# Clustering computation in pair approximation model

Let's consider a square lattice of cells. Each cell can be either occupied by a species (1 or 2) or be empty (0). Each cell can be either in state 1, 2 or 0.

In the pair approximation model, I would like to compute the clustering of the species, i.e. the clustering of the occupied cells ($$+$$). It is define as:

• $$C_{++} = \frac{q_{+|+}}{\rho_{+}} = \frac{\rho_{++}}{\rho_{+}^2}$$

Where

• $$q_{+|+}$$ is the conditional probability to find an occupied cell in the surrounding cells knowing that the focal cell is occupied
• $$\rho_+$$ is the density of occupied cells in the landscape, defines as: $$\rho_+ = \rho_1 + \rho_2$$
• $$\rho_{++}$$ is the density of occupied cell pairs in the landscape, defines as: $$\rho_{++} = \rho_{11} + \rho_{12} + \rho_{21} + \rho_{22}$$

• $$q_{i|j} = \frac{\rho_{ij}}{\rho_{i}}$$

One approach successfully describes the clustering but the second does not and I do not find why.

## A first approach (Works)

• $$q_{+|+} = \frac{\rho_{++}}{\rho_{+}}$$

• $$\rho_{++} = \rho_{11} + \rho_{12} + \rho_{21} + \rho_{22}$$

Knowing that $$\rho_{12} = \rho_{21}$$:

• $$q_{+|+} = \frac{\rho_{11} + 2\rho_{12} + \rho_{22}}{\rho_{1} + \rho_{2}}$$

Hence:

\begin{align} C_{++} & = \frac{\rho_{11} + 2\rho_{12} + \rho_{22}}{\rho_{1} + \rho_{2}} \times \frac{1}{\rho_{+}} \\ & = \frac{\rho_{11} + 2\rho_{12} + \rho_{22}}{(\rho_{1} + \rho_{2})^2} \end{align}

This approach works, I have checked it by running simulation of cellular automata.

## A second approach (Do not works)

Let's define $$q_{+|+}$$.

It is the probability for one cell of state $$1$$ to be surrounded by occupied cells + the probability for one cell of state $$2$$ to be surrounded by occupied cells. It means that:

$$q_{+|+} = q_{+|1} + q_{+|2}$$

and q_{+|1} and q_{+|2} can be defined as:

• $$q_{+|1} = q_{1|1} + q_{2|1}$$
• $$q_{+|2} = q_{1|2} + q_{2|2}$$

So:

\begin{align} q_{+|+} & = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2} \\ & = \frac{\rho_{11}}{\rho_{1}} \frac{\rho_{12}}{\rho_{1}} + \frac{\rho_{21}}{\rho_{2}} + \frac{\rho_{22}}{\rho_{2}} \\ & = \frac{\rho_{11} + \rho_{12}}{\rho_{1}} + \frac{\rho_{21} + \rho_{22}}{\rho_{2}} \\ & = \frac{\rho_{2}(\rho_{11} + \rho_{12})}{\rho_{1}\rho_{2}} + \frac{\rho_{1}(\rho_{21} + \rho_{22})}{\rho_{1}\rho_{2}} \\ & = \frac{\rho_{2}(\rho_{11} + \rho_{12}) + \rho_{1}(\rho_{21} + \rho_{22}) }{\rho_{1}\rho_{2}} \\ \end{align}

Obviously this formulation of $$q_{+|+}$$ is different for the first one. In simulation, this formulation gives $$C_{++}$$ twice higher than with a first approach.

I do not know what is the mistake in this approach but I guess that the following assertion is false:

$$q_{+|+} = q_{1|1} + q_{2|1} + q_{1|2} + q_{2|2}$$

I would be very helpful for me to know why.

$$q_{+|+} = \frac{\rho_{++}}{\rho_{+}} = \frac{\rho_{11} + \rho_{12} + \rho_{21} + \rho_{22}}{\rho_{1} + \rho_{2}} \ne \frac{\rho_{11} + \rho_{12}}{\rho_{1}} + \frac{\rho_{21} + \rho_{22}}{\rho_{2}} = \frac{\rho_{+1}}{\rho_{1}} + \frac{\rho_{+2}}{\rho_{2}} = q_{+|1} + q_{+|2}.$$
In particular, if $$\rho_{1} = \rho_{2}$$, then $$\rho_{1} + \rho_{2} = 2\rho_{1} = 2\rho_{2}$$, and so the left hand side will work out to exactly $$\frac12$$ times the right hand side.
(The same will also happen if $$q_{+|1} = q_{+|2}$$; in particular, that means that the assertion will always be off by exactly a factor of $$\frac12$$ for well mixed systems, where $$q_{a|b} = \rho_a$$ for all states $$a$$ and $$b$$. On the other extreme, if the states 1 and 2 are highly clustered such that $$\rho_{12} = \rho_{21} \approx 0$$, you can basically have $$q_{+|1}$$ and $$q_{+|2}$$ take any arbitrary values independently of each other and have $$q_{+|+}$$ be anywhere in between them, depending on the ratio of $$\rho_1$$ and $$\rho_2$$.)
The source of your confusion seems to be a basic misunderstanding of conditional probabilities. In particular, while it's true that $$\mathrm{Pr}[A \text{ or } B \mid C] = \mathrm{Pr}[A \mid C] + \mathrm{Pr}[B \mid C]$$ whenever $$A$$ and $$B$$ are mutually exclusive events, this additivity only holds when the probabilities are conditioned on the same event $$C$$. In deriving your incorrect formula, you've basically asserted that $$\mathrm{Pr}[C \mid A \text{ or } B] = \mathrm{Pr}[C \mid A] + \mathrm{Pr}[C \mid B]$$, which does not hold except in degenerate cases.