¿Cuántos rezagos usar en la prueba de Ljung-Box de una serie de tiempo?

Después de que un modelo ARMA se ajusta a una serie temporal, es común verificar los residuos a través de la prueba Ljung-Box portmanteau (entre otras pruebas). La prueba de Ljung-Box devuelve un valor p. Tiene un parámetro, h , que es el número de retrasos que se probarán. Algunos textos recomiendan usar h = 20; otros recomiendan usar h = ln (n); la mayoría no dice qué h usar.

En lugar de usar un solo valor para h , supongamos que hago la prueba de Ljung-Box para todo h <50, y luego elijo la h que da el valor p mínimo. ¿Es razonable ese enfoque? ¿Cuáles son las ventajas y desventajas? (Una desventaja obvia es el aumento del tiempo de cómputo, pero eso no es un problema aquí.) ¿Existe literatura sobre esto?

Para elaborar un poco .... Si la prueba da p> 0.05 para toda h , entonces obviamente las series de tiempo (residuales) pasan la prueba. Mi pregunta se refiere a cómo interpretar la prueba si p <0.05 para algunos valores de h y no para otros valores.

La respuesta definitivamente depende de: ¿Para qué están tratando de usar la prueba ?$Q$

La razón común es: tener más o menos confianza en la significación estadística conjunta de la hipótesis nula de no autocorrelación hasta el retraso (alternativamente, suponiendo que tiene algo cercano a un ruido blanco débil ) y construir un modelo parsimonioso , teniendo tan poco Número de parámetros como sea posible.$h$

Por lo general, los datos de series temporales tienen un patrón estacional natural, por lo que la regla práctica sería establecer al doble de este valor. Otro es el horizonte de pronóstico, si usa el modelo para las necesidades de pronóstico. Finalmente, si encuentra algunas desviaciones significativas en los últimos rezagos, trate de pensar en las correcciones (esto podría deberse a algunos efectos estacionales, o los datos no se corrigieron por valores atípicos).$h$

En lugar de usar un solo valor para h, supongamos que hago la prueba de Ljung-Box para todo h <50, y luego elijo h que da el valor p mínimo.

Es una prueba de significación conjunta , por lo que si la elección de depende de los datos, ¿por qué debería preocuparme por algunas pequeñas desviaciones (ocasionales?) En cualquier retraso menor que h , suponiendo que es mucho menor que n, por supuesto (el poder de la prueba que mencionaste). Buscando encontrar un modelo simple pero relevante, sugiero los criterios de información como se describe a continuación.$h$$h$$n$

Mi pregunta se refiere a cómo interpretar la prueba si para algunos valores de h y no para otros valores.$p<0.05$$h$

Por lo tanto, dependerá de cuán lejos del presente ocurra. Desventajas de desviaciones lejanas: más parámetros para estimar, menos grados de libertad, peor poder predictivo del modelo.

Intente estimar el modelo, incluidas las partes MA y \ o AR en el retraso donde se produce la partida Y, además, observe uno de los criterios de información (ya sea AIC o BIC, dependiendo del tamaño de la muestra), esto le brindará más información sobre qué modelo es más parsimonioso. Cualquier ejercicio de predicción fuera de la muestra también es bienvenido aquí.

Supongamos que especificamos un modelo AR (1) simple, con todas las propiedades habituales,

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

Denote la covarianza teórica del término de error como

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

Si pudiéramos observar el término de error, entonces la muestra de autocorrelación del término de error se define como

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

dónde

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

Pero en la práctica, no observamos el término de error. Por lo tanto, la muestra de autocorrelación relacionada con el término de error se estimará utilizando los residuos de la estimación, como

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

La estadística Q de Box-Pierce (la Ljung-Box Q es solo una versión asintóticamente neutra a escala)

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

Nuestro problema es exactamente si se puede decir que tiene asintóticamente una distribución de chi-cuadrado (bajo el nulo de no autocorellación en el término de error) en este modelo. Para que esto suceda, todos y cada uno de $Q_{BP}$
debe ser asintóticamente estándar normal. Una forma de verificar esto es examinar si$\sqrt n \hat\rho_j$ tiene la misma distribución asintótica como$\sqrt n \hat\rho$ (que se construye utilizando los errores verdaderos, y también tiene el comportamiento asintótico deseado bajo nulo).$\sqrt n \tilde\rho$

Tenemos eso

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

donde β es un estimador consistente. Entonces$\hat \beta$

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

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

Se supone que la muestra es estacionaria y ergódica, y se supone que existen momentos hasta el orden deseado. Puesto que el estimador de β es consistente, esto es suficiente para las dos sumas para ir a cero. Entonces concluimos$\hat \beta$

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

Esto implica que

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

Pero esto no garantiza automáticamente que converge a $\sqrt n \hat \rho_j$$\sqrt n\tilde \rho_j$(en distribución) (piense que el teorema de mapeo continuo no se aplica aquí porque la transformación aplicada a las variables aleatorias depende de). Para que esto suceda, necesitamos$n$

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

(el denominador -tilde o hat- convergerá con la varianza del término de error en ambos casos, por lo que es neutral para nuestro problema).$\gamma_0$

Tenemos

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

Entonces la pregunta es: ¿estas dos sumas, multiplicadas ahora por , ir a cero en probabilidad para que nos quedemos con$\sqrt n$asintóticamente?$\sqrt n \hat \gamma_j =\sqrt n\tilde \gamma _j$

Para la segunda suma tenemos

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

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).

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

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).

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).

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:

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:

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.

... 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

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.

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.

Actualización: Considerando la discusión alternativa en este hilo con respecto a ARMAX, otro incentivo para analizar hwwntest es la disponibilidad de una función de potencia teórica para una de las pruebas contra una hipótesis alternativa del modelo ARMA (p, q).