```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```



## Introduction


* Point processes are random scattering of points, randomly scattered with respect to some law of distribution.  

* Poisson processes are a fundamental example of a point process that are used to model a variety of different phenomena

* They serve as the basis of more complicated processes.  

* For example we may be interested in a model for the arrival times or total of number of customers $N(t)$ that have visited a high street shop up to an arbitrary time $t$.  


## continued

* Poisson processes are often used to develop probabilistic models for $N(t)$.

* In this context, the actual arrival times of the customers form a Poisson point process.  

* We will consider a few different ways of defining and constructing Poisson processes, all of which give a process with the same law, but each description will have certain advantages over the other


## Some review 


* We say that $X$ is  **Poisson random variable** with mean $\mu$ if it has probability mass function given by
$$ \mathbb{P}(X = n) = \frac{ e^{-\mu} \mu ^n}{n!} \  \text{ for $n = 0,1, 2, \ldots$}.$$

* We say that $Z$ is an **exponential random variable** with mean $\mu$ (rate $\lambda = 1/\mu$) if it is continuous random variable with probability density function given by
$$ z \mapsto  \tfrac{1}{\mu} e^{-\tfrac{ z} {\mu}} \  \text{ for  $z \geq 0$}.$$


## Exercises: Poisson and exponential random variables

* Check that the mean of a Poisson random variable with parameter $\mu$ does indeed have mean $\mu$.  Show that the variance is also $\mu$.

* Show that the minimal of two independent exponential random variables is again an exponential random variable.

* What is the memoryless property?  Prove that exponential random variables have this property.

## continued


* Let $n>0$ and $\mu >0$.  Let $W$ be a Poisson random variable with mean $\mu$.      Let $(X_i)_{i=1}^n$ be independent Bernoulli random variables with parameter $p=\lambda/n$.  Let $S_n = X_1 + \cdots + X_n$.  Show for all nonnegative integers $k$, we have $\lim_{n \to \infty} \mathbb{P}(S_n =k) = \mathbb{P}(W=k)$. 


* Let $\mu >0$, $p \in (0,1)$,   and $W$ be a Poisson random variable with mean $\mu$. Define the random variable $X$ in the following way:   if $W=n$, then we let $X$ be the sum of $n$ independent Bernoulli random variables with parameter $p$.  Show that $X$ is a Poisson random variable with mean $p\lambda$.


* Show that the sum of independent Poisson random variables is again a Poisson random variable.



## Generating a Poisson process as exponential inter-arrivals

We will start with the following construction which is also useful for simulations. 


*  Let $(X_i)_{i=1} ^{\infty}$ be a sequence of independent exponential random variables all with mean $\mu = 1/\lambda$. (Inter-arrival times)
  
*    Let $T_1 = X_1$. (Time of first arrival)
  
*    Let $T_{n+1} = X_{n+1} + T_n = \sum_{i=1} ^{n+1} X_i$ (Arrival times)
   
*    Set $N(t) = \sum_{n=1} ^{\infty} \mathbf{1}[ T_n \leq t]$ (Total number of arrivals by time $t$).


## continued
  
We say that $N$ is a **Poisson process on $[0, \infty)$ with rate $\lambda$**.  We call the set of random arrival times $\{T_n:  n \in \mathbb{Z}^{+}\}$ the corresponding **Poisson point process on $[0, \infty)$ of intensity $\lambda$**.


With R it is easy to simulate a Poisson process:

* Call $k$ independent exponential random variables of rate $\lambda >0$ (inter-arrival times).  

* Compute the partial sums (arrival times).

* To visualize: plot arrival times to see the resulting Poisson point process

## Code


