Uses of Distirbutions in GIScience

Point Processes

This week, we considered the binomial distirbution and saw that the distirbution can be used to model the probability of sequential, independent events, e.g. a coin toss. In the case of the coin toss, this was a discrete probability distribution that was visualized as a histogram. We can also identify a probability distirbution function, $f(x)$ that describes a continuous probability. In this case, we have some curve (described by $f(x)$) and using that curve can compute the probability of some even taking on some value.

What then, does this look like in the context of point processes? I will use an example drawn from Geographic Information Analysis (2nd Edition), written by David O'Sullivan and Mark Unwin.

Imagine that we have a study are $\mathbb{R^{2}}$ that is broken into 9 quadrants. We can refer to this space partioning as a lattice. In that lattice, we have 10 observations.

What is the probability that an independent observation will occur in any quadrant?

P(A) = P(area of the study area in the given quadrant) = $\dfrac{1}{9}$

This is because the total study area is encompassed by all of our observations and any given quadrant contains only $\dfrac{1}{9}$ of the total study area.

What then, is the probability that a given observation will occur in the shaded grid cell and no other observation will be observed in that frid cell? In other words, what is the probability that a given grid cell will contain one, and only one, observation?

$P(A_{alone}) = \dfrac{1}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9}$

In other words, given 10 independent observations the probability that a grid cell will contain one, and only one observation is equal to the probability that one observation will be in the cell ($\dfrac{1}{9}$) times the probabilities that all other observations will be in another grid cell $\dfrac{8}{9}$.

We can expand this concept and focus not just on a single grid cell, but on all grid cells. Specifically, we can ask the same question above ten times:

$P(A_{alone}) = 10 x \dfrac{1}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9} x \dfrac{8}{9}$

This then generalizes to

$P(\text{k events}) = \text{Number of possible combinations of k events} x (\dfrac{1}{9})^{k} \cdot (\dfrac{8}{9})^{\text{number of observations} -k}$, where

$k$ is the number of events observed per quadrant.

We learned last week that the number of possible combinations is simply:

$_{n}C_{r} = \dfrac{n!}{(n-r)!r!}$

so, we can substitute the variable k for the variable n ($n = k$) and notate

$P(\text{k events}) = _{k}C_{\text{number of observations}} \cdot (\dfrac{1}{9})^{k} \cdot (\dfrac{8}{9})^{\text{number of observations} - k}$

Knowing that can have between 0 and 10 observations per quadrant, we can compute the probability that $k=0$, $k=1$, $\ldots$, $k=10$ for a given quadrant. By doing this, we are get a probability distribution function that describes the probability of observing a given point pattern. This function is a binomial distribution!

We therefore have

$P(k, n, x) = _{k}C_{n} \cdot (\dfrac{1}{x})^{k} \cdot (\dfrac{x-1}{x})^{(n-k)}$,

where $k$ is the number of events observed per quadrant, $n$ is the total number of events, and $x$ is the total number of quadrants the study area has been partitioned into.

At this point, you may be wondering why any of this matters. What does having $P(k, n, x)$ tell us? Knowing $P(k,n,x)$ means that we can mathematically predict the probability seeing a point pattern with a given distribution over $\mathbb{R}^{2}$. We can compute the binomial distribution for $k$ and check whether the observered pattern follows our expected probability distribution. If we see that k=4 in two quadrants, we could determine that the observed pattern is does not meet the CSR (Complete Spatial Randomness) constraint.

Poisson Distribution

The above is all well and good. Until you have ~50 observations. At that point, calculating values of $n$ greater than 50 ($50!$) could become tricky. The poisson distribution is a good stand-in for the binomial distribution and many spatial point pattern analysis methods utilize the poisson distribution.

Network Constrained Point Processes

To highlight a bit more of the applicability of these methods, we can look to network constrained point patterns. That is, point patterns in $\mathbb{R}^{2}$ that are also constrained to a network (for example a road network). In this case, we can estimate a probability distribution function:

$f(x) = \dfrac{1}{|L|} for x \in L$,

where $|L|$ is the length of the network (the sum of all of the lengths of the roads using a street network as an example). Here, the above point process formula is applicable, with a simple modification. We no longer use quadrant area to compute the probability of observing some number of events in a given quadrant (in the above example, we used $\dfrac{1}{9}$ as the probability of event A being observed in one of nine quadrants). If we have a network, what metric is used instead of area ( in other words, what metric is the demoninator in the previous fraction)?

Knowing this probability distirbution function, we can then estimate the probability that a given road segment (a single street perhaps) contains $k$ observations and start to assess whether the observed pattern is random or not.