This session is concerned with Gibbs models for point patterns with interpoint interaction. The lecturer’s R script is available here (right click and save).
library(spatstat)
## Loading required package: spatstat.data
## Loading required package: methods
## Loading required package: nlme
## Loading required package: rpart
##
## spatstat 1.56-1.007 (nickname: 'Damn You Autocorrect')
## For an introduction to spatstat, type 'beginner'
In this question we fit a Strauss point process model to the swedishpines
data.
We need a guess at the interaction distance (R). Compute and plot the (L)-function of the dataset and choose the value (r) which maximises the discrepancy (\lvert L(r)-r \rvert).
We plot the above function which we want to maximize.
plot(Lest(swedishpines), abs(iso - r) ~ r, main = "")
As seen from the plot, the maximum lies around (r = 6.5) by eye. We find the optimum explicitly like follows:
discrep <- function(r) {
return(abs(as.function(Lest(swedishpines))(r) - r))
}
res <- optimise(discrep, interval = c(0.1, 20), maximum = TRUE)
print(res)
## $maximum
## [1] 6.984333
##
## $objective
## [1] 2.992058
R <- res$maximum
This corresponds nicely with the plot.
Fit the stationary Strauss model with the chosen interaction distance using
ppm(swedishpines ~ 1, Strauss(R))
where R
is your chosen value.
Interpret the printout: how strong is the interaction?
Plot the fitted pairwise interaction function using plot(fitin(fit))
.
As we have assigned R
, we simply write:
fit <- ppm(swedishpines ~ 1, Strauss(R))
print(fit)
## Stationary Strauss process
##
## First order term: beta = 0.0281221
##
## Interaction distance: 6.984333
## Fitted interaction parameter gamma: 0.1434456
##
## Relevant coefficients:
## Interaction
## -1.941799
##
## For standard errors, type coef(summary(x))
As seen, the (\gamma = 0.14) parameter is quite small. Thus there seems to be a strong negative association between points within distance R of each other. A (\gamma) of (0) implies the hard core process whereas (\gamma = 1) implies the Poisson process and thus CSR.
The pairwise interaction function become:
plot(fitin(fit))
In Question 1 we guesstimated the Strauss interaction distance parameter. Alternatively this parameter could be estimated by profile pseudolikelihood.
Look again at the plot of the (L)-function of swedishpines
and determine a plausible range of possible values for the interaction distance.
plot(Lest(swedishpines), main = "")
A conservative range of plausible interaction distances seems to be 3 to 15 meters.
Generate a sequence of values equally spaced across this range, for example, if your range of plausible values was ([0.05, 0.3]), then type
rvals <- seq(0.05, 0.3, by=0.01)
We generate the numbers between 3 and 12.
rvals <- seq(3, 12, by = 0.1)
Construct a data frame, with one column named r
(matching the argument name of Strauss
), containing these values. For example
D <- data.frame(r = rvals)
OK,
D <- data.frame(r = rvals)
Execute
fitp <- profilepl(D, Strauss, swedishpines ~ 1)
to find the maximum profile pseudolikelihood fit.
OK, let’s execute it:
fitp <- profilepl(D, Strauss, swedishpines ~ 1)
## (computing rbord)
## comparing 91 models...
## 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
## 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76,
## 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91.
## fitting optimal model...
## done.
Print and plot the object fitp
.
print(fitp)
## profile log pseudolikelihood
## for model: ppm(swedishpines ~ 1, interaction = Strauss)
## fitted with rbord = 12
## interaction: Strauss process
## irregular parameter: r in [3, 12]
## optimum value of irregular parameter: r = 9.8
plot(fitp)
Compare the computed estimate of interaction distance (r) with your guesstimate. Compare the corresponding estimates of the Strauss interaction parameter (\gamma).
(Ropt <- reach(as.ppm(fitp)))
## [1] 9.8
The (r = 9.8) is not totally inconsistent with the previous estimate of (7).
Extract the fitted Gibbs point process model from the object fitp
as
bestfit <- as.ppm(fitp)
OK, let’s do that:
bestfit <- as.ppm(fitp)
For the Strauss model fitted in Question 1,
Generate and plot a simulated realisation of the fitted model using simulate
.
s <- simulate(fit, drop = TRUE)
plot(s, main = "")
Plot the (L)-function of the data along with the global simulation envelopes from 19 realisations of the fitted model.
plot(envelope(fit, Lest, global = TRUE, nsim = 19, nsim2 = 100), main = "")
## Generating 119 simulated realisations of fitted Gibbs model (100 to
## estimate the mean and 19 to calculate envelopes) ...
## 1, 2, 3, 4.6.8.10.12.14.16.18.20.22.24.26.28.30.32.34.36.38.
## 40.42.44.46.48.50.52.54.56.58.60.62.64.66.68.70.72.74.76.78
## .80.82.84.86.88.90.92.94.96.98.100.102.104.106.108.110.112.114.116.
## 118 119.
##
## Done.
Read the help file for Geyer
.
