Proof for simulation of NHPP by thinning Background:
I'm trying to show equivalency between the density function for a non-homogenous exponential process (NHEP?), (i.e. the arrival times of events generated by a non-homogenous Poisson process with time-varying rate parameter $\lambda(t)$) and the density function for arrival times generated by the "thinning" method for simulating a NHEP.
I know that the PDF of an NHEP is given by:
$f_T(t) = \lambda(t) e^{-\int_0^t \lambda(t)dt}$
I also know that an NHPP can be simulated by thinning, where we generate homogeneous exponential arrival times with rate parameter $\lambda^* > \textrm{max}(\lambda(t))$, and for each such arrival time, accept that arrival time with probability $\frac{\lambda(t)}{\lambda^*}$.  
This process may be thought of as a combination of a standard exponential process and a Bernoulli random variable (call it $p$) that varies in time.
Question:
I'd like to somehow relate the thinning simulation method to the PDF of an NHPP; that is, show that the process produced by the thinning method is governed by the same underlying PDF.
Approach:
My idea is to show that the CDFs are the same.  The CDF for the NHEP is given by:
$\textrm{Pr}[T \leq t] = 1 - e^{-\int_0^t \lambda(t)dt}$ 
So the question is, how to show that the CDF for the thinning simulation is equivalent.
To start, define:


*

*$N^*_{T}$ as the number of events  generated by the "un-thinned," homogeneous exponential process over a given time-interval, $T$

*$p_t$ be the Bernoulli random variable at time $t$. If it $p_t=0$, the event at time $t$ (if present) is deleted.  If $p_t=1$, the event at time $t$ (if present) is not deleted


I tried to write the CDF by considering the probability that no event occurred at each infinitesimally small time-step ($\Delta t$) between $0$ and $T$.  At each such time-step, one of two conditions must be met:


*

*No event is generated from the homogenous exponential process or 

*The Bernoulli random variable is zero


The probabilities of these two events are:


*

*$\textrm{Pr}[N^*_{\Delta t} = 0] = e^{-\lambda^* \Delta t}$

*$\textrm{Pr}[p_t = 0] = 1 - \frac{\lambda(t)}{\lambda^*}$
So, the probability that either of those events occur at a given infinitesimal time-step can be written as (skipping a few lines of algebra):
$\textrm{Pr}[N^*_{\Delta t} = 0 \cup p_t=0] = 1 - \left(\frac{\lambda(t)}{\lambda^*}\left(1 - e^{-\lambda^* \Delta t}\right)\right)$
From there, we can say that the probability of the first arrival time being greater than $T$ is the product of the probability of no events occurring during each of the infinitesimal time-steps between zero and $T$, i.e.
$\textrm{Pr}[T > t] = \lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \textrm{Pr}[N_{i} = 0 \cup p_{i\Delta T}=0]$
$\textrm{Pr}[T > t] = \lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \left(\frac{\lambda(i\Delta T)}{\lambda^*}\left(1 - e^{-\lambda^* \Delta t}\right)\right)\right)$ 
So then, the final CDF for the arrival times generated by the thinning process would be
$\textrm{Pr}[T \leq t] = 1-\textrm{Pr}[T > t] = 1-\lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \left(\frac{\lambda(i\Delta T)}{\lambda^*}\left(1 - e^{-\lambda^* \Delta t}\right)\right)\right)$ 
But, this is as far as I have gotten.  I am at a loss as to how to show that this mess is equal to the NHEP CDF:
$\textrm{Pr}[T \leq t] = 1 - e^{-\int_0^t \lambda(t)dt}$
Or, perhaps this is the wrong approach altogether?
 A: I think I've found an answer that at least satisfies me, so I thought I'd post it in case anyone else is interested in this question. It's a bit "hand-wavy" in parts, but I think it more or less makes sense. Picking up from where the question left off:
$\textrm{Pr}[T \leq t] = 1-\lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \left(\frac{\lambda(i\Delta T)}{\lambda^*}\left(1 - e^{-\lambda^* \Delta t}\right)\right)\right)$
Now, this is one of the "hand-wavy" bits.  Based on the Taylor series for $e^x$, we can say:
$e^{-\lambda^* \Delta t} = 1 + (-\lambda^* \Delta t) + \frac{(-\lambda^* \Delta t)^2}{2!} + \frac{(-\lambda^* \Delta t)^3}{3!} + ...$ 
But, we also know that we're dealing with the limit as $\Delta t \to 0$.  So, I'm thinking we can neglect all but the first order terms.
$\lim_{\Delta t \to 0} e^{-\lambda^* \Delta t} = 1 -\lambda^* \Delta t$ 
Using this expression, we can rewrite the probability expression as:
$\textrm{Pr}[T \leq t] = 1-\lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \left(\frac{\lambda(i\Delta T)}{\lambda^*}\left(1 - 1 + \lambda^* \Delta t\right)\right)\right)\\
= 1-\lim_{\Delta T \to 0} \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \lambda(i\Delta T)\Delta t\right)\\
= 1-\exp\left(\lim_{\Delta T \to 0} \ln \prod_{i=0}^{\frac{t}{\Delta T}} \left(1 - \lambda(i\Delta T)\Delta t\right)\right)\\
= 1-\exp\left(\lim_{\Delta T \to 0} \sum_{i=0}^{\frac{t}{\Delta T}} \ln \left(1 - \lambda(i\Delta T)\Delta t\right)\right)$
Now, we can do the reverse approximation.  Specifically, $\ln(1+x) = x + O(x^2)$ as $x\to 0$, so we can say:
$\textrm{Pr}[T \leq t] = 1-\exp\left(\lim_{\Delta T \to 0} \sum_{i=0}^{\frac{t}{\Delta T}} \lambda(i\Delta T)\Delta t\right)$
But now the term in the integral is just the Riemann sum for the integral, i.e.
$\lim_{\Delta T \to 0} \sum_{i=0}^{\frac{t}{\Delta T}} \lambda(i\Delta T)\Delta t  = \int_0^T \lambda(t) dt$
So:
$\textrm{Pr}[T \leq t] = 1-e^{\int_0^T \lambda(t) dt}$
