R – Ties in a Two-Sample Kolmogorov-Smirnov Test

kolmogorov-smirnov testrties

This question is somehow connected to this one.
I am performing a two-sample KS test in R and I think I have not fully understood the issue of the ties.

Reading the help: The presence of ties always generates a warning, since continuous distributions do not generate them. If the ties arose from rounding the tests may be approximately valid, but even modest amounts of rounding can have a significant effect on the calculated statistic.
So, I understand this in the case of a single sample, but why do I get the same warning if the tie is represented by the same value present in the two vectors?

Example:

no ties case

set.seed(123)
x <- rnorm(50)
y <- runif(30)
ks.test(x, y)

Two-sample Kolmogorov-Smirnov test
data:  x and y
D = 0.52, p-value = 3.885e-05
alternative hypothesis: two-sided

case with ties

x <- c(0,1,1, rnorm(47)) # this vector has the value 1 repeated twice
y <- c(1,runif(29))
ks.test(x, y)

Two-sample Kolmogorov-Smirnov test
data:  x and y
D = 0.5, p-value = 0.0001696
alternative hypothesis: two-sided
Warning message:
In ks.test(x, y) : cannot compute exact p-value with ties

case I thought it shouldn't be tied, but in fact it is:

x <- c(0,1,1, rnorm(47))
y <- c(1,runif(29))
ks.test(unique(x), unique(y))

Two-sample Kolmogorov-Smirnov test
data:  unique(x) and unique(y)
D = 0.59184, p-value = 4.363e-06
alternative hypothesis: two-sided
Warning message:
In ks.test(unique(x), unique(y)) : cannot compute exact p-value with ties

Best Answer

The reason in the one-sample case is exactly the same reason in the two-sample case: in general, $Pr(X = c) = 0$ for some continuously distributed $X$ and some single value $c$. Ties (single sample or two sample) imply that $Pr(X = c) \ne 0$.

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