7610 Lab 4

Problem 1

See Figure 1.

Problem 2

A rejection sampling is used where the first \(N=50\) samples s.t. \(y > -x + 30\) is retained. See Figure 2.

Problem 3

We can still use rejection sampling in this problem. Assume we have a circle centered at \((0, 0)\) and have a radius of \(2\). To get the desired sample size \(N=30\), we use \(\lambda = 4\) to simulate the HPPP. See Figure 3.

Problem 4

See Figure 4. A mercator map projection is used.

Problem 5

Follwing the procedure in problem 3, the number of points \(X\) in that circle follows a Poisson distribution with mean and variance both equal to \(\pi r^2 \lambda\). Now we have \(r = 2\) and \(\lambda = 2\), so that \(E(X) = 8\pi\) and \(Var(X) = 8\pi\). Using the method of moment estimates, we can estimate \(\pi\) using the observed number of points in the circle. 1000 realizations is obtained in the following R chunk:

n.pts <- replicate(1000, expr = {
  df <- hppp.gen(c(-2, 2), c(-2, 2))
  sum(df$y < sqrt(4 - df$x^2) & df$y > -sqrt(4 - df$x^2))
})
mean(n.pts) / 8

## [1] 3.146625

The variance of that estimator is \(8\pi/n\), so an approximate 95% C.I. would be:

(mean(n.pts) + qnorm(c(.025, .975)) * sqrt(mean(n.pts) / 1000))/8

## [1] 3.107754 3.185496

(Note: this is not a raw output. This is a designed output related to the context and gave the exact answer of the question being asked.)

From the formula above we can see the estimator is \(\sqrt{N}\)-consistent.