library(spatstat)

Exercise 1

The command rpoispp(100) generates realisations of the Poisson process with intensity \(\lambda = 100\) in the unit square.

  1. Repeat the command plot(rpoispp(100)) several times to build your intuition about the appearance of a completely random pattern of points.

    Let’s plot it three times:

    replicate(3, plot(rpoispp(lambda = 100), main = ""))

    As can be seen, the points (unsurprisingly) are much more random that want one might think. “Randomly” drawing points on a piece of paper one would usually draw a point pattern that is more regular (i.e. the points are repulsive).

  2. Try the same thing with intensity \(\lambda = 1.5\).

    For brevity we only do it once here:

    plot(rpoispp(lambda = 1.5), main = "")

    Here we expect 1.5 points in the plot each time.

Exercise 2

Returning to the Japanese Pines data,

  1. Fit the uniform Poisson point process model to the Japanese Pines data

    ppm(japanesepines~1)

    We fit the Poisson process model with the given command and print the output:

    m.jp <- ppm(japanesepines ~ 1)
print(m.jp)
    ## Stationary Poisson process
## Intensity: 65
##             Estimate      S.E.  CI95.lo  CI95.hi Ztest     Zval
## log(lambda) 4.174387 0.1240347 3.931284 4.417491   *** 33.65499

  2. Read off the fitted intensity. Check that this is the correct value of the maximum likelihood estimate of the intensity.

    We extract the coeficient with the coef function, and compare to the straightforward estimate obtained by `intensity``:

    unname(exp(coef(m.jp)))
    ## [1] 65
    intensity(japanesepines)
    ## [1] 65

    As seen, they agree exactly.

Exercise 3

The japanesepines dataset is believed to exhibit spatial inhomogeneity.

  1. Plot a kernel smoothed intensity estimate.

    Plot the kernel smoothed intensity estimate selecting the bandwidth with bw.scott:

    jp.dens <- density(japanesepines, sigma = bw.scott)
plot(jp.dens)
plot(japanesepines, col = "white", cex = .4, pch = 16, add = TRUE)

  2. Fit the Poisson point process models with loglinear intensity (trend formula ~x+y) and log-quadratic intensity (trend formula ~polynom(x,y,2)) to the Japanese Pines data.

    We fit the two models with ppm:

    jp.m <- ppm(japanesepines ~ x + y)
jp.m2 <- ppm(japanesepines ~ polynom(x, y, 2) )

  3. extract the fitted coefficients for these models using coef.

    coef(jp.m)
    ## (Intercept)           x           y 
##   4.0670790  -0.2349641   0.4296171
    coef(jp.m2)
    ## (Intercept)           x           y      I(x^2)    I(x * y)      I(y^2) 
##   4.0645501   1.1436854  -1.5613621  -0.7490094  -1.2009245   2.5061569

  4. Plot the fitted model intensity (using plot(predict(fit)))

    par(mar=rep(0,4))
plot(predict(jp.m), main = "")

    plot(predict(jp.m, se=TRUE)$se, main = "")

    plot(predict(jp.m2), main = "")

    plot(predict(jp.m2, se=TRUE)$se, main = "")

  5. perform the Likelihood Ratio Test for the null hypothesis of a loglinear intensity against the alternative of a log-quadratic intensity, using anova.

    anova(jp.m, jp.m2)
    ## Analysis of Deviance Table
## 
## Model 1: ~x + y   Poisson
## Model 2: ~x + y + I(x^2) + I(x * y) + I(y^2)      Poisson
##   Npar Df Deviance
## 1    3            
## 2    6  3   3.3851

  6. Generate 10 simulated realisations of the fitted log-quadratic model, and plot them, using plot(simulate(fit, nsim=10)) where fit is the fitted model.

    par(mar=rep(0.5,4))
plot(simulate(jp.m2, nsim=10), main = "")
    ## Generating 10 simulated patterns ...1, 2, 3, 4, 5, 6, 7, 8, 9,  10.

