# Why do Markov processes have exponentially distributed waiting times?

Suppose I have a length of slab $$\Delta z,$$ consisting of two types of material. The process $$Y(z)$$ controls the type of material where the state space is $$S = (y_1,y_2).$$

In this small interval of slab the process switches to the other material with probability proportional to $$\Delta z,$$ say $$\lambda\Delta z$$ and remains the same with probability $$1-\lambda\Delta z,$$ where $$\lambda$$ is a positive constant so $$Y(z)$$ is homogeneous and symmetric in the two materials. The semigroup has the form

$$P_{\Delta z} = \begin{bmatrix} 1-\lambda\Delta z & \lambda\Delta z \\ \lambda\Delta z & 1 - \lambda\Delta z \end{bmatrix}$$ with infinitesimal generator

$$\mathcal{L} = \lambda\begin{bmatrix} -1 & 1 \\ 1 & -1 \end{bmatrix}.$$ The trajectories of this process are constructed via first introducing the increasing sequence

$$0 < Z_0 < Z_1 < Z_2 < ... < Z_n < ...$$

of successive random transition points, where the process switches material. In this model the layer sizes are

$$Z_1,Z_2 - Z_1,...,Z_n - Z_{n-1},...$$

and form a sequence of independent random variables with the common exponential distribution with parameter $$\lambda$$

$$\mathbb{P}[Z_n-Z_{n-1}\leq z] = 1 - e^{-\lambda z}$$

for $$n\geq 1.$$

I am really struggling how to see how this sequence has an exponential distribution. Could someone explain this to me or provide a reference?

• A Markov chain must comply with the memoryless propety (Markov property). Only two kinds of distributions are memoryless: geometric distributions of non-negative integers and the exponential distributions of non-negative real numbers. Your case deals with the real numbers. Oct 4, 2021 at 13:00
• @FeedbackLooper Thanks for clarifying that...so here $\mathbb{P}[Z_n - Z_{n+1}\leq z] = 1 - e^{-\lambda z}$ is the cumulative distribution function? Oct 4, 2021 at 14:55
• Yes, it is the cumulative distribution for $Z_n-Z_{n+1}$. Oct 4, 2021 at 16:34