# Why can expectation and ratio be exchanged in this case?

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 $E[H(X)] = \frac{E[R]}{E[r]}$?

Thanks!

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