# Why is $\lambda$ called the *instantaneous* rate of change in the exponential distribution?

In the following paramterisation of the exponential distribution

$${\displaystyle f(t;\lambda )={\begin{cases}\lambda e^{-\lambda t}&t\geq 0,\\0&t<0.\end{cases}}},$$

$$\lambda$$ is called the "rate" parameter. If $$T \sim \text{Exp}(\lambda)$$, I think I understand the intuition behind why it's called a (average) rate - because it's the average number of arrivals per unit time $$\left( \lambda = \frac{1}{\mathbb E(T)}\right)$$; On average, there is 1 arrival in $$\mathbb E (T)$$ amount of time.

However, in some places (for example, in continuous time markov chains), this $$\lambda$$ is called the instantaneous rate of change.

How is $$\lambda$$ an instantaneous rate of change (what makes it instantaneous?)?

• instantaneous = derivative? I mean, $df/dt = - \lambda f$ ? Jul 19, 2020 at 2:11
• @FormerMath Thanks! However, that seems to be a coefficient for the rate of change of the density, and I'm not sure how to interpret that. I was hoping to connect the explanation to the actual speed of arrivals, somehow, if that's possible Jul 19, 2020 at 2:15

Some motivation for the term "instantaneous" in this context is that the transition rate matrix for a continuous-time Markov chain is actually an infinitesimal generator. Let $$\{X(t):t\geqslant 0\}$$ be a CTMC. Define the jump times of the process by $$J_0=0$$ and $$J_{n+1} = \inf\{t>J_n: X_t\ne X_{J_n}\},\ n\geqslant 1,$$ the holding times by $$S_n = J_n-J_{n-1},\ n\geqslant 0,$$ and the jump process by $$Y_n = X_{J_n},\ n\geqslant 0.$$ We are mostly interested in CTMCs that have right-continuous sample paths, that is for any $$t\geqslant 0$$, there exists $$\varepsilon>0$$ such that $$X(t+s)=X(t)$$ for all $$0\leqslant s\leqslant\varepsilon$$. This ensures that the holding times are strictly positive. There is also the matter of "explosion," where there can exist a random time $$\xi$$ such that $$\xi:= \sup_n J_n =\sum_{n=1}^\infty S_n <\infty.$$ Note that this can only happen in CTMCs on countably infinite state spaces, as boundedness of the transition rates implies that $$\xi=+\infty$$. This is a rather pathological case, however, since it means there are infinite transitions in a finite amount of time - and it is not clear how to define the process after that time!

Now, for times $$s,t>0$$ and states $$i,j$$ we can write $$P_t:= \mathbb P(X(t+s)=j\mid X_s=i)$$ due to homogeneity. The collection of matrices $$\{P_t:t\geqslant 0\}$$ determine the transient behavior of the process and in fact form a semigroup, as $$P_{t+s}=P_tP_s$$ (a semigroup is a set with a binary operation that is associative). Morever, since $$P_\varepsilon\to P_0=I$$ (the identity matrix) as $$\varepsilon\downarrow0$$, this semigroup is right-continuous for all $$t$$.

Some important results are the following:

For any states $$i$$ and $$j$$, the the following limits exist and are nonnegative: \begin{align} q_i:&=\lim_{\varepsilon\downarrow0}\frac{(1-P_\varepsilon(i,i))}\varepsilon\\ q_{ij} :&= \lim_{\varepsilon\downarrow0}\frac{P_\varepsilon(i,j))}\varepsilon. \end{align}

Set $$q_{ii}=-q_i$$ and $$q_{ij}$$ as defined above, then the matrix $$A=(q_{ij})$$ is the infinitesimal generator of the semigroup. An interesting example of this is a discrete time Markov chain subordinated to a Poisson process. Let $$\{\hat X_n:n=0,1,\ldots\}$$ be a Markov chain with transition matrix $$Q$$ and $$\{N(t):t\geqslant0\}$$ an independent Poisson process with intensity $$\lambda>0$$. Define $$X_t := \hat X_{N_t},\ t\geqslant 0.$$ Then $$\{X_t:t\geqslant 0\}$$ is a continuous time Markov chain with generator $$A=\lambda(Q-I)$$.

The infinitesimal generator also happens to be the unique solution to the backward Kolmogorov differential equations $$P'(t)=AP(t),$$ where we can explicitly write $$P$$ as the matrix exponential of $$A$$: $$P(t) = e^{Qt} := \sum_{n=0}^\infty \frac{Q^n}{n!}.$$ It also turns out that explosiveness becomes an issue here - the backward equations are well-defined for any CTMC, but the analogous forward equation $$P'(t)=P(t)A$$ cannot be rigorously justified for explosive processes.

I hope this answer sheds some light as to why the transition rates in a continuous time Markov chain are called "instantaneous."

Consider a inhomogenous poisson process with rate function $$\lambda(t)$$. For any given interval $$[0,t]$$ the count distribution is given by $$P(N(t)=n) = \frac{\Lambda(t)^n}{n!}e^{-\Lambda(t)}$$, where $$\Lambda(t) = \int_0^t \lambda(t) \;dt$$

From this perspective, it is hopefully clearer why $$\lambda$$ is a rate. For continuous-time Markov chains, the probability of a transition from state $$i$$ to state $$j$$ after time interval $$\delta$$ is also a poisson process, with rate interpretation as above.

When dealing with a standard poisson process, the rate does not change and so $$\lambda$$ can be interpreted as an average accumulation rate: $$\Lambda(t) = \lambda t$$