Hello,
I would like to use modFit and modCost from the package FME to find
the optimal initial time t0 of a process. For simplicity, the process
is either "off" (value 0) or "on" (value h). So I have a
data vector
with some zeros followed by some h's, e.g.
> c(0,0,0,2,2,2,2,2,2,2)
[1] 0 0 0 2 2 2 2 2 2 2
(hence h=2 here). I want to find the best guess for the initial time
t0 when the process starts, in the example the process starts at t0 3 (in fact,
t0 is any number between 3 and 4. All we know is that the
process must have started before t=4). And I want to use modFit and
modCost for this (in this example it's of course trivial, it suffices
to find the first non-zero entry, but it become more challenging when
e.g. the data is noisy).
However, it seems that modFit can not do this, at least not in a
straight forward manner. Here is my minimal example:
###########
library(FME)
#function makeframe creates a vector with 10 entries, the first t0
entries are 0, the remaining entries are h. Output is a #data.frame
with "time" (1,2,...,10) and "M" (0,...,0,h,...,h)
makeframe <- function(Par){
with(as.list(c(Par)),{
t <- seq(1,10,by=1)
vec <- h*as.numeric(t>t0)
return(data.frame(time=t,M=vec))
})
}
#Let the "measurement" be (0,0,0,2,2,2,2,2,2,2)
fakeobs <- makeframe(c(t0=3,h=2))
#cost function provided by modCost
cost <- function(Par){
out <- makeframe(Par)
cost <- modCost(model=out,obs=fakeobs)
return(cost)
}
#Fit for best t0 and h value, initial guess t0=5.5, h=1.2
initial <- c(t0=5.5,h=1.2)
Fit <- modFit(f=cost,p=initial)
coef(Fit) #this should be t0=3 (or and value t0 in the interval [3,4))
and h=2.
#h=2 is fitted correctly, but t0 remains at the initial guess.
#plot results for illustration
plot(fakeobs)
lines(makeframe(initial),lty=2)
lines(makeframe(coef(Fit)))
legend("bottomright",c("data","initial","fitted"),lty=c(NA,
2,1),pch=c(1,NA,NA))
###########
> coef(Fit)
t0 h
5.5 2.0
As you can see, the height h is fitted correctly, but the initial time
t0 remains at the initial guess. I assume this is because the cost
function does not change for any t0 in the interval (5,6), as
makeframe(c(t0=5.5,h=...)) is the same as makeframe(c(t0=5,h=...)).
Hence the numerical solver is at a local minimum for any value of t0
in (5,6), when h=2. So the x-values can not be fitted because they are
discrete; the y-values can be fitted because they are continuous.
Is there a common "trick" used in such a situation? How can I find the
best t0 (more precisely: a best t0)?
Thank you and kind regards!