Exercise 4

The bei dataset gives the locations of trees in a survey area with additional covariate information in a list bei.extra.

  1. Fit a Poisson point process model to the data which assumes that the intensity is a loglinear function of terrain slope and elevation (hint: use data = bei.extra in ppm).

    We fit the log-linear intensity model with the following:

    bei.m <- ppm(bei ~ elev + grad, data = bei.extra)

  2. Read off the fitted coefficients and write down the fitted intensity function.

    The coefficents are extraced with coef:

    coef(bei.m)
    ## (Intercept)        elev        grad 
## -8.56355220  0.02143995  5.84646680

    Hence the model is \(log\lambda(u) = -8.55 + 0.02\cdot E(u) + 5.84 G(u)\) where \(E(u)\) and \(G(u)\) is the elevation and gradient, respectively, at \(u\).

  3. Plot the fitted intensity as a colour image.

    plot(predict(bei.m), main = "")
plot(bei, cex = 0.3, pch = 16, cols = "white", add = TRUE)

  4. extract the estimated variance-covariance matrix of the coefficient estimates, using vcov.

    We call vcov on the fitted model object:

    vcov(bei.m)
    ##               (Intercept)          elev          grad
## (Intercept)  0.1163586583 -7.774771e-04 -0.0354792767
## elev        -0.0007774771  5.234331e-06  0.0001992266
## grad        -0.0354792767  1.992266e-04  0.0654239289

  5. Compute and plot the standard error of the intensity estimate (see help(predict.ppm)).

    From the documentation the argument se will trigger the computation of the standard errors. These are then plotted in the standard manner.

    std.err <- predict(bei.m, se = TRUE)$se
plot(std.err, main = "")

Exercise 5

The dataset ants is a multitype point pattern representing the locations of nests of two species of ants.

  1. plot the data.

    plot(ants)

  2. Fit the model ppm(ants ~ marks) and interpret the result. Compare the result with summary(ants) and explain the similarities.

    fit1 <- ppm(ants ~ marks)

    This is a Poisson model with a separate constant intensity for each mark. The fitted intensities are:

    exp(coef(fit1)[1])
    ##  (Intercept) 
## 6.762949e-05
    exp(coef(fit1)[1] + coef(fit1)[2])
    ##  (Intercept) 
## 0.0001585795

    This agrees perfectly with the output of summary(ants):

    summary(ants)
    ## Marked planar point pattern:  97 points
## Average intensity 0.0002261486 points per square unit (one unit = 0.5 feet)
## 
## Coordinates are integers
## i.e. rounded to the nearest unit (one unit = 0.5 feet)
## 
## Multitype:
##             frequency proportion    intensity
## Cataglyphis        29  0.2989691 6.761144e-05
## Messor             68  0.7010309 1.585372e-04
## 
## Window: polygonal boundary
## single connected closed polygon with 11 vertices
## enclosing rectangle: [-25, 803] x [-49, 717] units
##                      (828 x 766 units)
## Window area = 428922 square units
## Unit of length: 0.5 feet
## Fraction of frame area: 0.676

  3. Fit the model ppm(ants ~ marks + x) and write down an expression for the fitted intensity function.

    fit2 <- ppm(ants ~ marks + x)
(co <- coef(fit2))
    ##   (Intercept)   marksMessor             x 
## -9.5243832518  0.8522118655 -0.0002041438

    Intensity for the reference type (Cataglyphis):

    \[\lambda( (x,y) ) = \exp(-9.5243833 + -2.0414381\times 10^{-4} \cdot x)\]

    Intensity for the other type (Messor):

    \[\lambda( (x,y) ) = \exp(-9.5243833 + 0.8522119 + -2.0414381\times 10^{-4} \cdot x)\]

  4. Fit the model ppm(ants ~ marks * x) and write down an expression for the fitted intensity function.

    fit3 <- ppm(ants ~ marks * x)
(co <- coef(fit3))
    ##   (Intercept)   marksMessor             x marksMessor:x 
## -9.605698e+00  9.676854e-01  1.107981e-05 -3.071343e-04

