question about the results given by the Box.test?
Hopefully the test is the same no matter what software you are using. A small p-value is an indication that there is structure in the data. So in your case there is no indication of autocorrelation up to lag 5, but it appears that there might be something going on at around lags 6 to 9. "How small is too small?" is not a reasonable question to ask in general. It depends on how likely it is that there is structure near that lag, how much data you have, how important it is to capture all of the structure in your model versus the harm of overfitting, ... Patrick Burns patrick at burns-stat.com +44 (0)20 8525 0696 http://www.burns-stat.com (home of S Poetry and "A Guide for the Unwilling S User")
oliver wee wrote:
Hello, I am using the Ljung Box test in R to compute
if the resiudals of my fitted model is random or not.
I am not sure though what the results mean, I have
looked at various sources on the internet and have
come up with contrasting explanations (mainly because
these info deal with different program languages, like
SAS, SPSS, etc).
I know that my residuals should appropriate white
noise( is random) since a check of its ACF shows it to
be so (signifant correlation only at lag 1, decays
very quickly to zero).
But I am not sure how to interpret the ljung-box
result given by R.
To check for randomness of residuals, should the
p-value be small or large? How small and how large?
And at what lags should I check for the randomness of
the residuals? Is a p-value > 0.05 (or < 0.05) enough?
What if I have a very large p-value of 0.9796 at lag
1, but its value is 0.0139 at lag 8?
For example, here's what I got for the first 10 lags
of the residuals I'm testing:
-------------------
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =1)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 7e-04, df = 1, p-value = 0.9796
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =2)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 0.1088, df = 2, p-value = 0.947
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =3)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 1.4179, df = 3, p-value = 0.7014
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =4)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 3.866, df = 4, p-value = 0.4244
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =5)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 6.0251, df = 5, p-value = 0.3038
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =6)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 12.11, df = 6, p-value = 0.05956
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =7)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 13.0307, df = 7, p-value = 0.07137
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =8)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 19.1766, df = 8, p-value = 0.01394
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =9)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 19.6753, df = 9, p-value = 0.02003
Box.test(SP500DataSetFitMA2$residuals, type =
"Ljung", lag =10)
Box-Ljung test
data: SP500DataSetFitMA2$residuals
X-squared = 19.7124, df = 10, p-value = 0.03209
--------------
I know this is not really a programming question, so I
apologize if it is inappropriate or if the question is
too elementary.
Thank you very much for your help.
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