Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

From Simulation and the Monte Carlo Method by Reuven Y. Rubinstein, Dirk P. Kroese

Let $\{X_t\}$ be a regenerative process with regeneration times $T_0,T_l,\dots$ Let $r_i = T_i - T_{i-1} i = 1,2,\dots$ be the cycle lengths. Depending on whether $\{X_t\}$ is a discrete-time or continuous-time process, define, for some real-valued function $H$, $$ R_i= \sum_{t=T_{i-1}}^{T_i-1} H(X_t), (4.21) $$ or $$R_i= \int_{T_{i-1}}^{T_i} H(X_t)dt, (4.22)$$ respectively, for $i = 1, 2, \dots$. We assume for simplicity that $T_0 = 0$. We also assume that in the discrete case the cycle lengths are not always a multiple of some integer greater than 1. Let $r = T_i$ and $R = R_i$.

If $E[r] < \infty$, then, under mild conditions, the process $\{X_t\}$ has a limiting (or steady-state) distribution, in the sense that there exists a random variable $X$, such that $$\lim_{t\to \infty} P(X_t \leq x) = P(X \leq x) .$$ In the discrete case, no extra condition is required. In the continuous case a sufficient condition is that the sample paths of the process are right-continuous and that the cycle length distribution is non-lattice — that is, the distribution does not concentrate all its probability mass at points $n \delta, n \in \mathbb{N}$, for some $\delta > 0$.

If the above conditions hold, then the steady-state expected value is given by $$ E[H(X)] = \frac{E[R]}{E[r]}. (4.23)$$

If I am correct, $H(X) = R/r$. But how can expectation and ration exchange their order in $ E[H(X)] = \frac{E[R]}{E[r]}$?


share|cite|improve this question
Where do you see that H(X)=R/r? – Did Nov 13 '12 at 21:06
@did: I messed it up. It may not be true. See my answer please. Thanks! – Tim Nov 13 '12 at 21:42
A remarkable euphemism, if ever there was one. – Did Nov 13 '12 at 21:47
I beg your pardon, if I can? – Tim Nov 13 '12 at 21:50

It is not necessarily true that $H(X) = R/r$.

In the discrete case $$ \frac{E[R]}{E[r]} = \frac{E[\sum_{t=0}^{r - 1} H(X_t)]}{E[r]} = \frac{E[ E(\sum_{t=0}^{T - 1} H(X_t) | r=T)]}{E[r]} = \frac{E[ r E(H(X))]}{E[r]} = E(H(X)). $$

In the continuous case, use Fubini's theorem to prove it. See here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.