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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?)?

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  • $\begingroup$ instantaneous = derivative? I mean, $df/dt = - \lambda f$ ? $\endgroup$
    – FormerMath
    Jul 19, 2020 at 2:11
  • $\begingroup$ @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 $\endgroup$
    – user523384
    Jul 19, 2020 at 2:15

2 Answers 2

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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."

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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: $\Lambda(t) = \lambda t$

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