Consider a homogeneous Markov Chain $X$ on a countable state space, ie a jump process. It is said to be regular (does not explode) if there are only a finite number of jumps in every finite interval. $$X(t) < \infty\quad \forall t > 0$$

What can you say the expected number of jumps in some interval? Is it finite too for regular processes in general? Is there a $a > 0$ so that

$$E[X(a)] < \infty$$


Consider for example a Poisson process with paramter $\lambda$. It is regular, and the expected number of jumps in a unit time is $\lambda$.

Especially I'd like to find out if the expected number of jumps of a regular pure birth-process is finite in some interval. That is a process process that jumps from $k$ to $k+1$ with rate $\lambda_k$, where for the rates $$\sum_{k=1}\frac{1}{\lambda_k} = \infty$$ holds. By Reuter's criterion this is sufficient for the pure-birth process to be regular.


The answer is: not necessarily.

We begin with some partial positive results. Recall that for every pure birth-process $(X_t)_{t\geqslant0}$ with positive rates $(\lambda(k))_{k\geqslant0}$ and for every suitable function $u$, $$ \frac{\mathrm d}{\mathrm dt}\mathrm E(u(X_t))=\mathrm E((u(X_t+1)-u(X_t))\cdot\lambda(X_t)). $$ In particular, the expectation $\mathrm E(X_t)$, if it exists, solves the differential equation $$ \frac{\mathrm d}{\mathrm dt}\mathrm E(X_t)=\mathrm E(\lambda(X_t)). $$ If $\lambda(k)\leqslant\lambda_0(k)$ for every integer $k$, for some positive concave function $\lambda_0$, then Jensen inequality yields $\mathrm E(\lambda(X_t))\leqslant\mathrm E(\lambda_0(X_t))\leqslant\lambda_0(\mathrm E(X_t))$. Integrating this yields $t\geqslant M(\mathrm E(X_t))$, where $$ M(x)=\int_{\mathrm E(X_0)}^{x}\frac{\mathrm dz}{\lambda_0(z)}. $$ If the integral of the function $1/\lambda_0$ diverges at infinity, $M$ is unbounded and this proves that $\mathrm E(X_t)\leqslant M^{-1}(t)$ is finite for every $t$.

Here is another integrability result, always valid. Consider $$ \Lambda(k)=\sum_{i=0}^{k-1}\frac1{\lambda(i)}. $$ Then the derivative of the function $t\mapsto\mathrm E(\Lambda(X_t))$ is $1$, hence $\mathrm E(\Lambda(X_t))=\Lambda(X_0)+t$ for every $t$ and in particular $\mathrm E(\Lambda(X_t))$ is finite for every $t$.

But of course, nothing guarantees that such a concave function $\lambda_0$ exists nor that $\Lambda(k)$ would be equivalent to a multiple of $k$ when $k\to\infty$...

...Which brings us to the negative result. Consider an infinite increasing positive integer sequence $(K(i))_{i\geqslant0}$, to be chosen later on. The sites in the set $\mathcal K=\{K(i)\,;\,i\geqslant0\}$ are slow sites and the other sites are fast sites, in the sense that one assumes that $\lambda(k)=1$ for every $k$ in $\mathcal K$ and that $$ \sum\limits_{k\notin \mathcal K}\frac1{\lambda(k)}\ \text{is finite}. $$ Then, Reuter criterion holds thanks to the infinitely many slow sites hence $(X_t)_{t\geqslant0}$ is regular. However, the total time $T$ spent at fast sites is almost surely finite, hence, on $[T\leqslant t]$, $$ X_t\geqslant K(Y_{t-T(t)})\geqslant K(Y_{t-T}), $$ where $T(t)$ is the time spent at fast sites up to time $t$, and $(Y_t)_{t\geqslant0}$ is a pure birth-process with constant rate $1$, independent on $T$. For every $t\gt0$, $[T\leqslant t]$ has positive probability and $$ \mathrm E(X_{2t}:T\leqslant t)\geqslant\mathrm E(K(Y_{2t-T}):T\leqslant t)\geqslant\mathrm E(K(Y_{t}):T\leqslant t)=\mathrm E(K(Y_{t}))\cdot\mathrm P(T\leqslant t). $$ Since $Y_t$ is Poisson with parameter $t$, the choice $K(i)=(i!)^2$ yields $$ \mathrm E(K(Y_t))=\mathrm e^{-t}\sum_{i=0}^{+\infty}i!\,t^i, $$ which diverges, for every positive $t$. Thus, $\mathrm E(X_{2t})\geqslant\mathrm E(X_{2t}:T\leqslant t)$ is infinite for every positive $t$.

  • $\begingroup$ Thanks! I think i understood the first example. To get the estimation involving M() you would need to make use the differential equation however, right? But i thought for that to hold you would need the existence of the expectation in first place. $\endgroup$ – Haderlump Aug 2 '12 at 12:52
  • $\begingroup$ I skipped this step in the answer because it is standard: consider $X^n_t=\min(X_t,n)$, this is a pure-birth process with rates $\lambda^n(k)=\lambda(k)$ if $k\lt n$ and $\lambda^n(k)=0$ otherwise, estimate $\mathrm E(X^n_t)$ (up to then, everything is bounded), finally let $n\to\infty$ then use that $\mathrm E(X^n_t)\to\mathrm E(X_t)$ for every fixed $t$. $\endgroup$ – Did Aug 2 '12 at 13:24
  • $\begingroup$ Ah, I see. Hm I hoped that regularity would suffice for the process I'm dealing with in my work. I opened a new question that includes all the assumptions that I have for $X$. $\endgroup$ – Haderlump Aug 2 '12 at 19:18

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