```{r}
inter = rexp(15, 1) 
arr = cumsum(inter)
one = rep(1, times = length(arr))
plot(arr, one, yaxt = 'n', ann=FALSE)
```

##  Experimenting

We can make use of R and simulate many realizations of Poisson point processes.  In particular, pay attention to the following points:

* Simulate Poisson processes with different intensities and number of arrivals
  
* Are the points clumped sometimes are evenly spaced?


  
* Simulate say $50$ arrivals with intensity $1$.  Does it appears as though the arrival time of the last point is always around $50$?  Explain.


## continued
  
* If someone gave you a (single) sample realization of a Poisson process on a large interval, how would you estimate the intensity as unknown parameter?
    
*  If you are using a Poisson process of intensity $\lambda$ to model the arrival of customers, what should the *units* be?




##  What's Poisson got to do with it?

* It is not obvious what Poisson random variables have to do with Poisson processes from its construction as exponential inter-arrival times.  


*  In the following more mathematical description, the word *Poisson* appears from the outset.

## continued

* Let $N$ be a Poisson process on $[0, \infty)$ with rate $\lambda$.  It is possible to show that:

* For every $t >s>0$, $N(t) - N(s)$ is a Poisson random variable with mean $\lambda(t-s)$ (Stationary increments distributed as a Poisson)

* For any finite collection of disjoint intervals $(s_1, t_1), \ldots, (s_n, t_n)$ the random variables $N(t_1) - N(s_1), \ldots, N(t_n) - N(s_n)$ are independent.

These two properties also characterize a Poisson process and can be used to define spatial Poisson processes in higher dimensions. 


##  A weak version of the first property.


**Lemma:**  *Let $N$ be a Poisson process constructed with exponential inter-arrival times of rate $\lambda>0$.  Then $N(t)$ is a Poisson random variable with mean $t \lambda$.*  


*Proof*:

Let $N$ be given by exponential inter-arrival times, so that $X_i$ are the inter-arrival times, and $T_k$ is the time of the $k$ arrival.     Observe that
$$\mathbb{P}(N(t) = 0)  = \mathbb{P}( X_1 >t) = e^{-\lambda t}.$$ 


## continued

For $k >0$, we have
$$
\begin{eqnarray*}
\mathbb{P}(N(t) = k) &=&  \mathbb{P}( T_k \leq t, T_{k+1} >t)  \\
&=& \mathbb{P}( T_k \leq t, T_{k} + X_{k+1} >t)  \\
&=&  \int_0 ^t \int_{t-y} ^{\infty} g(y) f(x) dx dy,
\end{eqnarray*}
$$
where $g$ is the pdf for $T_k$ and $f$ is the pdf for $X_{k+1}$; here we need to appeal the independence of these two random variables. Thus 
$$\mathbb{P}(N(t) = k)  = \int_0 ^t g(y) e^{-(t-y) \lambda } dy$$

## continued


We recall that the sum of independent exponentials has the law of a gamma distribution; specifically
$$ g(y) =  \frac{\lambda^k}{\Gamma(k)} y^{k-1} e^{-\lambda y}.$$
Thus 
$$
\begin{eqnarray*}
 \mathbb{P}(N(t) = k) &=& \frac{\lambda^k e^{-t \lambda} }{\Gamma(k)} \int_0 ^ t y^{k-1} dy \\
 &=&  \frac{\lambda^k e^{-t \lambda} t^k}{k\Gamma(k) } \\
 &=&  \frac{(\lambda t)^k e^{-t \lambda} }{k! }
\end{eqnarray*}
$$


## Exercises

* 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 $\lambda$.  The shop keeper has records of how many customers arrive each day for $n$ days given by $x = (x_1, \ldots, x_n)$ and opens everyday for $6$ hours.  Find an estimate for $\lambda$.  Carefully justify why this is a reasonable estimate.


* Suppose the shop keeper computes the mean the variance for her data and finds that the variance is much smaller than the mean.  Would you re-evaluate whether a Poisson process is a good model?  Explain.


## continued

* Suppose the shop is really high-end and on some days has no customers.  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.  Can you still estimate $\lambda$?


* Suppose that arrivals to a shop are modeled by a Poisson process.  Suppose that you are told that there is exactly one arrival in the time interval $[0,1]$; let $U$ be the time of this arrival.  What is distribution of $U$?


## R code


We demonstrate the last exercise by simulations in the following way. 

* We simulate Poisson processes, and record the position of the first arrival if it occurs in $[0,1]$ and the second arrival occurs outside this interval.   

* We plot a histogram of the recorded occurrences  

## continued


```{r}
    record <- function(){
 inter = rexp(2,1)
 arr = cumsum(inter)
 r = 2
 if ( (arr[1] < 1) & (arr[2] >1)){
 r <- arr[1]
}
r
}
```

## continued

```{r}
x = replicate(10000, record())
y = setdiff(x, 2)
hist(y, prob=T)
```


##  A characterization via modelling assumptions

* Poisson processes are good models for arrivals only if it is reasonable  to assume that one also has the memoryless property that comes with the exponential distribution.  

* In addition, if  an arrival process $N(t)$ satisfies the following mild conditions, then it can be shown that it is a Poisson process of intensity $\lambda$.  


## continued


* (Stationarity)  The number of arrives in an interval of time depends only on the length of time.)  For all $t >s$ we have that  
$N(t) - N(s)$ has the same distribution as  $N(t-s)$.

* (Independent increments) For any finite collection of disjoint intervals $(s_1, t_1), \ldots, (s_n, t_n)$ the random variables $N(t_1) - N(s_1), \ldots, N(t_n) - N(s_n)$ are independent.


## continued


* (Orderliness:  two customers do not arrive at the same time)
$$\lim_{h \to 0} \frac{1}{h}\mathbb{P}( N(h) \geq 2) = 0.$$


* (Rate: In a small interval time, the probability that a customer arrives is proportional to $\lambda$.)
$$ \lim_{h \to 0} \frac{1}{h}\mathbb{P}( N(h) =1) = \lambda >0.$$




## Paritions into small bins


* We can sketch a proof of our previous lemma which connected Poisson random variables with Poisson processes.  

* Let $t >0$, and partition the interval $[0,t]$ into $n$ intervals of size $t/n$, where $n$ is large.  

* By orderliness condition with stationarity,  we can  assume that in each interval there is at most one arrival.   

* Let $p = \mathbb{P}( N(t/n) =1))$.   

## continued


* By the independent increments and  rate conditions, we have that probability that there are $k$ arrives is  given by 


$$
\begin{eqnarray*}
&& \mathbb{P}(N(t) = k) = \\
&& {n \choose k }p^k (1-p)^{n-k} =  \\
&& \frac{n!}{(n-k)! k!}\frac{(\lambda t)^k}{n^k}\Big(1- \frac{\lambda t}{n}\Big)^{n-k} + g(n),
\end{eqnarray*}
$$
where $g(n) \to 0$ as $n \to \infty$.  Thus the desired limit follows. 

##  Exercises
    
* Would a Poisson process be a good model for the number of arrivals to a sandwich shop for an entire day of business?   Discuss.
  
* How reasonable is the orderliness assumption for arrivals to a restaurant?  Discuss.
    
* Show that Poisson processes constructed as exponential inter-arrial times satisfy orderliness.



##  Poisson processes as uniform random variables

Another way to simulate Poisson point processes is given by the following characterization on finite volumes.  To simulate a Poisson point process of intensity $\lambda$ on an interval $[s,t]$:
  
* Call a Poisson random variable  $M$ with mean $\lambda(t-s)$.

* If $M=m$, then place $m$ independent random variables in $[s,t]$ that are uniformly distributed.  

* If a Poisson point process is desired on $[0, \infty)$, then repeat this procedure independently on each interval $[n, n+1)$, for every nonnegative integer $n$. 


## continued


This characterization is my favourite and also has the advantage that it can easily be generalized to higher dimensions and other spaces.  


The following code outputs the a sequence of arrivals of a Poisson process of intensity $2$ in the interval $[0,1]$

