Testing CSR

Monte Carlo Tests

Let $T$ be any test statistic where larger $T$ cast doubt on the null hypothesis.
Let $t_1$ be the value of $T$ calculated from dataset.
For convenience, assume that the null sampling distribution of $T$ is continuous.

Let $t_2,\ldots,t_s$ be the values of $T$ calculated from $s-1$ independent simulations of $H_0$ . Then under $H_0$ , the values $t_1,\ldots,t_s$ are exchangeable, i.e.

$P(t_i=t_{(j)})=\frac{1}{s}, j=1,\ldots,s$

Hence, if $R$ denotes the number of $t_i>t_1$ then $P(R\leq r) = \frac{r+1}{s}.$ Which means that the p-value of Monte Carlo test is $(r+1)/s$ .

Inter-event distance based Test

Let $T$ be the distance between two events independently and uniformly distributed in $A$ .
For a unit square of $A$
$H(t) &= P(T\leq t) \\ &= \begin{cases} \pi t^2 - {8\over 3}t^3+{1\over 2}t^4,\quad 0\leq t\leq 1 \\ {1\over3}-2t^2-{1\over2}t^4+{4\over3}(t^2-1)^{1\over2}(2t^2+1)+2t^2\sin^{-1}(2t^{-2}-1),\quad 1\leq t\leq \sqrt{2} \end{cases} \end{aligned}$$$
For a circle of unit radius $A$ ,
$H(t) = 1+\pi^{-1}[2(t^2-1)\cos^{-1}({t\over2})-t(1+{t^2\over2})\sqrt{1-{t^2\over4}}],\quad 0\leq t\leq2 \end{aligned}$$$
Consider empirical distribution function(EDF) of inter-event distances as:

$\hat{H}_1(t)={2\over n(n-1)}\sum_{i<j}I(t_{ij}\leq t)$

where $t_{ij}$ are observed inter-event distances from data.
Monte Carlo-based approach is used for this test.
Generating Monte-Carlo Samples

Generate $s-1$ times of $n$ events in $A$ under CSR assumption

Calculate $\hat{H}_i(t), i=2,\ldots,s$

Calculate envelopes:

$\begin{aligned} U(t) &=\max_{2\leq i\leq s}\lbrace \hat{H}_i(t)\rbrace \\ L(t) &=\min_{2\leq i\leq s}\lbrace \hat{H}_i(t)\rbrace \end{aligned}$

Two common MC test approaches
1. Choose appropriate $t_0$ and define $u_i=\hat{H}_i(t_0)$ under CSR. Note that under $H_0$ at MC test,
  
  $P(u_1=u_{(j)})=1/s$
  
  If $u_1$ ranks $k$ th largest or higher than $k$ th, the test that rejects CSR based on that gives an exact one-sided test of size $k/s$ .
  
  example : $k=5, s=100, u_1\geq u_{(5)}$ then size = 0.05
2. Define
  
  $u_i = \int(\hat{H}_i(t)-H(t))^2 dt.\tag{*}$
  
  Then proceed to a test based on the rank of $u_1$ .
Note that the approach 2 is more objective but known to have weak power.

Nearest neighbor distance based Test

Let $Y$ be the nearest neighbor distance under CSR when there are $n$ events in a region $A$
Theorical distribution of $Y$ is quite difficult, instead use an approximation.
Note that an event being within distance $y$ from known(specified) event is

$\frac{\pi y^2}{|A|}$

Then, the CDF can be approximated by as follows:
$G(y)=P(Y\leq y)&\approx 1-(1-\pi y^2|A|^{-1})^{n-1} \\ &\approx 1-\exp(-\lambda\pi y^2),\quad y\geq 0 \end{aligned}$$ where (2) is a further approximation with large $n$ with $\lambda=n\vert A\vert^{-1}$.$
Empirical CDF is gien as:

$\hat{G}_1(y) = \frac{1}{n}\sum_i I(y_i\leq y)$
Let $\bar{G}_i(y)=\frac{1}{s-1}\sum_{j\neq i}\hat{G}_j(y)$ then the MC test is given as same as (*) where

$u_i = \int(\hat{G} _i(y)-\bar{G}_i(y))^2 dt.$