Re: [R] help with ARIMA and predict

From: Spencer Graves <spencer.graves_at_pdf.com>
Date: Sat 09 Jul 2005 - 15:03:09 EST

          What class is "newx" when you get the error message? Is it a vector or a 1xK array? If the former, force it to be an array. (Hint: "array(1:4, dim=c(2,2))[1,]" is a vector of length 2, while "array(1:4, dim=c(2,2))[1,,drop=FALSE]" is a 1 x 2 matrix..(

          If this does not solve your problem, then I suggest you make up a few very simple examples where you know the the answers under the most plausible thoughts about what it's doing, and try those. I've also learned a lot by copying the arima function into a script file and walking through it line by line.

          If none of this work for you, I suggest you read the posting guide! "http://www.R-project.org/posting-guide.html". People often answer their own questions from following suggestions in in the posting guide.   If that fails, they seem more likely to get useful answers from the questions they post after following this guide.

          spencer graves

Brian Scholl wrote:

> I'm trying to do the following out of sample
> regression with autoregressive terms and additional x
> variables:
>
> y(t+1)=const+B(L)*y(t)+C(1)*x_1(t)...+C(K)*x_K(t)
>
> where:
> B(L) = lag polynom. for AR terms
> C(1..K) = are the coeffs. on K exogenous variables
> that have only 1 lag
>
> Question 1:
> -----------
>
> Suppose I use arima to fit the model:
>
> df.y<-arima(yvec,order=c(L,0,0),xreg=xmat[,(1:K)],n.cond=maximum.lag)
>
>
> Now suppose I want to do a 1-period ahead prediction
> based on the results of this regression, using
> predict:
>
> predict(df.y,newxreg=newx,n.ahead=1)
>
> I'm expecting newx to be 1X3. After all, I just want
> to predict 1 value of y, so in my mind I should just
> need 1 time period's observation of x (i.e. #
> rows=n.ahead). I'm sort of expecting predict to grab
> the last two values of yvec to use as y(t),y(t-1) in
> prediction. If I make such a pass, I get:
>
> Error in predict.Arima(df.y, newxreg = newx) :
> 'xreg' and 'newxreg' have different numbers of columns
>
> If I try passing 2+ rows of x, predict accepts the
> call and I get:
>
> Time Series:
> Start = 41
> End = 42
> Frequency = 1
> [1] -0.03165 -0.03165 (for simplicity I passed two
> identical rows of x)
>
> $se
> Time Series:
> Start = 41
> End = 41
> Frequency = 1
> [1] 0.02707
>
> So I'm puzzled as to what I'm doing wrong. When I
> have n.ahead rows in newxreg, I get an error, but by
> passing a second row in it is accepted. But what am I
> predicting in the latter case? Is R requiring another
> row so that it can form a prediction of y(t) to use in
> forecasting y(t+1) (this is not what I want to do), or
> have I simply goofed in some other way?
>
> Is there a better way to do this? I've also attempted
> something similar using lm, but I'm unclear how to
> interpret the "predicted" time series it returns.
> The obvious alternative is to construct the forecast
> using df.y$coef and a relevant data vector.
>
> Q2:
> ---
>
> Suppose I want to select the autoregressive order
> using AIC. If I have understood, in the excellent
> MASS text comments (p415) that comparisons are only
> valid if n.cond is the same for each model. Yet, when
> I set n.cond=maximum.lag (say =5), I get df.y$n.cond
> =0. So I'm unclear if the AICs are comparable for
> different models (i.e. different L's and different
> K's).
>
> ______________________________________________
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> PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html

-- 
Spencer Graves, PhD
Senior Development Engineer
PDF Solutions, Inc.
333 West San Carlos Street Suite 700
San Jose, CA 95110, USA

spencer.graves@pdf.com
www.pdf.com <http://www.pdf.com>
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Received on Sat Jul 09 15:09:37 2005

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