2.6 Example Exam Question

(Taken from 2010 Exam)

An infectious disease spreads as follows. An initial outbreak of cases (1st generation) forms a homogeneous Poisson process with intensity $\mathrm{\lambda }$. Subsequently, the $\mathrm{(}\mathrm{i}\mathrm{+}\mathrm{1}\mathrm{)}$th generation is formed by individuals infected by the $\mathrm{i}$th generation: $\mathrm{i}$th generation cases each infect one or no extra cases, with probabilities ${\mathrm{p}}_{\mathrm{i}}$ and $\mathrm{1}\mathrm{-}{\mathrm{p}}_{\mathrm{i}}$ respectively, and then become uninfectious. The location of each case (except those in the 1st generation) is uniformly distributed in a circle of radius $\mathrm{r}$ centred on the location of the case that infected it. A point process is formed by the superposition of all cases.

1. 1.

   Consider the case in which $p_i=p$ for each $i$, where , giving rise to a process $X$.

   1. (a)

      Show that the expected total number of cases in future generations that arise from an individual case in the 1st generation is $p/(1-p)$. The following result may be helpful: if , ${\sum }_{i=1}^{\mathrm{\infty }}i{a}^{i-1}=1/{(1-a)}^2$

   2. (b)

      Hence find the intensity of $X$.

2. 2.

   Consider now the case in which $p_1=p$, where , and $p_2=0$, giving rise to a process $Y$.

   1. (a)

      Write down the intensity, $\eta$, of $Y$.

   2. (b)

      Define the $K$-function of a stationary, isotropic spatial point process.

   3. (c)

      Clearly stating any theorems used, show that, for $s\le r$, the $K$-function of $Y$ is

      $$K(s)=\frac{2p}{\eta (1+p)}.\frac{s^2}{r^2}+\pi s^2.$$

   4. (d)

      Write down $K(s)$ for $s>r$.

Solution:

1. 1.
   1. (a)

      Let $Z$ be the random variable denoting the number of offspring arising from a single case. We can write down the probability mass function of $Z$ as follows:

      $\mathbb{P}(Z=0)$	$\mathbb{P}(Z=1)$	$\mathbb{P}(Z=2)$	$\mathbb{P}(Z=3)$	$\mathrm{\cdots }$
      $1-p$	$p(1-p)$	$p^2(1-p)$	$p^3(1-p)$	$\mathrm{\cdots }$

      so the expected number of offspring, $𝔼(Z)$ is by definition

      	$𝔼(Z)$	$=$	${\displaystyle \sum _{z=0}^{\mathrm{\infty }}}z\mathbb{P}(Z=z)$
      		$=$	${\displaystyle \sum _{z=0}^{\mathrm{\infty }}}z[p^z(1-p)]$
      		$=$	$(1-p){\displaystyle \sum _{z=0}^{\mathrm{\infty }}}z{p}^z$
      		$=$	$p(1-p){\displaystyle \sum _{z=0}^{\mathrm{\infty }}}z{p}^{z-1}$
      		$=$	${\displaystyle \frac{p(1-p)}{{(1-p)}^2}}$
      		$=$	${\displaystyle \frac{p}{(1-p)}}$

   2. (b)

      This is a homogeneous process, so the intensity is the expected number of points per unit area. We calculated the expected number of ‘children’ arising from a single ‘parent’ in the first part, so the intensity of children is $\lambda p/(1-p)$. To get the intensity of the parents and children combined, we add the intensities of the parents and children i.e., the intensity of $X$ is $\lambda +\lambda p/(1-p)=\lambda /(1-p)$

2. 2.
   1. (a)

      The intensity of $Y$ is the sum of the intensities of the parents and the children: $\eta =\lambda +\lambda p=\lambda (1+p)$.

   2. (b)

      The answer to this part just requires a statement of Definitions 2.6 and 2.8; make sure you define all terms.

   3. (c)

      We will use Theorem 2.1 to answer this part of the question, which we re-state below for clarity. Theorem: For a stationary, isotropic, orderly process of intensity $\mathrm{\eta }$, we have $\mathrm{\eta }\mathsf{}\mathrm{K}\mathsf{}\mathrm{(}\mathrm{s}\mathrm{)}\mathrm{=}\mathrm{E}\mathsf{}\mathrm{[}\mathsf{\text{number of further events within distance }}\mathrm{s}\mathsf{\text{ of an arbitrary event}}\mathrm{]}$

      The points in the process $Y$ are either ‘parents’, which follow a homogeneous Poisson process with rate $\lambda$, or ‘offspring’, which arise as described in the question. We will split this expectation up, considering the contributions from the parents and offspring separately. In order to do this, we will use the tower law of expectation, which states that for random variables $A$ and $B$, ${𝔼}_A(A)={𝔼}_B[{𝔼}_{A|B}(A|B)]$, see Theorem A.1 for further details. Put $E=\eta K(s)$, then using the tower law, we expand this expectation as

      The probability that a randomly chosen point is a parent is $\lambda /(\lambda +\lambda p)=1/(1+p)$ and the probability that a randomly chosen point is an offspring is $\lambda p/(\lambda +\lambda p)=p/(1+p)$. We next split the expectations above into contributions from ‘related’ and ‘unrelated’ points.

      Since $Y$ is a homogeneous Poisson process with rate $\eta$, the number of unrelated points within $s$ of either a parent or an offspring is $\eta \pi s^2$.

      The expected number of related points within $s$ of a parent is:

      Similarly, since each offspring definitely belongs to a parent, the expected number of related points within $s$ of an offspring is:

      Putting all this together, for $s\le r$,

      	$E$	$=$	$\left[\eta \pi s^2+{\displaystyle \frac{p{s}^2}{r^2}}\right]\left[{\displaystyle \frac{1}{1+p}}\right]+\left[\eta \pi s^2+{\displaystyle \frac{s^2}{r^2}}\right]\left[{\displaystyle \frac{p}{1+p}}\right],$
      		$\mathrm{⋮}$
      		$=$	$\eta \pi s^2+{\displaystyle \frac{2p}{1+p}}{\displaystyle \frac{s^2}{r^2}},$

      and dividing both sides by $\eta$, we are done.

   4. (d)

      Going through the algebra for the case $s>r$ yields

      $$K(s)=\pi s^2+\frac{2p}{\eta (1+p)}.$$