We know that a homogeneous Poisson process is a process with a constant intensity $\lambda$. That is, for any time interval $[t, t+\Delta t]$, $P\left \{ k \text{ events in } [t, t+\Delta t] \right \}=\frac{\exp(-\lambda \, \Delta t)(\lambda \, \Delta t)^k}{k!}$.
And therefore, event count in $[0, T]$ follows a Poisson distribution with rate $\lambda T$. That is, $P\left \{ N(T)=k\right \}=\frac{\exp(-\lambda T)(\lambda T)^k}{k!}$. ($N$ 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 $[0,T]$, $P\left \{ k \text{ events in } [t, t+\Delta t] \right \}=\frac{\exp(-\lambda \, \Delta t)(\lambda \, \Delta t)^k}{k!}$.
Now we look at the problem, we have
Given $m$ events in $[0,T]$,
\begin{align} & P\left \{ k \text{ events in } [t, t+\Delta t] \right \}\\ = {} & \sum^\infty_{m=k} P\left \{ k \text{ events in } [t, t+\Delta t],m \;\text{events in}\; [0,T]\right \}\\ = {} & \sum^\infty_{m=k} P\left\{ k \text{ events in } [t, t+\Delta t] \mid m \text{ events in } [0,T]\right \}\cdot P\left \{ m \text{ events in } [0,T] \right \}\\ = {} & \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{\exp(-\lambda T)(\lambda T)^m}{m!} \end{align}
So in order to prove the result, we should have
$$\sum_{m=k}^\infty\binom{m}{k} \left(\frac{\Delta t}{T}\right)^k \left(\frac{T-\Delta t}{T}\right)^{m-k} \cdot \frac{\exp(-\lambda T)(\lambda T)^m}{m!}=\frac{\exp(-\lambda \,\Delta t)(\lambda \,\Delta t)^k}{k!} \tag{$*$} $$
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.
For simplicity of notation, assume $T=1$. Then, the left hand side of your expression is: \begin{align*} \sum_{m=k}^\infty {m \choose k}(\Delta t)^k (1-\Delta t)^{m-k} \frac{e^{-\lambda} \lambda^m}{m!} &= \frac{e^{-\lambda \Delta t} (\lambda \Delta t)^k}{k!} \sum_{m=k}^\infty {m \choose k}(\Delta t)^k (1-\Delta t)^{m-k} \frac{e^{-\lambda} \lambda^m}{m!} \frac{k!}{e^{-\lambda \Delta t}(\lambda \Delta t)^k} \\ &\text{(multiply and divide by the expression we want)} \\ &\\ &= \frac{e^{-\lambda \Delta t} (\lambda \Delta t)^k}{k!} \sum_{m=k}^\infty e^{-\lambda (1-\Delta t)} \frac{(\lambda (1-\Delta t))^{m-k}}{(m-k)!} \\ &\text{(simplify the expression within the summation)} \\ &\\ &= \frac{e^{-\lambda \Delta t} (\lambda \Delta t)^k}{k!} \sum_{s=0}^{\infty} e^{-\lambda (1-\Delta t)} \frac{(\lambda (1-\Delta t))^s}{s!} \\ &\text{(substitute $s=m-k$)} \\ &\\ &= \frac{e^{-\lambda \Delta t} (\lambda \Delta t)^k}{k!} \\ &\text{(the summation is the sum of a poisson pdf, and hence equals 1)} \end{align*}