[R] Bootstrapping the prediction intervals for regression

[Rui Barradas]

Hello,

The jackknife is used as a bias reduction technique, and since linear 
regression estimates are unbiased I don't see why you should use it.

Rui Barradas

Em 15-05-2014 19:21, varin sacha escreveu:
> Thanks Bert for your suggestion that is working.
>
> To answer to your question, I can say that some econometricians say that
> using bootstrap techniques on a linear regression model when the sample
> size N is small, one of the most interesting purpose is on the
> prediction intervals which is better fitted.
> For me, the bootstrap will give results that are at least as good as the
> normal approximation.
>
> When the sample size N is small, are bootstrap or jackknife better than
> the standard methods for estimating the best fit parameters ?
> What is your opinion about that question ?
>
> Best
>
> Le Jeudi 15 mai 2014 14h44, Bert Gunter <gunter.berton at gene.com> a écrit :
> Please note that this can (and should) be considerably sped up by
> taking advantage of the fact that lm() will work on a matrix of
> responses. Also, some improvement in speed can usually be obtained by
> generating all samples at once rather than generating the sample each
> time within a loop.
>
> something like (untested):
>
> nr <- nrow(dat)
> ymat <- matrix(with(dat,sample(y,100*nr,rep=TRUE)), nrow =nr)
> out <- lm(ymat ~x1+x2,data=dat)
> b <- predict(out)
>
> However, why do this anyway, as for linear models the standard closed
> form solutions are available?
>
> Cheers,
> Bert
>
> Bert Gunter
> Genentech Nonclinical Biostatistics
> (650) 467-7374
>
> "Data is not information. Information is not knowledge. And knowledge
> is certainly not wisdom."
> H. Gilbert Welch
>
>
>
>
> On Thu, May 15, 2014 at 4:01 AM, Rui Barradas <ruipbarradas at sapo.pt
> <mailto:ruipbarradas at sapo.pt>> wrote:
>  > Hello,
>  >
>  > Try to follow the example below and see if you can adapt it to your
> needs.
>  > Since you don't provide us with a dataset example, I start by making up
>  > some.
>  >
>  >
>  > # make up some data
>  > n <- 22
>  > set.seed(8873)
>  > dat <- data.frame(x1 = rnorm(n), x2 = rnorm(n))
>  > dat$y <- x1 + x2 + rnorm(n)
>  >
>  >
>  > B <- 100  # number of bootstrap samples
>  > result <- array(dim = c(n, 3, B), dimnames = list(NULL, c("fit", "upr",
>  > "lwr"), NULL))
>  > for(i in 1:B){
>  >        s <- sample(nrow(dat), n, replace = TRUE)
>  >        lm.tmp <- lm(y ~ x1 + x2, data = dat[s, ])
>  >        result[,,i] <- predict(lm.tmp, interval = "prediction")
>  > }
>  >
>  >
>  > Then you can do whatever you want with 'result', including computing
> the min
>  > and max values.
>  >
>  >
>  > Hope this helps,
>  >
>  > Rui Barradas
>  >
>  > Em 15-05-2014 10:37, varin sacha escreveu:
>  >>
>  >> Dear experts,
>  >>>
>  >>>
>  >>>
>  >>> I have done a multiple linear regression on a small sample size (n=22).
>  >>> I have computed the prediction intervals (not the confidence
> intervals).
>  >>>
>  >>>
>  >>> Now I am trying to bootstrap the prediction intervals.
>  >>>
>  >>>
>  >>> I didn't find any package doing that.
>  >>> So I decide to create my own R function, but it doesn't work !
>  >>>
>  >>>
>  >>> Here are my R codes :
>  >>>
>  >>>
>  >>> LinearModel.1 <- lm(GDP.per.head ~ Competitivness.score +Quality.score,
>  >>> data=Dataset)
>  >>> summary(LinearModel.1)
>  >>> predict(LinearModel.1, interval = "prediction")
>  >>>
>  >>>
>  >>> HERE IS MY R FUNCTION WHERE I HAVE TRIED TO BOOTSTRAP THE PREDICTION
>  >>> INTERVALS
>  >>>
>  >>>
>  >>> pred.min<-rep(nrow(Dataset), na.rm=F)
>  >>> pred.max<-rep(nrow(Dataset), na.rm=F)
>  >>> for(i in 1:n)
>  >>> {s<-sample(1:nrow(Dataset),size=22)
>  >>> reg<-lm(GDP.per.head ~ Competitivness.score +
>  >>> Quality.score,data=Dataset[s])
>  >>> pred.min<-pmin(reg,pred.min)
>  >>> pred.max<-pmax(reg,pred.max)}
>  >>>
>  >>>
>  >>> Thanks for your precious help.
>  >>>
>  >>>
>  >>        [[alternative HTML version deleted]]
>  >>
>  >>
>  >>
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>