    Intensity for the reference type (Cataglyphis):

    \[\lambda( (x,y) ) = \exp(-9.605698 + 1.1079805\times 10^{-5} \cdot x)\]

    Intensity for the other type (Messor):

    \[\lambda( (x,y) ) = \exp(-9.605698 + 0.9676854 + (1.1079805\times 10^{-5} + 0.9676854) \cdot x)\]

  5. Compute the fitted intensities of the three models fitted above using predict and plot the results.

    pred <- c(predict(fit1), predict(fit2), predict(fit3))
plot(as.solist(pred), ncols = 2, main = "")

  6. Explain the difference between the models fitted by ppm(ants ~ marks + x) and ppm(ants ~ marks * x) .

    For the additive model the effect of the x coordinate is the same for both types of ants, while the effect of x differs in the multiplicative model.

Exercise 6

The study region for the ants’ nests data ants is divided into areas of “scrub” and “field”. We want to fit a Poisson model with different intensities in the field and scrub areas.

The coordinates of two points on the boundary line between field and scrub are given in ants.extra$fieldscrub. First construct a function that determines which side of the line we are on:

fs <- function(x,y) {
  ends <- ants.extra$fieldscrub
  angle <- atan(diff(ends$y)/diff(ends$x))
  normal <- angle + pi/2
  project <- (x - ends$x[1]) * cos(normal) + (y - ends$y[1]) * sin(normal)
  factor(ifelse(project > 0, "scrub", "field"))
}

Now fit the models:

ppm(ants ~ marks + side, data = list(side=fs))
ppm(ants ~ marks * side, data = list(side=fs))

and interpret the results.

fit1 <- ppm(ants ~ marks + side, data = list(side=fs))
fit2 <- ppm(ants ~ marks * side, data = list(side=fs))

In the first model the fitted intensity is lower in the scrub than in the field (but this effect is not significant).

In the second model the fitted intensity of Cataglyphis is lower in the scrub than the intensity of Cataglyphis in the field, where as it is the other way around for Messor. When we allow for the different effect between ant types the scrub/field covariate is significant.

Exercise 7

The update command can be used to re-fit a point process model using a different model formula.

  1. Type the following commands and interpret the results:

    fit0 <- ppm(japanesepines ~ 1)
fit1 <- update(fit0, . ~ x)
fit1
    ## Nonstationary Poisson process
## 
## Log intensity:  ~x
## 
## Fitted trend coefficients:
## (Intercept)           x 
##   4.2895587  -0.2349362 
## 
##               Estimate      S.E.   CI95.lo   CI95.hi Ztest       Zval
## (Intercept)  4.2895587 0.2411952  3.816825 4.7622926   *** 17.7845936
## x           -0.2349362 0.4305416 -1.078782 0.6089098       -0.5456759
    fit2 <- update(fit1, . ~ . + y)
fit2
    ## Nonstationary Poisson process
## 
## Log intensity:  ~x + y
## 
## Fitted trend coefficients:
## (Intercept)           x           y 
##   4.0670790  -0.2349641   0.4296171 
## 
##               Estimate      S.E.    CI95.lo   CI95.hi Ztest       Zval
## (Intercept)  4.0670790 0.3341802  3.4120978 4.7220602   *** 12.1703167
## x           -0.2349641 0.4305456 -1.0788181 0.6088898       -0.5457357
## y            0.4296171 0.4318102 -0.4167154 1.2759495        0.9949211

