Wie viele Verzögerungen müssen im Ljung-Box-Test einer Zeitreihe verwendet werden?


20

Nachdem ein ARMA-Modell in eine Zeitreihe eingepasst wurde, ist es üblich, die Residuen (unter anderem) über den Ljung-Box-Portmanteau-Test zu überprüfen. Der Ljung-Box-Test liefert einen p-Wert. Der Parameter h gibt die Anzahl der zu testenden Verzögerungen an. Einige Texte empfehlen die Verwendung von h = 20; andere empfehlen die Verwendung von h = ln (n); die meisten sagen nicht , was h zu verwenden.

Anstatt einen einzelnen Wert für h zu verwenden , nehme ich an, dass ich den Ljung-Box-Test für alle h <50 durchführe, und wähle dann das h, das den minimalen p-Wert ergibt. Ist dieser Ansatz vernünftig? Was sind die Vor- und Nachteile? (Ein offensichtlicher Nachteil ist die erhöhte Rechenzeit, aber das ist hier kein Problem.) Gibt es Literatur dazu?

Um es etwas zu erläutern ... Wenn der Test für alle h p> 0,05 ergibt , besteht offensichtlich die Zeitreihe (Residuen) den Test. Meine Frage betrifft, wie der Test zu interpretieren ist, wenn p <0,05 für einige Werte von h und nicht für andere Werte.


1
@ user2875, ich habe meine Antwort gelöscht. Tatsache ist, dass der Test für große h nicht zuverlässig ist. So ist die Antwort hängt wirklich für die h , p<0.05 . Was ist der genaue Wert von p ? Wenn wir den Schwellenwert auf senken 0.01, ändert sich das Ergebnis des Tests? Persönlich suche ich bei widersprüchlichen Hypothesen nach anderen Indikatoren, ob das Modell gut ist oder nicht. Wie gut passt das Modell? Wie vergleicht sich das Modell mit alternativen Modellen? Haben die alternativen Modelle die gleichen Probleme? Bei welchen anderen Verstößen lehnt der Test die Null ab?
mpiktas

1
@mpiktas, Der Ljung-Box-Test basiert auf einer Statistik, deren Verteilung asymptotisch (wenn h groß wird) im Chi-Quadrat ist. Wenn h relativ zu n groß wird, nimmt die Leistung des Tests jedoch auf 0 ab. Daher besteht der Wunsch, h groß genug zu wählen, dass die Verteilung nahe am Chi-Quadrat liegt, aber klein genug, um eine nützliche Leistung zu haben. (Ich weiß nicht, was das Risiko eines falschen Negativs ist, wenn h klein ist.)
user2875

@ user2875, dies ist das dritte Mal, dass Sie die Frage geändert haben. Zuerst fragen Sie über die Strategie der Kommissionierung h mit kleinstem Wert, dann , wie der Test , wenn zu interpretieren p<0.05 für einige Werte von h , und jetzt , was die optimalen h zu wählen. Alle drei Fragen haben unterschiedliche Antworten und können je nach Kontext des jeweiligen Problems sogar unterschiedliche Antworten haben.
mpiktas

@mpiktas, die Fragen sind alle gleich, nur unterschiedliche Betrachtungsweisen. (Wie bereits erwähnt, wenn p> 0,05 für alle h ist, dann wissen wir, wie wir das kleinste p interpretieren können. Wenn wir das optimale h kennen - wir tun es nicht - dann würden wir uns nicht darum kümmern, das kleinste p zu wählen.)
user2875

Antworten:


9

Die Antwort hängt definitiv davon ab, wofür der Test tatsächlich verwendet werden soll.Q

Der häufigste Grund ist, mehr oder weniger sicher zu sein , ob die Nullhypothese ohne Autokorrelation eine gemeinsame statistische Signifikanz von bis h hat (alternativ unter der Annahme, dass Sie etwas haben, das einem schwachen weißen Rauschen nahe kommt ), und ein sparsames Modell zu erstellen , das so wenig wie möglich ist Anzahl der Parameter wie möglich.h

In der Regel weisen Zeitreihendaten ein natürliches saisonales Muster auf, sodass die praktische Faustregel darin besteht, auf das Doppelte dieses Werts festzulegen. Ein anderer ist der Prognosehorizont, wenn Sie das Modell für Prognoseanforderungen verwenden. Wenn Sie bei letzteren Abweichungen signifikante Abweichungen feststellen, versuchen Sie, über die Korrekturen nachzudenken (könnte dies an saisonalen Effekten liegen oder die Daten wurden nicht für Ausreißer korrigiert).h

Anstatt einen einzelnen Wert für h zu verwenden, nehme ich an, dass ich den Ljung-Box-Test für alle h <50 durchführe, und wähle dann das h, das den minimalen p-Wert ergibt.

