Exercise 1 (Deterministic colouring) Let \(\Pi\) a Poisson point process on \([0, \infty)\). Suppose we colour the first arrival blue, and then next arrival red, and continue colouring the points in this alternating fashion. Consider the point processes \(\Gamma\) formed by considering only the blue points. Is this a Poisson point process? Explain.

Solution. No. The interarrival times between two blue points is given by the sum of two exponential random variables, and is no longer an exponential.

Exercise 2 (Shop keeper) Suppose we model the number of customers that arrive at a high street shop on at particular day by a Poisson process of intensity \(\lambda >0\), where \(\lambda\) is measured in customers per hour. We wish to estimate \(\lambda\). Suppose the shop is really high-end and on some days has no customers, on its \(6\) hours of operations. The shop keeper only keeps track of whether she had has any customers are not; that is, her records \(x = (x_1, \ldots, x_n)\) are a binary sequence. Find a consisent estimtor for \(\lambda\)


Solution. Let \(X_i\) be the binary Bernoulli random variables that corresponds to whether a customer arrived on the \(i\)th day. Note that \(\mathbb{P}(X_i = 0) = e^{-6\lambda}\) and \(\mathbb{P}(X_i = 1) = 1 - e^{-6\lambda}\). Assume that these are independent random variables, the law of large numbers gives that \[ n^{-1}(X_1 + \cdots X_n) \to \mathbb{E}X_1 = 1 - e^{-6\lambda},\] as \(n \to \infty\). Thus setting \[T_n := -\frac{1}{6}\log( 1- n^{-1}(X_1 + \cdots+ X_n)),\] we also know that \(T_n \to \lambda\), as \(n \to \infty\), since \(\log\) is continuous.
Exercise 3 Suppose we \(\Pi\) is a Poisson point process on \([0, \infty)\) of intensity \(\lambda\). Using the construction of \(\Pi\) as exponential inter-arrival times, prove that conditioned on the event that the unit interval contains exactly one point, the distribution of the its location is uniform.

Solution. Let \(X_1\) and \(X_2\) be independent exponential random variables of rate \(\lambda\) representing the inter-arrival times. Let \(U\) be the location of the point in question. We are asked to compute \[\mathbb{P}(U \leq x | X_1 < 1, X_2+X_1 >1)=\mathbb{P}(X_1 \leq x | X_1 < 1, X_2+X_1 >1)\] for \(0 < x <1\).
Since \(X_1\) and \(X_2\) are independent, we have \[\begin{eqnarray*} \mathbb{P}(X_1 < x, X_2+X_1 >1) &=& \int_0 ^x \lambda e^{- \lambda x_1} \Big( \int_{1-x_1} ^{\infty} \lambda e^{ -\lambda x_2} dx_2 \Big)dx_1 \\ &=&\int_0 ^x \lambda e^{-\lambda x_1} e^{-\lambda(1-x_1)}dx_1 \\ &=& x \lambda e^{-\lambda}. \end{eqnarray*}\] Hence, using this formula in the numerator, and also in the denominator with \(x=1\), we have \[\mathbb{P}(U \leq x | X_1 < 1, X_2+X_1 >1) = \frac{x \lambda e^{-\lambda}}{ \lambda e^{-\lambda}} =x\] as desired.
Exercise 4 (Random deletion) Simulate a Poisson process of intensity \(\lambda=2\), say with \(10000\) arrivals. Delete each arrival independently with probability \(p=\tfrac{1}{2}\) to from a new thinned process. Plot a histogram of the inter-arrival times of the thinned process. What should you see? Why?
Solution. The histogram approximates the exponential density with rate \(1\), since we know that the resulting thinned process is still a Poisson process with intensity \(p \lambda=1\). 
inter=rexp(10000, 2)
arr = cumsum(inter)
coin = rbinom(length(arr), 1, 0.5)
for (i in 1:length(arr)) if (coin[i]==0){arr[i]<- 0}
arr<-setdiff(arr, 0)
interthin = arr[1]
for (i in 1: length(arr)-1){interthin <- c(interthin, arr[i+1] - arr[i]) }
x = seq(0, max(interthin)+1, by=0.1)
hist(interthin, prob=T, x)
curve(dexp(x), add=T)

Exercise 5 (Poisson on a disc) Simulate a Poisson point process of intensity \(100\) on a disc.

Solution. We simulate a single point on the disc, by simulating a uniform point on a square, and keep it if lands inside an inscribed disc, or repeat otherwise. Then we implent the characterization of Poisson point processes as a Poisson number of uniform points. Note a disc has area \(\pi r^2\).

point <- function(){
  z=2
  while(length(z)==1){
  x = 2*runif(1) -1
  y = 2*runif(1) -1
if (x^2 + y^2 <1) { z<-c(x,y)}
  }
  z
}
 re=replicate(rpois(1,100*pi), point())
 plot(re[1,], re[2,], xlim=c(-1.1,1.1), ylim=c(-1.1,1.1), asp=1)