```{r}
M = rpois(1, 2) 
P = runif(M) 
sort(P, decreasing=F)
```

## Exercises

* Using the characterization of Poisson point processes as a Poisson number of uniform random variables, show by simulations that the first arrival is distributed as an exponential random variable.


##  Other spaces and spatial Poisson point processes


* In more general spaces, it is more natural to consider the point process (the location of the arrivals) rather than the counting process (the  total number  arrivals).   

* Here we think of the point process $\Pi$  as a random subset of $\mathbb{R}^d$, and we let $\Pi(A)$ to be the number of $\Pi$-point in $A \subset \mathbb{R}^d$.      

* We say that $\Pi$ is a **Poisson point process on $\mathbb{R}^d$ with intensity $\lambda$** if: 


## continued

* For every subset $A \subset \mathbb{R}^d$ of finite volume $|A|$, the random variable is $\Pi(A)$ is a Poisson random variable with mean $\lambda |A|$.

* For any finite collection of disjoint sets  $A_1, \ldots, A_n$ with finite volume the random variables $\Pi(A_1), \ldots, \Pi(A_n)$ are independent.


## continued

This characterization is also lends itself to easy simulations.  To simulate a Poisson point process $\Pi$ of intensity $\lambda$ on a set $A \subset \mathbb{R}^d$:
    
* Call a Poisson random variable  $M$ with mean $\lambda |A|$.

* If $M=m$, then place $m$ independent random variables in $A$ that are uniformly distributed.  

* If a Poisson point process is desired on $\mathbb{R}^d$, then consider a partition of $\mathbb{R}^d = \bigcup_{i=1} ^{\infty} A_i$ into sets of finite volume and   repeat this procedure independently on each set $A_i$, for every $i \in \mathbb{Z}^{+}$.  


## R code

The following R code gives a Poisson point process on a square.

```{r}
M= rpois(1,10)
x = runif(M)
y = runif(M)
```

## continued

```{r}
plot(x,y, xlim=c(0,1), ylim=c(0,1))
```

## Some other nice properties


* (Superposition) The sum of two independent Poisson point processes is again a Poisson point process.

* (Colouring)  Given a Poisson point process if we colour the points red or blue independently with probability $p \in (0,1)$, the resulting blue and red point processes are Poisson and independent.


* (Scaling)  Given a Poisson point process $\Pi$ on $\mathbb{R}^d$ of intensity $\lambda$, the scaled point process $c \Pi$ formed by multiplying each $\Pi$-point by $c >0$, is a Poisson point process of intensity $c^{-1} \lambda$.


## Exercises
    
    
* Simulate a Poisson point process on a disc.   
  
   
*   Show that the law of a Poisson point process is invariant under rotations; that is, if $\Pi$ is a Poisson point process on a disc, and $\theta$ is a rotation, the point process $\theta \Pi$ formed by rotating all the point by $\theta$ is still a Poisson point process on the disc.
   
    
* Show that the addition of two independent Poisson point on a disc is a Poisson point process on a disc. 

* Prove the scaling property for Poisson point processes on $[0, \infty)$



## Summary

* We reviewed the basics of Poisson point processes from the persepective of: inter-arrival times, modelling, and uniform random variables.  

* We saw how to simulate Poisson point processes on an interval and higher dimensions   

* We thought about how about how all these characterizations are equivalent, and the suitability of the Poisson process for modelling.   

##  Version: `r format(Sys.time(), '%d %B %Y')`
  * [Rmd Source](https://tsoo-math.github.io/ucl/poisson-slides.Rmd)





