# p-value for complete spatial randomness tests

I am using the Kolmogorov-Smirnov test to study the homogeneity of a point pattern in 3D using the distance transform to a structure as a covariate.

I am using the spatstat book, where it is explained (Section 10.5) that the Kolmogorov-Smirnov test statistic ($D$) can be calculated as the maximum vertical separation between the cumulative distribution function of the spatial covariate evaluated at the data points ($F_0(z)$) and at all locations ($\hat{F}(z)$). It is:

$D = \max\limits_z|\hat{F}(z)-F_0(z)|$

However, I do not understand how to obtain the value $D$ for my null hypothesis (that the points are homogeneously distributed) so that I can compare them and know whether I can reject the null hypothesis.

If there is also a formula to calculate the critical p-value would be excellent, since I read how to calculate it when you compare your data with a defined distribution like Poisson, but in this case I understand that I should use $\hat{F}(z)$ as the distribution to compare.

Furthermore, I could only use this test in spatstat for 2D data, but not for 3D. If will also appreciate any hint about how to do it.

EDIT: Following the example in the spatstat book, what I am doing is:

elev <- bei.extra$elev cdf.test(bei,elev)  Which produces this result: Spatial Kolmogorov-Smirnov test of CSR in two dimensions data: covariate ‘elev’ evaluated at points of ‘bei’ and transformed to uniform distribution under CSR D = 0.10634, p-value < 2.2e-16 alternative hypothesis: two-sided  As far as I understand,$D$is calculated with the formula above, however that says nothing about the rejection of the null hypothesis. Now I would like to know how the p-value can be calculated. As I want to do this test in 3D data, I apply the distance transform in 3D to the structure of interest (covariate), and I compare its result evaluated at all locations$\hat{F}(z)$and at the observed points ($F_0(z)$). Then, I can compare the distributions as in this figure: At this point I know how to calculate$D$, but again, that does not tell if the null hypothesis can be rejected. In this figure it looks obvious that the the points are not homogeneously distributed, but I would like to calculate with which p-value I can reject the null hypothesis. EDIT 2: Two-Sample Kolmogorov-Smirnov test. Up to now I have been using the two-sample Kolmogorov-Smirnov test as Ege Rubak mentioned in his answer. I was using the MATLAB function kstest2 that calculates the p-value based on the number of samples in$x$(length(x)) and$y$(length(y)), where$x$is a vector with the data points and$y$the covariate values at the grid points. The problem is that the size of$y$corresponds to the number of pixels in the image and are not ‘real’ samples. This means that if my image has a small pixel size or I use a smaller grid (independently of the real volume of the sample) it will be easier to reject the null hypothesis. As far as I understand (please correct me if I am wrong) the test should only consider the size of$x$, since it contains the real samples, and not length(y). Now I was checking the R code that Ege Rubak proposed to understand what it is actually doing, and I found that the ks.test function calls the function psmirnov2x in C code (which I could not found). The inputs of this function are$D$(maximum distance between CDFs), length(x) and length(y), so I understand that in this function the p-value is also based on the lengths of both vectors. Am I wrong thinking that this is not the correct approach? Or should I find a different way of doing this test? • It would be helpful if you could provide an example of what you have done so far. If you can't share your real data use some of the built-in datasets of spatstat or generate them artificially. Commented Mar 19, 2017 at 15:46 • Thank you for the interest. I have included more details, I hope that helps. Commented Mar 20, 2017 at 13:08 ## 1 Answer The code in spatstat calls ks.test() which is a default R function (in stats package), so the actual calculation of the p-value under the null hypothesis is done there. In the spatstat code a few tricks are done (e.g. transformation to uniform distribution using the inverse CDF), so it might not be easy to follow that code. I see two different options: 1. Quick and dirty with two sample KS test. 2. Transform to uniform distribution and use one sample KS test. ## Option 1 I think you can simply calculate the two sample KS test and get what you want. Simply put the covariate values at the data points in a vector x and the covariate values at the grid points in a vector y and then do ks.test(x,y). In principle the more grid points you can afford the better. As you add more grid points I expect the p-value converges to the optimal you can get from a two sample KS test. It may however still be different (probably conservative) than what the p-value a one sample KS test would give. Following the example from the book: require(spatstat, quietly = TRUE) #> #> spatstat 1.49-0.011 (nickname: 'Instinctively Correct') #> For an introduction to spatstat, type 'beginner' # Covariate image elev <- bei.extra$elev
# Data values:
x <- elev[bei]
# Grid values (simply extracted from image):
y <- as.matrix(elev)
# Test
test <- ks.test(x,y)
#> Warning in ks.test(x, y): p-value will be approximate in the presence of
#> ties
test
#>
#>  Two-sample Kolmogorov-Smirnov test
#>
#> data:  x and y
#> D = 0.10568, p-value < 2.2e-16
#> alternative hypothesis: two-sided


The issue with tied values is solved in spatstat by jittering values, but I will leave that to you.

## Option 2

If you want to use a one sample test you can use that under the null hypothesis Z(xi) (covariate values at data points) has theoretical CDF FZ (true spatial CDF of covariate Z) and then F−1(Z(xi)) is uniformly distributed, so you can make this transformation and do ks.test(x, "punif"), with the transformed values in x. This is effectively what is happening in spatstat. I don't expect you will get a huge difference from the two sample test, but it might depend on the number of data points you have.

• Please see "EDIT 2: Two-Sample Kolmogorov-Smirnov test" in my question. Thank you again. Commented Mar 21, 2017 at 19:38
• Hi @alvgom I have now edited the question. Does that help? Commented Mar 23, 2017 at 6:04
• Yes, it helped a lot. I tried both options and you were right, the critical p-value is almost the same when there are no ties. In the presence of ties I found the first option is more robust, which makes sense considering the effect of ties when calculating the inverse CDF. Thank you for your help. Commented Mar 25, 2017 at 20:13