Definition of Poisson Process Through Exponential Distribution
Definition
- Let’s define $\tau_{1} , \tau_{2} , \cdots \sim \text{exp} ( \lambda )$.
$\lambda$ is referred to as Intensity.
- $\displaystyle s_{n}:= \sum_{k=1}^{n} \tau_{k}$ is called the Arrival Time.
- A stochastic process defined as $N_{t}:= \begin{cases} 0 , & 0 \le t < s_{1} \\ k , & s_{k} \le t < s_{k+1} \end{cases}$, $\left\{ N_{t} \right\}_{t = 0}^{\infty}$ is called a Poisson Process.
Basic Properties
- [1]: $\displaystyle p (N_{t} = k ) = {{ ( \lambda t )^{t} e^{ - \lambda t} } \over { k! }}$
- [2]: $\displaystyle p \left( ( N_{t} - N_{s} ) = k \right) = {{ ( \lambda (t - s ) )^{t} e^{ - \lambda ( t - s )} } \over { k! }}$
- [3]: $E ( N_{t} - N_{s} ) = \lambda (t - s)$
- [4]: $\operatorname{Var} ( N_{t} - N_{s} ) = \lambda ( t - s)$
Explanation
Since $\displaystyle E ( \tau_{i} ) = {{1} \over {\lambda }}$, saying that the intensity is high means that $\tau_{i}$ is shorter and events occur more frequently. Compared to the definition of a Poisson process through a differential matrix, it’s less apparent that the arrival time follows a continuous Markov chain, instead the exponential distribution is directly visible.
Understanding why this is called a Poisson process might be difficult with the definition alone, but it becomes relatively easy if one knows about the relationship between the exponential distribution and the Poisson distribution. Their proofs are essentially the same.
The proofs of [3] and [4] are essentially the same as those of the mean and variance of the Poisson distribution.
Example
The Poisson process is useful in a wide range of fields such as inventory management and epidemiological investigation. For example, consider the situation of an epidemic outbreak.
The government and relevant agencies need to come up with countermeasures as quickly as possible, and naturally, understanding the current situation is the top priority. Typically, epidemics have an incubation period, meaning besides the ‘patients’ who are already showing symptoms, there are ‘carriers’ who have not yet shown symptoms. Failing to accurately assess the number of these carriers could eventually lead to an inability to meet the demand for vaccines.
Now, let’s say the incubation period $\tau$ is a random variable with a cumulative density function $Q ( r ):= p ( \tau \le r )$, and $N_{1;t}$ is the current number of patients, and $N_{2;t}$ is the current number of carriers. $N_{1;t}$, $N_{2;t}$ are both Poisson processes, and the total number of infected individuals will still be represented as a Poisson process $N_{t}:= N_{1;t} + N_{2;t}$.
Addition of Poisson Distributions: If $X_i \sim \text{Poi}( m_{i} )$ then $$\sum_{i=1}^{n} X_{i} \sim \text{Poi} \left( \sum_{i=1}^{n} m_{i} \right)$$
If we call the moment an individual first comes into contact with the virus $s$ and the present moment $t$, then the symptoms will appear when $(t - s) > \tau$. Since $Q ( t - s ) = p ( \tau \le t - s )$, the expected number of patients at this moment is $$ n: = E ( N_{1;t} ) = \lambda \int_{0}^{t} Q (t - s ) ds $$ and the expected number of carriers can be represented as $$ m: = E ( N_{2;t} ) = \lambda \int_{0}^{t} \left[ 1 - Q (t - s ) \right] ds $$ Now, substituting with $x:= t -s$, it can be neatly summarized as $$ n = \lambda \int_{0}^{t} Q ( x) dx \\ m = \lambda \left( t - \int_{0}^{t} Q ( x) dx \right) $$ Since we already know $n$, we can use $\displaystyle \hat{\lambda}:= {{ n } \over { \int_{0}^{t} Q ( x) dx }}$ as an estimate for $\lambda$. Then, $$ n = \lambda \int_{0}^{t} Q ( x) dx \\ m = \lambda \left( t - \int_{0}^{t} Q ( x) dx \right) $$ and if we know $Q’(x)$, which will be the probability density function for the incubation period, we can approximately find the number of carriers.