See help(Geyer)
Fit a stationary Geyer saturation process to swedishpines
, with the same interaction distance as for the Strauss model computed in Question 2, and trying different values of the saturation parameter sat = 1, 2, 3
say.
ppm(swedishpines ~ 1, Geyer(r = Ropt, sat = 1))
## Stationary Geyer saturation process
##
## First order term: beta = 0.07472669
##
## Interaction distance: 9.8
## Saturation parameter: 1
## Fitted interaction parameter gamma: 0.1871555
##
## Relevant coefficients:
## Interaction
## -1.675815
##
## For standard errors, type coef(summary(x))
ppm(swedishpines ~ 1, Geyer(r = Ropt, sat = 2))
## Stationary Geyer saturation process
##
## First order term: beta = 0.04707047
##
## Interaction distance: 9.8
## Saturation parameter: 2
## Fitted interaction parameter gamma: 0.5242884
##
## Relevant coefficients:
## Interaction
## -0.6457134
##
## For standard errors, type coef(summary(x))
ppm(swedishpines ~ 1, Geyer(r = Ropt, sat = 3))
## Stationary Geyer saturation process
##
## First order term: beta = 0.07603509
##
## Interaction distance: 9.8
## Saturation parameter: 3
## Fitted interaction parameter gamma: 0.5261429
##
## Relevant coefficients:
## Interaction
## -0.6421823
##
## For standard errors, type coef(summary(x))
Fit the same model with the addition of a log-quadratic trend.
gfit <- ppm(swedishpines ~ polynom(x, y, 2), Geyer(r = Ropt, sat = 3))
Plot the fitted trend and conditional intensity.
Here we use the log scale to be able to see the discs in the conditional intensity.
par(mfrow=c(1,2))
plot(gfit, log = TRUE, pause = FALSE)
Modify question 1 by using the Huang-Ogata approximate maximum likelihood algorithm (method="ho"
) instead of maximum pseudolikelihood (the default, method="mpl"
).
fit.mpl <- ppm(swedishpines ~ 1, Strauss(R), method = "mpl")
fit.ho <- ppm(swedishpines ~ 1, Strauss(R), method = "ho")
## Simulating... 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
## 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76,
## 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100.
## Done.
print(fit.ho)
## Stationary Strauss process
##
## First order term: beta = 0.03114674
##
## Interaction distance: 6.984333
## Fitted interaction parameter gamma: 0.139353
##
## Relevant coefficients:
## Interaction
## -1.970745
##
## For standard errors, type coef(summary(x))
print(fit.mpl)
## Stationary Strauss process
##
## First order term: beta = 0.0281221
##
## Interaction distance: 6.984333
## Fitted interaction parameter gamma: 0.1434456
##
## Relevant coefficients:
## Interaction
## -1.941799
##
## For standard errors, type coef(summary(x))
The fits are very similar.
Repeat Question 2 for the inhomogeneous Strauss process with log-quadratic trend. The corresponding call to profilepl
is
fitp <- profilepl(D, Strauss, swedishpines ~ polynom(x,y,2))
fitp2 <- profilepl(D, Strauss, swedishpines ~ polynom(x,y,2))
## (computing rbord)
## comparing 91 models...
## 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38,
## 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76,
## 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91.
## fitting optimal model...
## done.
print(fitp)
## profile log pseudolikelihood
## for model: ppm(swedishpines ~ 1, interaction = Strauss)
## fitted with rbord = 12
## interaction: Strauss process
## irregular parameter: r in [3, 12]
## optimum value of irregular parameter: r = 9.8
print(fitp2)
## profile log pseudolikelihood
## for model: ppm(swedishpines ~ polynom(x, y, 2), interaction = Strauss)
## fitted with rbord = 12
## interaction: Strauss process
## irregular parameter: r in [3, 12]
## optimum value of irregular parameter: r = 9.8
Repeat Question 3 for the inhomogeneous Strauss process with log-quadratic trend, using the inhomogeneous (L)-function Linhom
in place of the usual (L)-function.
fit2 <- as.ppm(fitp2)
plot(envelope(fit2, Linhom, global = TRUE, nsim = 19, nsim2 = 100), main = "")
## Generating 119 simulated realisations of fitted Gibbs model (100 to
## estimate the mean and 19 to calculate envelopes) ...
## 1, 2, 3, 4.6.8.10.12.14.16.18.20.22.24.26.28.30.32.34.36.38.
## 40.42.44.46.48.50.52.54.56.58.60.62.64.66.68.70.72.74.76.78
## .80.82.84.86.88.90.92.94.96.98.100.102.104.106.108.110.112.114.116.
## 118 119.
##
## Done.