- #1

i_a_n

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**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.