 at 20:50
  • $\begingroup$ @mpiktas, Oops, you're right. I assumed this was an R question. As for the second part of your comment, there are several R packages that use Ljung-Box stats. So, it's a good idea to make sure the user understands what the package's "lag" means. $\endgroup$ – bill_080 Jan 22 '11 at 22:29
  • $\begingroup$ Thanks--I am using R, but the question is a general one. Just to be safe, I was doing the test with the LjungBox function in the portes package, as well as Box.test. $\endgroup$ – user2875 Jan 23 '11 at 16:16
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The thread "Testing for autocorrelation: Ljung-Box versus Breusch-Godfrey" shows that the Ljung-Box test is essentially inapplicable in the case of an autoregressive model. It also shows that Breusch-Godfrey test should be used instead. That limits the relevance of your question and the answers (although the answers may include some generally good points).

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  • $\begingroup$ The trouble with LB test is when autoregressive models have other regressors, i.e. ARMAX not ARM models. OP explicitly states ARMA not ARMAX in the question. Hence, I think that your answer is incorrect. $\endgroup$ – Aksakal Apr 2 '16 at 21:50
  • $\begingroup$ @Aksakal, I clearly see from Alecos Papadopoulos answer (and comments under it) in the above-mentioned thread that Ljung-Box test is inapplicable in both cases, i.e. pure AR/ARMA and ARX/ARMAX. Therefore, I cannot agree with you. $\endgroup$ – Richard Hardy Apr 3 '16 at 8:27
  • $\begingroup$ Alecos Papadopoulos's answer is good, but incomplete. It points out to Ljung-Box test's assumption of strict exogeneity but it fails to mention that if you're fine with the assumption, then L-B test is Ok to use. B-G test, which he and I favor over L-B, relies on weak exogeneity. It's better to use tests with weaker assumptions in general, of course. However, even B-G test's assumptions are too strong in many cases. $\endgroup$ – Aksakal Apr 3 '16 at 17:27
  • $\begingroup$ @Aksakal, The setting of this question is quite definite -- it considers residuals from an ARMA model. The important thing here is, L-B does not work (as shown explicitly in Alecos post in this as well as the above-cited thread) while B-G test does work. Of course, things can happen in other settings (even B-G test's assumptions are too strong in many cases) -- but that is not the concern in this thread. Also, I did not get what the assumption is in your statement if you're fine with the assumption, then L-B test is Ok to use. Is that supposed to invalidate Alecos point? $\endgroup$ – Richard Hardy Apr 3 '16 at 19:14
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Escanciano and Lobato constructed a portmanteau test with automatic, data-driven lag selection based on the Pierce-Box test and its refinements (which include the Ljung-Box test).

The gist of their approach is to combine the AIC and BIC criteria --- common in the identification and estimation of ARMA models --- to select the optimal number of lags to be used. In the introduction of they suggest that, intuitively, ``test conducted using the BIC criterion are able to properly control for type I error and are more powerful when serial correlation is present in the first order''. Instead, tests based on AIC are more powerful against high order serial correlation. Their procedure thus choses a BIC-type lag selection in the case that autocorrelations seem to be small and present only at low order, and an AIC-type lag section otherwise.

The test is implemented in the R package vrtest (see function Auto.Q).

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The two most common settings are $\min(20,T-1)$ and $\ln T$ where $T$ is the length of the series, as you correctly noted.

The first one is supposed to be from the authorative book by Box, Jenkins, and Reinsel. Time Series Analysis: Forecasting and Control. 3rd ed. Englewood Cliffs, NJ: Prentice Hall, 1994.. However, here's all they say about the lags on p.314: enter image description here

It's not a strong argument or suggestion by any means, yet people keep repeating it from one place to another.

The second setting for a lag is from Tsay, R. S. Analysis of Financial Time Series. 2nd Ed. Hoboken, NJ: John Wiley & Sons, Inc., 2005, here's what he wrote on p.33:

Several values of m are often used. Simulation studies suggest that the choice of m ≈ ln(T ) provides better power performance.

This is a somewhat stronger argument, but there's no description of what kind of study was done. So, I wouldn't take it at a face value. He also warns about seasonality:

This general rule needs modification in analysis of seasonal time series for which autocorrelations with lags at multiples of the seasonality are more important.

Summarizing, if you just need to plug some lag into the test and move on, then you can use either of these setting, and that's fine, because that's what most practitioners do. We're either lazy or, more likely, don't have time for this stuff. Otherwise, you'd have to conduct your own research on the power and properties of the statistics for series that you deal with.

