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.


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.


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:

  1. No event is generated from the homogenous exponential process or
  2. The Bernoulli random variable is zero

The probabilities of these two events are:

  1. $\textrm{Pr}[N^*_{\Delta t} = 0] = e^{-\lambda^* \Delta t}$
  2. $\textrm{Pr}[p_t = 0] = 1 - \frac{\lambda(t)}{\lambda^*}$

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?


1 Answer 1


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: $\textrm{Pr}[T \leq t] = 1-e^{\int_0^T \lambda(t) dt}$


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