    OK, let’s do that:

    fit0 <- ppm(japanesepines ~ 1)
fit1 <- update(fit0, . ~ x)
fit1
    ## Nonstationary Poisson process
## 
## Log intensity:  ~x
## 
## Fitted trend coefficients:
## (Intercept)           x 
##   4.2895587  -0.2349362 
## 
##               Estimate      S.E.   CI95.lo   CI95.hi Ztest       Zval
## (Intercept)  4.2895587 0.2411952  3.816825 4.7622926   *** 17.7845936
## x           -0.2349362 0.4305416 -1.078782 0.6089098       -0.5456759
    fit2 <- update(fit1, . ~ . + y)
fit2
    ## Nonstationary Poisson process
## 
## Log intensity:  ~x + y
## 
## Fitted trend coefficients:
## (Intercept)           x           y 
##   4.0670790  -0.2349641   0.4296171 
## 
##               Estimate      S.E.    CI95.lo   CI95.hi Ztest       Zval
## (Intercept)  4.0670790 0.3341802  3.4120978 4.7220602   *** 12.1703167
## x           -0.2349641 0.4305456 -1.0788181 0.6088898       -0.5457357
## y            0.4296171 0.4318102 -0.4167154 1.2759495        0.9949211

  2. Now type step(fit2) and interpret the results.

    The backwards selection is done with the code:

    step(fit2)
    ## Start:  AIC=-407.96
## ~x + y
## 
##        Df     AIC
## - x     1 -409.66
## - y     1 -408.97
## <none>    -407.96
## 
## Step:  AIC=-409.66
## ~y
## 
##        Df     AIC
## - y     1 -410.67
## <none>    -409.66
## 
## Step:  AIC=-410.67
## ~1
    ## Stationary Poisson process
## Intensity: 65
##             Estimate      S.E.  CI95.lo  CI95.hi Ztest     Zval
## log(lambda) 4.174387 0.1240347 3.931284 4.417491   *** 33.65499

    First, given two models the preferred model is the one with the minimum AIC value. In step 1, the removal of x results in the least AIC and is hence deleted. In step 2, removing y results in a lower AIC than not deleing anything and is thus deleted. This results in the constant model.

Exercise 8

Fit Poisson point process models to the Japanese Pines data, with the following trend formulas. Read off an expression for the fitted intensity function in each case.

Trend formula Fitted intensity function
~1 \(\log\lambda(u) = 4.17\)
~x \(\log\lambda(u) = 4.28 - 0.23x\)
~sin(x) \(\log\lambda(u) = 4.29 - 0.26\sin(x)\)
~x+y \(\log\lambda(u) = 4.07 - 0.23x + 0.42y\)
~polynom(x,y,2) \(\log\lambda(u) = 4.06 + 1.14x - 1.56y - 0.75x^2 - 1.20xy + 2.51y^2\)
~factor(x < 0.4) \(\log\lambda(u) = 4.10 + 0.16\cdot I(x < 0.4)\)

(Here, \(I(\cdot)\) denote the indicator function.)

The fitted intensity functions have been written into the table based on the follwing model fits:

coef(ppm1 <- ppm(japanesepines ~ 1)) 
## log(lambda) 
##    4.174387
coef(ppm2 <- ppm(japanesepines ~ x))
## (Intercept)           x 
##   4.2895587  -0.2349362
coef(ppm3 <- ppm(japanesepines ~ sin(x)))
## (Intercept)      sin(x) 
##   4.2915935  -0.2594537
coef(ppm4 <- ppm(japanesepines ~ x + y))
## (Intercept)           x           y 
##   4.0670790  -0.2349641   0.4296171
coef(ppm5 <- ppm(japanesepines ~ polynom(x, y, 2)))
## (Intercept)           x           y      I(x^2)    I(x * y)      I(y^2) 
##   4.0645501   1.1436854  -1.5613621  -0.7490094  -1.2009245   2.5061569
coef(ppm6 <- ppm(japanesepines ~ factor(x < 0.4)))
##         (Intercept) factor(x < 0.4)TRUE 
##           4.1048159           0.1632665

Exercise 9

Make image plots of the fitted intensities for the inhomogeneous models above.

Again, we use plot(predict()):

plot(predict(ppm1), main = "")

plot(predict(ppm2), main = "")

plot(predict(ppm3), main = "")

plot(predict(ppm4), main = "")

plot(predict(ppm5), main = "")

plot(predict(ppm6), main = "")