A math proof within a question about homogeneous Poisson process

In summary, a homogeneous Poisson process is a mathematical model that represents random events occurring at a constant rate and independently of one another. It can be used in various fields such as finance, insurance, and telecommunications, and is also applied in queuing theory. It differs from a non-homogeneous Poisson process in that it assumes a constant rate and independent events. However, there are limitations to its use, such as its inability to model dependent or varying events and its assumption of a constant rate.
  • #1
i_a_n
83
0
We know that a homogeneous Poisson process is a process with a constant intensity $\lambda$. That is, for any time interval $[t, t+\Delta t]$, $P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$.

And therefore, event count in $[0, T]$ follows a Poisson distribution with rate $\lambda T$. That is, $P\left \{ N(T)=k\right \}=\frac{\text{exp}(-\lambda T)(\lambda T)^k}{k!}$. ($N$ is the count.)

The problem is:

Prove that the following simulation generates a homogeneous Poisson process with rate $\lambda$ on $[0, T]$: Step 1: Sample $m$ from Poisson distribution with mean $\lambda T$. Step 2: Sample $s_1, \cdots,s_m$ i.i.d. from uniform $[0, T]$. That is, demonstrate that for any time interval $[t, t+\Delta t]$ in $[0,T]$, $P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$.

Now we look at the problem, we have

Given $m$ events in $[0,T]$,

$P\left \{ k \;\text{events in}\; [t, t+\Delta t] \right \}\\
=\Sigma ^{\infty }_{m=k}P\left \{ k \;\text{events in}\; [t, t+\Delta t],m \;\text{events in}\; [0,T]\right \}\\
=\Sigma ^{\infty }_{m=k}P\left \{ k \;\text{events in}\; [t, t+\Delta t] | m \;\text{events in}\; [0,T]\right \}\cdot P\left \{ m \;\text{events in}\; [0,T] \right \}\\
=\sum_{m=k}^{\infty }\binom{m}{k}(\frac{\Delta t}{T})^k(\frac{T-\Delta t}{T})^{m-k} \cdot \frac{\text{exp}(-\lambda T)(\lambda T)^m}{m!}$

So in order to prove the result, we should have

$\sum_{m=k}^{\infty }\binom{m}{k}(\frac{\Delta t}{T})^k(\frac{T-\Delta t}{T})^{m-k} \cdot \frac{\text{exp}(-\lambda T)(\lambda T)^m}{m!}=\frac{\text{exp}(-\lambda \Delta t)(\lambda \Delta t)^k}{k!}$ $(*)$

and this should hold. But my question is how to derive $(*)$ mathematically? How to show the two sides are equal in $(*)$? Can you show it?
Thanks in advance.
 
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  • #2
I'll give a detailed solution. There's no hard mathematics involved. We have the LHS:

$$\sum_{m=k}^{\infty }\binom{m}{k}\left(\frac{\Delta t}{T}\right)^k\left(\frac{T-\Delta t}{T}\right)^{m-k} \cdot \frac{\text{exp}(-\lambda T)(\lambda T)^m}{m!}$$
Using the fact that $\binom{m}{k} = \frac{m!}{k!(m-k)!}$, we obtain
$$ = \frac{e^{-\lambda T}}{k!} (\Delta t)^k\sum_{m=k}^{\infty} \frac{(T-\Delta t)^{m-k}}{(m-k)!} \lambda^{m}$$
Write $\lambda^m = \lambda^{(m-k)+k}$ then the previous is equal to
$$\frac{e^{-\lambda T}}{k!} (\Delta t)^k \lambda^k \sum_{m=k}^{\infty} \frac{(\lambda(T-\Delta t))^{m-k}}{(m-k)!}$$
Set $m-k = j$ then we finally have
$$\frac{e^{-\lambda T}}{k!}(\Delta t)^k \lambda^k \sum_{j=0}^{\infty} \frac{(\lambda (T-\Delta t))^j}{j!} = \frac{e^{-\lambda T}}{k!}(\Delta t)^k \lambda^k e^{\lambda T- \lambda \Delta t} = \frac{e^{-\lambda \Delta t}(\lambda \Delta t)^k}{k!}$$
which is the desired result.
 

Related to A math proof within a question about homogeneous Poisson process

1. What is a homogeneous Poisson process?

A homogeneous Poisson process is a mathematical model used to represent the occurrence of random events over time. It assumes that events occur at a constant rate and independently of one another.

2. Can you explain the concept of a math proof within a question about homogeneous Poisson process?

A math proof within a question about homogeneous Poisson process refers to using mathematical techniques and principles to prove the properties or relationships within the process. This can involve using formulas, equations, and other mathematical methods to demonstrate the validity of a statement or hypothesis.

3. What are some real-life applications of homogeneous Poisson process?

Homogeneous Poisson process is commonly used in fields such as finance, insurance, and telecommunications to model the arrival of customers, claims, or calls, respectively. It is also used in queuing theory, where it helps to estimate waiting times and system performance.

4. How is a homogeneous Poisson process different from a non-homogeneous Poisson process?

A homogeneous Poisson process assumes that the rate of events is constant over time and independent of previous events. In contrast, a non-homogeneous Poisson process allows the rate to vary over time and may be dependent on past events.

5. Are there any limitations to using a homogeneous Poisson process?

Yes, there are limitations to using a homogeneous Poisson process. It may not be suitable for modeling events that are not independent or occur at varying rates. Additionally, it assumes a constant rate, which may not accurately reflect real-life situations.

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