ks.test with left truncated weibull - r

Question

The Distribution I accept my values to follow is a left truncated weibull distribution. I do know the parameters a, shape and scale of this distribution for using the ptrunc command:
require(truncdist);
ptrunc(x,"weibull",a=a,scale=b,shape=c)
so I want the ks.test command (see below) to use the described left truncated weibull distribution instead of the "normal weibull".
myvalues<-c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4);
a<-7;
scale<-36.37516;
shape<-9.437013;
So I do know, that in this case it is not necessary to do the left-side truncation. But in others it will be.
ks.test(myvalues,"pweibull",scale=b,shape=c) #for normal weibull
but
ks.test(myvalues,ptrunc(x,"weibull",a=a,scale=b,shape=c)) # for leftruncated
gives a wrong result.

Answer 1

First of all, ptrunc should be replaced by rtrunc. ptrunc gives a vector of probability values. But by the documentation of ks.test we need a sample, and this is what rtrunc gives us. If the argument a of rtrunc is set to -Inf, there is no truncation and the result with a=-Inf is indeed the same as with a=7:
library(truncdist)
myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)
a <- 7
scale<-36.37516
shape <- 9.437013
set.seed(1)
y1 <- rtrunc(myvalues,"weibull",a=-Inf,scale=scale,shape=shape)
set.seed(1)
y2 <- rtrunc(myvalues,"weibull",a=a,scale=scale,shape=shape)
set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape )
set.seed(1)
ks1 <- ks.test( myvalues, y1 )
set.seed(1)
ks2 <- ks.test( myvalues, y2 )
.
> ks1
Two-sample Kolmogorov-Smirnov test
data: myvalues and y1
D = 0.21429, p-value = 0.2898
alternative hypothesis: two-sided
> ks2
Two-sample Kolmogorov-Smirnov test
data: myvalues and y2
D = 0.21429, p-value = 0.2898
alternative hypothesis: two-sided
But still the result of ks.test( myvalues, "pweibull",scale=scale,shape=shape ) is different:
> ks0
One-sample Kolmogorov-Smirnov test
data: myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided
The reason is that myvalues is too small. If we make it larger in the call of rtrunc (not ks.test), ks0, ks1, and ks2 are almost the same:
library(truncdist)
myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)
myManyValues <- c(outer((0:9999)/100000,myvalues,"+"))
a <- 7
scale<-36.37516
shape <- 9.437013
set.seed(1)
y1 <- rtrunc(myManyValues,"weibull",a=-Inf,scale=scale,shape=shape)
set.seed(1)
y2 <- rtrunc(myManyValues,"weibull",a=a,scale=scale,shape=shape)
set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape )
set.seed(1)
ks1 <- ks.test( myvalues, y1 )
set.seed(1)
ks2 <- ks.test( myvalues, y2 )
.
> ks0
One-sample Kolmogorov-Smirnov test
data: myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided
> ks1
Two-sample Kolmogorov-Smirnov test
data: myvalues and y1
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided
> ks2
Two-sample Kolmogorov-Smirnov test
data: myvalues and y2
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided
Now let's see what happens when we do truncate the distribution:
library(truncdist)
myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4)
myManyValues <- c(outer((0:9999)/100000,myvalues,"+"))
a <- 29
scale<-36.37516
shape <- 9.437013
set.seed(1)
y1 <- rtrunc(myManyValues,"weibull",a=-Inf,scale=scale,shape=shape)
set.seed(1)
y2 <- rtrunc(myManyValues,"weibull",a=a,scale=scale,shape=shape)
set.seed(1)
ks0 <- ks.test( myvalues, "pweibull",scale=scale,shape=shape )
set.seed(1)
ks1 <- ks.test( myvalues, y1 )
set.seed(1)
ks2 <- ks.test( myvalues, y2 )
.
> ks0
One-sample Kolmogorov-Smirnov test
data: myvalues
D = 0.15612, p-value = 0.2576
alternative hypothesis: two-sided
> ks1
Two-sample Kolmogorov-Smirnov test
data: myvalues and y1
D = 0.15655, p-value = 0.2548
alternative hypothesis: two-sided
> ks2
Two-sample Kolmogorov-Smirnov test
data: myvalues and y2
D = 0.2059, p-value = 0.05683
alternative hypothesis: two-sided

Answer 2

You are using the ptrunc function incorrectly (I assume), it needs to be fed a sequence of quantiles. Below I calculate the mean and standard deviation of the Weibull based on your scale and shape parameters, then sample from 5 standard deviations above and below to produce a comparison set.
require(truncdist);
myvalues <- c(37.5, 35.4, 27.1, 32.9, 35.9, 35.1, 34.1, 32.5, 35.5, 31.5, 38.2, 36.1,29.9, 30.1, 34.7, 38.7 ,32.3, 38.0, 34.9, 44.2, 35.8, 30.8, 39.3, 26.0, 34.2, 40.0, 36.1 ,41.5 ,32.8, 31.9, 41.3 ,30.5, 39.9, 35.0 ,31.2 ,35.0, 30.3, 29.0, 34.4, 35.7, 34.1, 35.4);
a <- 7;
scale <- 36.37516;
shape <- 9.437013;
# Calculate standard deviation of the weibull
weib_mean <- scale * gamma(1 + 1/shape)
weib_sd <- sqrt((scale^2) * (gamma(1 + 2/shape) - (gamma(1 + 1/shape))^2))
# Get a sample
quant <- seq(weib_mean - 5 * weib_sd, weib_mean + 5 * weib_sd, length.out = 1E5)
weibull_samp <- ptrunc(quant, "weibull", a = a, scale = scale, shape = shape)
# Take a look
plot(weibull_samp ~ quant)
# Use with test
> ks.test(sort(myvalues), weibull_samp)
Two-sample Kolmogorov-Smirnov test
data: sort(myvalues) and weibull_samp
D = 1, p-value < 2.2e-16
alternative hypothesis: two-sided