A math proof within a question about homogeneous Poisson process

[i_a_n]

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.

 

[Siron]

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.