[R] P-values Kolmogorov–Smirnov test

Boo G. g|@n|uc@@boo @end|ng |rom @oton@@c@uk
Thu Sep 5 21:09:11 CEST 2019

```Hello again.

I have tied this before but I see two problems:

1) According to the documentation I could read (including the ks.test code), the ks statistic would be max(abs(x - y)) and if you plot this for very low sample sizes you can actually see that this make sense. The results of ks.test(x, y) yields very different values.

2) Also in this case the p-values don’t make much sense, according to my previous interpretation.

Again, I could be wrong in my interpretation.

Hello,

I'm sorry, but apparently I missed the point of your problem.

But you can use ks.test, just in a different way than what I wrote previously.

Corrected code:

#simulation
for (i in 1:1000) {
#sample from the reference distribution
m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
m_2 <-m_2[order(m_2\$d_1),]
d_2 <- m_2\$d_1
p_2 <- m_2\$p_1

#weighted ecdf for the reference distribution and the sample
f_d_1 <- ewcdf(d_1, normalise=F)
f_d_2 <- ewcdf(d_2, 1/p_2, normalise=F, adjust=1/length(d_2))

#kolmogorov-smirnov distance
x <- f_d_1(d_2)
y <- f_d_2(d_2)
ht <- ks.test(x, y)
d_stat[i, 2] <- ht\$statistic
d_stat[i, 3] <- ht\$p.value
}

Hope this helps,

Às 19:29 de 05/09/19, Rui Barradas escreveu:
Hello,
I don't have the algorithms at hand but the KS statistic calculation is more complicated than your max/abs difference.
Anyway, why not use ks.test? it's not that difficult:
set.seed(1234)
#reference distribution
d_1 <- sort(rpois(1000, 500))
p_1 <- d_1/sum(d_1)
m_1 <- data.frame(d_1, p_1)
#data frame to store the values of the simulation
d_stat <- data.frame(1:1000, NA, NA)
names(d_stat) <- c("sample_size", "ks_distance", "p_value")
#simulation
for (i in 1:1000) {
#sample from the reference distribution
m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
d_2 <- m_2\$d_1
ht <- ks.test(d_1, d_2)
#kolmogorov-smirnov distance
d_stat[i, 2] <- ht\$statistic
d_stat[i, 3] <- ht\$p.value
}
hist(d_stat[, 2])
hist(d_stat[, 3])
Note that d_2 is not sorted, but the results are equal in the sense of function identical(), meaning they are *exactly* the same. Why shouldn't they?
Hope this helps,
Às 17:06 de 05/09/19, Boo G. escreveu:
Hello,

I am trying to perform a Kolmogorov–Smirnov test to assess the difference between a distribution and samples drawn proportionally to size of different sizes. I managed to compute the Kolmogorov–Smirnov distance but I am lost with the p-value. I have looked into the ks.test function unsuccessfully. Can anyone help me with computing p-values for a two-tailed test?

Below a simplified version of my code.

Gianluca

library(spatstat)

#reference distribution
d_1 <- sort(rpois(1000, 500))
p_1 <- d_1/sum(d_1)
m_1 <- data.frame(d_1, p_1)

#data frame to store the values of the siumation
d_stat <- data.frame(1:1000, NA, NA)
names(d_stat) <- c("sample_size", "ks_distance", "p_value")

#simulation
for (i in 1:1000) {
#sample from the reference distribution
m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
m_2 <-m_2[order(m_2\$d_1),]
d_2 <- m_2\$d_1
p_2 <- m_2\$p_1

#weighted ecdf for the reference distribution and the sample
f_d_1 <- ewcdf(d_1, normalise=F)
f_d_2 <- ewcdf(d_2, 1/p_2, normalise=F, adjust=1/length(d_2))

#kolmogorov-smirnov distance
d_stat[i,2] <- max(abs(f_d_1(d_2) - f_d_2(d_2)))
}

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