Es ist ein gemeinsamer Signifikanztest. Wenn also die Wahl von datenbasiert ist, warum sollte ich mich dann für einige kleine (gelegentliche?) Abweichungen mit einer Verzögerung von weniger als h interessieren, vorausgesetzt, es ist natürlich viel weniger als n (die Leistung) des von Ihnen erwähnten Tests). Um ein einfaches, aber relevantes Modell zu finden, schlage ich die unten beschriebenen Informationskriterien vor.hhn

Meine Frage betrifft, wie der Test zu interpretieren ist, wenn für einige Werte von h und nicht für andere Werte.p<0.05h

Es kommt also darauf an, wie weit es von der Gegenwart entfernt ist. Nachteile weit entfernter Abweichungen: mehr zu schätzende Parameter, weniger Freiheitsgrade, schlechtere Vorhersagekraft des Modells.

Versuchen Sie, das Modell einschließlich der MA- und / oder AR-Teile an der Verzögerung zu schätzen, an der die Abweichung auftritt, UND sehen Sie sich zusätzlich eines der Informationskriterien an (entweder AIC oder BIC, je nach Stichprobengröße). Dadurch erhalten Sie mehr Einsichten darüber, welches Modell besser ist sparsam. Alle Vorhersagetests außerhalb der Stichprobe sind hier ebenfalls willkommen.


+1, das ist, was ich ausdrücken wollte, aber nicht in der Lage war :)
mpiktas

8

Angenommen, wir geben ein einfaches AR (1) -Modell mit allen üblichen Eigenschaften an,

yt=βyt1+ut

Bezeichnen Sie die theoretische Kovarianz des Fehlerbegriffs als

γjE(ututj)

Wenn wir den Fehlerterm beobachten könnten, dann ist die Beispielautokorrelation des Fehlerterms definiert als

ρ~jγ~jγ~0

woher

γ~j1nt=j+1nututj,j=0,1,2...

In der Praxis wird der Fehlerbegriff jedoch nicht beachtet. Daher wird die Autokorrelation der Stichprobe in Bezug auf den Fehlerterm unter Verwendung der Residuen aus der Schätzung geschätzt

γ^j1nt=j+1nu^tu^tj,j=0,1,2...

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

QBP=nj=1pρ^j2=j=1p[nρ^j]2d???χ2(p)

Our issue is exactly whether QBP 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 nρ^jnρ^nρ~

Wir haben das

u^t=ytβ^yt1=ut(β^β)yt1

where β^ is a consistent estimator. So

γ^j1nt=j+1n[ut(β^β)yt1][utj(β^β)ytj1]

=γ~j1nt=j+1n(β^β)[utytj1+utjyt1]+1nt=j+1n(β^β)2yt1ytj1

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

γ^jpγ~j

This implies that

ρ^jpρ~jpρj

But this does not automatically guarantee that nρ^j converges to nρ~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

nγ^jdnγ~j

(the denominator γ0 -tilde or hat- will converge to the variance of the error term in both cases, so it is neutral to our issue).

We have

nγ^j=nγ~j1nt=j+1nn(β^β)[utytj1+utjyt1]+1nt=j+1nn(β^β)2yt1ytj1

So the question is : do these two sums, multiplied now by n, go to zero in probability so that we will be left with nγ^j=nγ~j asymptotically?

For the second sum we have

1nt=j+1nn(β^β)2yt1ytj1=1nt=j+1n[n(β^β)][(β^β)yt1ytj1]

Since [n(β^β)] converges to a random variable, and β^ is consistent, this will go to zero.

For the first sum, here too we have that [n(β^β)] converges to a random variable, and so we have that

1nt=j+1n[utytj1+utjyt1]pE[utytj1]+E[utjyt1]

The first expected value, E[utytj1] 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 nρ^j won't have the same asymptotic distribution as nρ~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).


1
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.
Aksakal

3
@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.
Alecos Papadopoulos

2
@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.
Richard Hardy

3

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


@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.
mpiktas

@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.
bill_080

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.
user2875

2

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).


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.
Aksakal

@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.
Richard Hardy

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.
Aksakal

@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?
Richard Hardy

1

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).


1

The two most common settings are min(20,T1) and lnT 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 xt:

yt=xtβ+ϕ(L)yt+ut

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 xt and εs in any event, so the two tests are asymptotically equivalent. On the other hand, because it does not condition on xt , the Box–Pierce test is less powerful than the LM test when the null hypothesis is false, as intuition might suggest.


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)?
Richard Hardy

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.
Richard Hardy

0

... 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


0

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: min(n22,40). 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.


0

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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