I am studying stochastic processes by myself with the textbook written by Sheldon M. Ross.
Because of my short knowledge, I have been faced with some difficulties to understand....
My question is about the alternating renewal processes in the chapter renewal theory.
$\{X_n, n=1,2,\cdots\}$ is a sequence of nonnegative independent random variables with a common distribution $F$, interpreting $X_n$ as the time between the $(n-1)$st and $n$th event.
$S_n$ is the time of the $n$th event. \begin{align} \displaystyle S_n= \begin{cases} \sum_{i=1}^n{X_i}& n\ge1 \\ 0 &n=0 \end{cases} \end{align}
$Y(t)$ is called the excess or residual life at $t$.
$Y(t)=S_{N(t)+1}-t$
$A(t)$ is called the age at $t$.
$A(t)=t-S_{N(t)}$
Theorem 3.4.4
If $E[Z_n+Y_n]<\infty$ and $F$ is nonlattice, then $$\lim_{t\to\infty}P(t)=\frac{E[Z_n]}{E[Z_n]+E[Y_n]}$$ where \begin{align} X_n&=Z_n+Y_n\\ Z_n& \mbox{ is on time.}\\ Y_n& \mbox{ is off time.} \end{align}
Suppose we want to derive $P\{A(t)\le x\}$. To do so let an on-off cycle correspond to a renewal and say that the system is "on" at time $t$ if the age at $t$ is less than or equal to $x$. In other words, the system is "on" the first $x$ units of a renewal interval and "off" the remaining time. Then, if the renewal distribution is not lattice, we have by Theorem 3.4.4 that \begin{align} \lim_{t\to\infty}P\{A(t)\le{x}\}&=\frac{E[\min(X, x)]}{E[X]}\\ &=\frac{1}{E[X]}\int_0^\infty P\{\min(X, x) \gt y\}dy\\ &=\frac{1}{\mu}\int_0^x{1-F(y)}dy \end{align}
Similarly to obtain the limiting value of $P\{Y(t)\le x\}$, say that the system is "off" the last $x$ units of a renewal cycle and "on" otherwise. Thus the off time in a cycle is min(x, X), and so \begin{align} \lim_{t\to\infty}P\{Y(t)\le{x}\}&=\lim_{t\to\infty}P\{\mbox{off at }t\}\\ &=\frac{E[\min(x, X)]}{E[X]}\\ &=\frac{1}{\mu}\int_0^x{1-F(y)}dy \end{align}
My question is How $\min$ emerged? I think $x$ should be always less than $X$ because if $x$ became larger than $X$, the renewal wouldn't happen. I understood like the following picture. Can someone correct my faults?
Upper one:
Lower one:
Thank you for reading my long question.
