[R] P-values Kolmogorov–Smirnov test

Rui Barradas ru|pb@rr@d@@ @end|ng |rom @@po@pt
Thu Sep 5 21:21:14 CEST 2019 

Hello,

Inline.

Às 20:09 de 05/09/19, Boo G. escreveu:
> 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.

The problem is that the distribution of Dn is very difficult to compute. 
 From the reference [1] in the help page ?ks.test:

  	Kolmogorov's goodness-of-fit measure, Dn , for a sample CDF has 
consistently been set aside for methods such as the D+n or D-n of 
Smirnov, primarily, it seems, because of the difficulty of computing the 
distribution of Dn . As far as we know, no easy way to compute that 
distribution has ever been provided in the 70+ years since Kolmogorov's 
fundamental paper. We provide one here, a C procedure that provides 
Pr(Dn < d) with 13-15 digit accuracy for n ranging from 2 to at least 16000.


That is why I used ks.test and its Dn and p-values. Note that n >= 2, 
size = 1 is not covered (p-value == 1).

Also, the p-values distribution seem to become closer to a uniform with 
increasing sizes. Try

hist(d_stat[801:1000, 3])


[1]https://www.jstatsoft.org/article/view/v008i18


Hope this helps,

Rui Barradas

> 
> 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.
> 
>> On 5 Sep 2019, at 20:46, Rui Barradas <ruipbarradas using sapo.pt 
>> <mailto:ruipbarradas using sapo.pt>> wrote:
>>
>> Hello,
>>
>> I'm sorry, but apparently I missed the point of your problem.
>> Please do not take my previous answer seriously.
>>
>> 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$p.value
>> }
>>
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>> Às 19:29 de 05/09/19, Rui Barradas escreveu:
>>> Hello,
>>> I don't have the algorithms at hand but the KS statistic calculation 
>>> is more complicated than your max/abs difference.
>>> Anyway, why not use ks.test? it's not that difficult:
>>> set.seed(1234)
>>> #reference distribution
>>> d_1 <- sort(rpois(1000, 500))
>>> p_1 <- d_1/sum(d_1)
>>> m_1 <- data.frame(d_1, p_1)
>>> #data frame to store the values of the simulation
>>> d_stat <- data.frame(1:1000, NA, NA)
>>> names(d_stat) <- c("sample_size", "ks_distance", "p_value")
>>> #simulation
>>> for (i in 1:1000) {
>>> #sample from the reference distribution
>>> m_2 <-m_1[(sample(nrow(m_1), size=i, prob=p_1, replace=F)),]
>>> d_2 <- m_2$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,
>>> Rui Barradas
>>> 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.
>>>>
>>>> Thanks in advance.
>>>> 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)))
>>>> }
>>>>
>>>>
>>>>     [[alternative HTML version deleted]]
>>>>
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