Hello,
No, the Ljung-Box test wouldn't be inappropriate in that case. First you
detrend the series and then test for serial independence. It's even
usual to do so. I would use the default values for lag and fitdf. But
use type="Ljung", the Box-Pierce test is nowadays seldom used in
pratice, if at all. It does have great historical and pedagogic
interess, the Ljung-Box test statistic follows it and corrects its
variance estimation's bias.
The parameter fitdf is relevant if you test the residuals of a fitted
ARMA(p, q) model, which isn't the case, so keep it equal to zero.
In that case, lag is chosen such that lag > fitdf. 1 will do.
Oh, and please, it's Rui, not Mr.
Rui Barradas
Em 27-06-2012 16:55, Steven Winter escreveu:
Dear Mr. Barradas,
Thank you for your help. Let's say I have the yearly standard deviation
of temperatures over New York City for the past 24 years. So, there are
24 data points. I would like to put a linear/quadratic/some kind of
model on top of the data to show that there might be a trend in the data
over time. But to do so, I have to test the time independence of the
residuals. Would you say the Ljung-Box test is inappropriate in this
case? If so, what would be my values for "lag" and "fitdf" that I plug
into the Box.test function in R?
Thank you,
Steven
------------------------------------------------------------------------
*From:* Rui Barradas <ruipbarra...@sapo.pt>
*To:* Steven Winter <stevenwinte...@yahoo.com>
*Cc:* r-help@r-project.org
*Sent:* Tuesday, June 26, 2012 3:13 PM
*Subject:* Re: [R] Ljung-Box test (Box.test)
Hello,
That's a statistics question, but it's also about using an R function.
The Ljung-Box test isn't supposed to be used in such a context, to test
the residuals of an ols y = bX + e. It is used to test time independence
of the original series or of the residuals of an ARMA(p, q) fit.
In both cases you are right, 'x' is a series.
'lag' can be explained as follows: you have a time series and want to
know if the value observed today depends on what was observed in the
past. Then, a linear regression of "today" on "yesterday" could be
X[t] = b[1]*X[t-1] + e[t], e ~ Normal(0, sigma^2)
A linear regression on two time units in the past would be
X[t] = b[1]*X[t-1] + b[2]*X[t-2] + e[t], e ~ Normal(0, sigma^2)
etc. This is a regression of the series on itself lagged by a certain
number of time units, the present is regressed on the past. Function
ar() fits this kind of model to a time series. In the first case, the
order is p=1, in the second, p=2.
Now, in the first case, is there second order serial correlation? Test
the residuals with lag=2, fitdf=1, the value of p. Third order? lag=3,
fitdf=p=1, etc.
You are NOT fitting this type of model, so the Ljung-Box test is
misused. Test the original series with default parameters, lag=1. If
there is serial correlation, fit an AR (Auto-Regressive) model with
ar(). See the help page ?ar. And see a statiscian with experience in
time series. It's a world on its own, I haven't even mentioned
seasonality. And almost everything else about time series.
Do ask someone near you.
Hope this helps,
Rui Barradas
Em 26-06-2012 19:01, Steven Winter escreveu:
> I fit a simple linear model y = bX to a data set today, and that
produced 24 residuals (I have 24 data points, one for each year from
1984-2007). I would like to test the time-independence of the residuals
of my model, and I was recommended by my supervisor to use the Ljung-Box
test. The Box.test function in R takes 4 arguments:
>
> x a numeric vector or univariate time series.
> lag the statistic will be based on lag autocorrelation
> coefficients.
> type test to be performed: partial matching is used.
> fitdf number of degrees of freedom to be subtracted if x is a series
of residuals.
>
> Unfortunately, I never took a statistics class where I learned the
Ljung-Box test, and information about it online is hard to find. What
does "lag" mean, and what value would you guys recommend I use for the
test? Also, what does "fitdf" represent, and what would the value for
that parameter be in my case? Finally, the value of x is a vector of my
24 residuals, correct?
>
> Thank you all so much. I apologize for the basic nature of the question.
>
> Steven
> [[alternative HTML version deleted]]
>
>
>
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