UPDATE.

Here's my answer to Richard Hardy's comment and his answer, which refers to another thread on CV started by him. You can see that the exposition in the accepted (by Richerd Hardy himself) answer in that thread is clearly based on ARMAX model, i.e. the model with exogenous regressors $x_t$:$$y_t = \mathbf x_t'\beta + \phi(L)y_t + u_t$$

However, OP did not indicate that he's doing ARMAX, to contrary, he explicitly mentions ARMA:

After an ARMA model is fit to a time series, it is common to check the residuals via the Ljung-Box portmanteau test

One of the first papers that pointed to a potential issue with LB test was Dezhbaksh, Hashem (1990). “The Inappropriate Use of Serial Correlation Tests in Dynamic Linear Models,” Review of Economics and Statistics, 72, 126–132. Here's the excerpt from the paper:

enter image description here

As you can see, he doesn't object to using LB test for pure time series models such as ARMA. See also the discussion in the manual to a standard econometrics tool EViews:

If the series represents the residuals from ARIMA estimation, the appropriate degrees of freedom should be adjusted to represent the number of autocorrelations less the number of AR and MA terms previously estimated. Note also that some care should be taken in interpreting the results of a Ljung-Box test applied to the residuals from an ARMAX specification (see Dezhbaksh, 1990, for simulation evidence on the finite sample performance of the test in this setting)

Yes, you have to be careful with ARMAX models and LB test, but you can't make a blanket statement that LB test is always wrong for all autoregressive series.

UPDATE 2

Alecos Papadopoulos's answer shows why Ljung-Box test requires strict exogeneity assumption. He doesn't show it in his post, but Breusch-Gpdfrey test (another alternative test) requires only weak exogeneity, which is better, of course. This what Greene, Econometrics, 7th ed. says on the differences between tests, p.923:

The essential difference between the Godfrey–Breusch and the Box–Pierce tests is the use of partial correlations (controlling for X and the other variables) in the former and simple correlations in the latter. Under the null hypothesis, there is no autocorrelation in εt , and no correlation between $x_t$ and $\varepsilon_s$ in any event, so the two tests are asymptotically equivalent. On the other hand, because it does not condition on $x_t$ , the Box–Pierce test is less powerful than the LM test when the null hypothesis is false, as intuition might suggest.

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  • $\begingroup$ I suppose that you decided to answer the question as it was bumped to the top of the active threads by my recent answer. Curiously, I argue that the test is inappropriate in the setting under consideration, making the whole thread problematic and the answers in it especially so. Do you think it is good practice to post yet another answer that ignores this problem without even mentioning it (just like all the previous answers do)? Or do you think my answer does not make sense (which would justify posting an answer like yours)? $\endgroup$ – Richard Hardy Apr 2 '16 at 14:25
  • $\begingroup$ Thank you for an update! I am not an expert, but the argumentation by Alecos Papadopoulos in "Testing for autocorrelation: Ljung-Box versus Breusch-Godfrey" and in the comments under his answer suggests that Ljung-Box is indeed inapplicable on residuals from pure ARMA (as well as ARMAX) models. If the wording is confusing, check the maths there, it seems fine. I think this is a very interesting and important question, so I would really like to find agreement between all of us here. $\endgroup$ – Richard Hardy Apr 3 '16 at 8:21
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... h should be as small as possible to preserve whatever power the LB test may have under the circumstances. As h increases the power drops. The LB test is a dreadfully weak test; you must have a lot of samples; n must be ~> 100 to be meaningful. Unfortunately I have never seen a better test. But perhaps one exists. Anyone know of one ?

Paul3nt

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There's no correct answer to this that works in all situation for the reasons other have said it will depend on your data.

That said, after trying to figure out to reproduce a result in Stata in R I can tell you that, by default Stata implementation uses: $\mathrm{min}(\frac{n}{2}-2, 40)$. Either half the number of data points minus 2, or 40, whichever is smaller.

All defaults are wrong, of course, and this will definitely be wrong in some situations. In many situations, this might not be a bad place to start.

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Let me suggest you our R package hwwntest. It has implemented Wavelet-based white noise tests that do not require any tuning parameters and have good statistical size and power.

Additionally, I have recently found "Thoughts on the Ljung-Box test" which is excellent discussion on the topic from Rob Hyndman.

Update: Considering the alternative discussion in this thread regarding ARMAX, another incentive to look at hwwntest is the availability of a theoretical power function for one of the tests against an alternative hypothesis of ARMA(p,q) model.

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