[R-SIG-Finance] Range intersections

[Enrico Schumann]

On Sat, 27 Jul 2013, Mark Knecht <markknecht at gmail.com> writes:

> For the sake of asking here assume I have 5 models that attempt to
> predict a pair of future values like tomorrow's high & low. The
> predictions are placed in a matrix with the low prediction in column 1
> and the high prediction in column 2. _IF_ there is an intersection of
> all 5 predictions then I can test for that using the simple equation
> in Range1. However if there is a prediction that's a complete outlier
> as is Pred2[4,] then the simple equation fails and says the high is
> less than the low.
>
> Are there any R packages that automatically handle simple intersection
> problems like this? My real goals include things like probability
> distributions across the ranges, but for now I'm just looking for
> what's out there?
>
> Thanks,
> Mark
>
>
> Pred1 = matrix(data=NA, nrow=5, ncol=2)
> Pred1[1,] = c(100,110)
> Pred1[2,] = c(88,125)
> Pred1[3,] = c(92,120)
> Pred1[4,] = c(25,105)
> Pred1[5,] = c(91,121)
>
> Range1 = c(max(Pred1[,1]), min(Pred1[,2]))
> Range1
>
> Pred2 = matrix(data=NA, nrow=5, ncol=2)
> Pred2[1,] = c(100,110)
> Pred2[2,] = c(88,125)
> Pred2[3,] = c(92,120)
> Pred2[4,] = c(25,70) #outlier - the high is too low
> Pred2[5,] = c(91,121)
>
> Range2 = c(max(Pred2[,1]), min(Pred2[,2]))
> Range2
>

Yes, taking the intersection "fails" because in the second case the
intersection is empty.  So /you/ will have to decide how you want to
handle such cases.  Perhaps remove the outlier, or use some majority
rule to find a "most likely" outcome (according to your preditions).
For instance, you could count how often a given price falls in the range
of any of the models.


ind <- function(x, min, max) {
    s <- numeric(length(x))
    for (i in seq_along(min))
        s <- s + as.integer(x >= min[i] & x <= max[i])
    s
}

x <- seq(0, 130, by = 0.5)
s <- ind(x, Pred2[ ,1], Pred2[ ,2])
plot(x, s, type = "S")

range(ca <- x[s == max(s)])            ## "consensus area"
all.equal(max(diff(ca)), diff(x)[1L])  ## single area?
 

Regards,
        Enrico

-- 
Enrico Schumann
Lucerne, Switzerland
http://enricoschumann.net