Also look at zoo's rollapply.
MW
On Apr 2, 2013, at 13:51, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> Hello,
>
> The error comes from NAs where you would expect coefficients. Try the
> following.
>
>
>
> set.seed(7511) # Make the example reproducible
>
> N <- 100
> Time <-1:N
> Price <- rnorm(N, 8, 2)
>
> estim <- numeric(N) # It's better to allocate the results
> error <- numeric(N) # vectors in advance
> for (i in Time) {
> regr <- lm(Price[1:i] ~ Time[1:i])
> estim[i] <- coef(regr)[2]
> if(is.na(coef(regr)[2]))
> error[i] <- NA
> else
> error[i] <- summary(regr)$coefficients[2,2]
>
> }
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 02-04-2013 17:44, triskell4-umbre...@yahoo.fr escreveu:
>> Hello,
>>
>> Some context: let's say I have a data series (let's call it PRICE, for
>> simplicity), sample size N. I have a desire to regress that on TIME, and
>> then to use the TIME and intercept coefficients to predict the price in the
>> next period and to use the standard error to calculate a confidence
>> interval. This is all very easy.
>>
>> However, what I need help for is to calculate a confidence interval for each
>> point in time: imagining that at the end of the 10th period I have 10 data
>> points, and wish to regress them on the 10 periods to create a confidence
>> interval for the next 'predicted' price. And so on from TIME[10:100]. So
>> the first regression would be of PRICE[1:10] on TIME[1:10], the second of
>> PRICE[1:11] on TIME[1:11], the third of PRICE[1:11] on TIME[1:11], and so on
>> to PRICE[1:N] and TIME[1:N]. I'd like to be able to vary the starting point
>> (so it would need to be an argument in the function, in this case it would
>> be 10). The ultimate output of the code would be to save the TIME
>> coefficients and standard errors it generates to two vectors, say TIME.coef
>> and TIME.SE.
>>
>> I'm not sure if lapply() can be bent to my will, or if a for loop would be
>> too inefficient, or what. I'm not new to R, but I'm fairly new to this kind
>> of programming.
>>
>> This is a bungled mess of a narrative, and I apologize. Please feel free to
>> use TIME=1:100 and PRICE=rnorm(100,8,2).
>>
>> Here's an attempt, which has failed for reasons I can only imagine. Any
>> help getting this to work would be greatly appreciated. Any help doing this
>> without loops would be even better.
>>
>>
>>> Time=1:100
>>> Price=rnorm(100,8,2)
>>>
>>> estim=0 #I'm hoping this will be the Time coefficient
>>> error=0 #I'm hoping this will be the standard error of the Time
>>> coefficient
>>> for (i in Time) {
>> + regr=lm(Price[1:i]~Time[1:i])
>> + estim=c(estim,coef(summary(regr))[2,1])
>> + error=c(error,coef(summary(regr))[2,1])
>> + }
>> Error: subscript out of bounds
>>
>>
>> Many, many thanks in advance.
>>
>> Mendi
>> [[alternative HTML version deleted]]
>>
>>
>>
>> ______________________________________________
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
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