7610 Homework 5

Problem 1

The dataset contains 120 event location in Uganda. Here is a map of the polygon and events.

From the map of the events, we would guess that there is some sort of clustering in the events. Most of the events are close to the center and there are not many events on the edge, so the events are not uniformly distributed on the map.

We can use the K-function and L-function to visually examine if its from a CSR process. Since the maximum distance between two polygon points in the domain is \(3978.2\), the values of lag \(h\) are chosen to be between \((0, 3000)\).

The theretical K-function and L-funtion is added to the plot. We can see from the figure that both lines fit perfectly well when \(h < 2000\) and some slight departure from the theoretical functions when \(h > 2000\). I would guess the departure is because of the edge effect. Overall these plots indicate that this process is from a CSR process.

We can further inspect the L-function by the envelop. The envelop of the L-function is computed by the Monte Carlo method with \(301\) generated CSR point patterns.

We can see from the plot that the estimated L-function is outside the 95% Monte Carlo confidence interval when \(h\) is within \((100, 500)\). This suggests that within this distance, the process is not CSR.

We can pick a \(h\) value within that range and do a hypothesis test. Here we choose \(h = 300\). From the envelop we can see that it’s above the 97.5% quantile line, so an upper test is used.

The upper tail p-value is calculated to be 0.0033113. Thus we can reject the null hypothesis of CSR when \(h=300\).

Problem 2

The map shows the data of locations of homes of juvenile offenders. From the map we can see that the point pattern shows some clustering too.

The estiamted K-function and L-function as well as the theoretical ones are plotted. Here only those \(h < 50\) is considered. We can see from the L-function that the process is clearly not a CSR.

We can justify that using the Monte Carlo quantiles. Here another \(101\) CSR point patterns is sampled and the quantiles are computed and plotted on the L-function.

The L-function is above the 97.5% upper quantile between 5 to 30. We can further compute the p-value of this when \(h = 15\).

The upper tail p-value is 0.0098039. Thus we can reject the null hypothesis of CSR when \(h=15\).

Problem 3

Amacrine Cell Data: These data record a bivariate pattern of chlinergic amacrine cells within the retina of a rabbit. I acquired them from Prof A Hughes (Melbourne). For a general description, see chapter 1 of Diggle (2003, Statistical Analysis of Spatial Point Patterns, second edition). The observation window for the data is the rectangle (0,a) X (0,1) where the unit of distance is 662um. There is some doubt about the precise value of a. A reasonable choice in my opinion would be 1060/662, or aproximately 1.6. There is some old evidence that 1070/662 SHOULD be correct, but this would leave a suspiciously empty border along one edge of the rectangle.

The data is included within many spatial packages like splancs. Here the cells with on signal is used.

From the data shown above, we would guess it’s a inhibition process. Again we choose h values from 0 to 1.5, and plot its estimated K-function and L-funtion.

The K-function and L-funtion looks quite CSR, so a simulation of \(h = 181\) generated and a Monte Carlo quantile is plotted onto the L-function.

We can know from the quantiles that when h is between 0 and 0.2, the process is not CSR. And the estimated line is below the lower 2.5% quantile indicates its a inhibition process. We can test the hypothesis of CSR when \(h=0.1\).

The lower tail p-value is 0.0065789. Thus we can reject the null hypothesis of CSR when \(h=.1\). This process being an inhibition process is reasonalbe because we would expect there should be uniformaly spaced off signal on the retina.