Why is the time to the next reaction in a stochastic simulation an exponential disbtribution?

Question

The title basically says it all, I am reading about stochastic simulation and each article/book I read says the time until the next reaction is given as and $Exp(h_i(x,c_i))$ where $h_i$ is our stochastic rate law (hazard), $c_i$ is the stochastic rate constant and $x$ is the current marking of the system.

When reading about this the literature just says that the time to next reaction is given as an exponential distribution, but I do not understand why. I have seen some things mention poisson process, but I don't know why I just assume that the system is a poisson process.

Answer 1

I am not familiar with the field that you are describing (from what I found on a quick web search, it is chemical kinetics), however the process as described defines an in-homogeneous Poisson process. Whilst I cannot concretely help with your context, perhaps the below helps to motivate the definition.

The exponential distribution is a canonical choice of distribution for waiting times between points (reactions as you call them above). This is because exponential waiting times have the memorylessness property.

In short, this says that the time left to wait for a reaction to occur is independent of the time already waited. In notation, if $T$ denotes the time between two reactions then this says

$$\mathbf P (T > t \, | T > s) = \mathbf P( T > t - s ), \qquad t > s.$$

The only continuous distribution with this property is the exponential distribution.

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Two ways to define the Poisson Process

One angle that might help is to note that there are two ways to define a (homogeneous) rate $\lambda > 0$ Poisson process.

1. Via the Poisson Distribution. This approach determines the Poisson point process as being the unique distribution of points $X = \{x_i\}_{i=1}^\infty \subset \mathbf R$ such that the number of points in any (measurable) set $A \subset \mathbf R$ satisfies $$\mathbf P( \# \{X \cap A \} = n) = e^{-\lambda |A|} \frac{(\lambda |A|)^n}{n!},$$ that is, the number of points in $A$ is poisson distributed with rate $\lambda |A|$, where $|A|$ denotes the lebesgue measure of $A$; so if $A = [a,b]$ is an interval then $|A| = b- a$.
2. Via Exponential Waiting Times. The second approach defines the points $X = (x_i)_{i=1}^\infty$ by setting $$x_i = J_i + x_{i-1},$$ where $x_0 = 0$ (which is not a point of $X$), and each $J_i \sim \text{Exp}(\lambda)$ is an independent exponential variable.

The point here is, if you think that your source is referring to the exponential waiting times as a property (rather than the definition), then perhaps the first definition will be clearer from the text.

Proving that these two definitions are equivalent takes a bit of work, but is a standard exercise in probability.