Exercise 1

In this question we fit a Strauss point process model to the swedishpines data.

  1. We need a guess at the interaction distance \(R\). Compute and plot the \(K\)-function of the dataset and choose the value \(r\) which maximises the discrepancy \(|K(r)-\pi r^2|\).

  2. Fit the stationary Strauss model with the chosen interaction distance using

    ppm(swedishpines ~ 1, Strauss(R))

    where R is your chosen value.

  3. Interpret the printout: how strong is the interaction?

  4. Plot the fitted pairwise interaction function using plot(fitin(fit)).

Exercise 2

In Question 1 we guesstimated the Strauss interaction distance parameter. Alternatively this parameter could be estimated by profile pseudolikelihood.

  1. Look again at the plot of the \(L\)-function of swedishpines and determine a plausible range of possible values for the interaction distance.

  2. Generate a sequence of values equally spaced across this range, for example, if your range of plausible values was \([5, 12]\), then type

    rvals <- seq(5, 12, by=0.5)

  3. 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)

  4. Execute

    fitp <- profilepl(D, Strauss, swedishpines ~ 1)

    to find the maximum profile pseudolikelihood fit.

  5. Print and plot the object fitp.

  6. Compare the computed estimate of interaction distance \(r\) with your guesstimate. Compare the corresponding estimates of the Strauss interaction parameter \(\gamma\).

  7. Extract the fitted Gibbs point process model from the object fitp as

    bestfit <- as.ppm(fitp)

Exercise 3

Modify Question 1 by using the Huang-Ogata approximate maximum likelihood algorithm (method="ho") instead of maximum pseudolikelihood (the default, method="mpl").

Exercise 4

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))

Exercise 5