lecture-08.tex

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\title{Lecture 08: Key Renewal Theorem and Applications}
\author{}

\begin{document}
\maketitle

\section{Key Renewal Theorem}

%\subsection{Directly Riemann Integrable}
%\begin{defn}%[Directly Riemann Integrable] 
For each $\delta > 0$ and $n \in \N$, we define intervals $I(n,\delta) = [(n-1)\delta, n\delta)$ that partition the positive axis $\R_+ = [0, \infty)$. 
Let $h: \R_+ \mapsto \R$ be a function bounded over finite intervals, denoting 
\begin{xalignat*}{3}
&\underline{m}(h,n,\delta) = \inf\{h(u): u \in I(n,\delta)\} &&\overline{m}(h,n,\delta) = \sup\{h(u): u \in I(n,\delta)\}.
\end{xalignat*}
A function $h: \R_+ \mapsto \R$ is \textbf{directly Riemann integrable} and denoted by $h \in \D$ if the partial sums obtained by summing the infimum and supremum of $h$, 
taken over intervals obtained by partitioning the positive axis, are finite and both converge to the same limit, for all finite positive interval lengths. That is,
\begin{equation*}
\overline{\sigma}_{\delta} \triangleq \lim_{\delta \to 0} \delta \sum_{n \in \N}\overline{m}(h,n,\delta)=\lim_{\delta \to 0} \delta \sum_{n \in \N} \underline{m}(h,n,\delta) \triangleq \underline{\sigma}_{\delta}.  
\end{equation*}   
 If both limits exist and are equal, then the integral value is equal to the limit. 
%\end{defn}
\begin{shaded*}
We compare the definitions of directly Riemann integrable and Riemann integrable functions. 
For a finite positive $M$, a function $g: [0, M] \to \R $ is Riemann integrable if 
\begin{align*}
\lim_{\delta \to 0} \delta \sum_{n \leq M/\delta}\overline{m}(g,n,\delta) &=\lim_{\delta \to 0} \delta \sum_{n \leq M/\delta}\underline{m}(g,n,\delta). 
\end{align*} 
In this case, the limit is the value of the integral. 
For $h$ defined on $\R_+$, 
\begin{align*}
\int_{u \in \R_+}h(u)du &= \lim_{M \to \infty}\int_{0}^{M}h(u)du, 
\end{align*}
if the limit exists. For many functions, this limit may not exist.

%\begin{rem}
A directly Riemann integrable function over $\R_+$ is also Riemann integrable, but the converse need not be true. 
For instance, consider the following Riemann integrable function
\begin{align*}
h(t) &= \sum_{n\in\N}1\left\{t \in \left[n-\frac{1}{(2n^2)},\,n+\frac{1}{(2n^2)}\right]\right\}
\end{align*} is Riemann integrable, but $\delta \sum_{n \in \N}\overline{m}(h,n,\delta)$ is always infinite for every $\delta>0.$
%\end{rem}  
\end{shaded*}
\begin{prop}%[Sufficiency for Directly Riemann Integrable] 
Following are sufficient conditions for a function $h$ to be directly Riemann integrable.
  \begin{enumerate}[(a)]
  \item If $h$ is non-negative, continuous, and has finite support. 
  \item If $h$ is non-negative, continuous, bounded, and $\overline{\sigma}_{\delta}$ is bounded for some $\delta$.  
  \item If $h$ is non-negative, monotone non-increasing, and Riemann integrable. 
  \item If $h$ is non-negative and bounded above by a directly Riemann integrable function.
  \end{enumerate}
\end{prop}
\begin{prop}[Tail Property] If $h$ is non-negative, directly Riemann integrable, and has bounded integral value, then 
\begin{equation*}
\lim_{t \to \infty} h(t)=0.
\end{equation*}
\end{prop}

\begin{thm}[Key Renewal Theorem] 
Consider a renewal process with renewal function $m(t)$, and the mean and the distribution of inter-renewal times being denoted by $\mu$ and $F$ respectively. 
If $F$ is non-lattice and $F(\infty) = 1$, then for any $h \in \D$, we have 
%Denoting mean of inter-renewal time by $\mu$, we have for non-lattice $F$
%Let $N(t)$ be a renewal process having mean $m(t)$, and \textit{iid} inter-arrival times with mean $\mu$ and distribution function $F$. If $F$ is non-lattice, and if a function $h(t)$ is directly Riemann integrable, then
\begin{equation}
\label{eqn:Key Renewal Theorem}
\lim_{t \to \infty} \int_{0}^{\infty}h(t-x)dm(x)=\frac{1}{\mu}\int_{0}^{\infty}h(t)dt. 
\end{equation}
If $F$ is lattice with period $d$ and $\sum_{k \in \N_0}h(t+kd)$ converges, then 
\begin{align*}
\lim_{n \to \infty} \int_{0}^{\infty}h(t+nd-x)dm(x)=\frac{d}{\mu}\sum_{n \in \N_0}h(t+kd). 
\end{align*}
%where 
%\begin{xalignat*}{3}
%&m(t) = \sum_{n \in \N}F_n(t),&& \mu= \int_{0}^{\infty}\bar{F}(t).
%\end{xalignat*}
\end{thm}
%\begin{proof}
%We know that $m \ast F = m - 1$, so $m \ast (1- F) = 1$. 
%Since the function $1-F$ is monotone, 
%\begin{align*}
%1 &= \int_0^tdm(s)[1-F(t-s)] \geq [m(t) - m(t-b)](1-F(b)),
%\end{align*}
%where $b$ is chosen so that $F(b) < 1$. Hence, we have
%\begin{align*}
%\sup_{t}[m(t) - m(t-b)] = \beta_b < \infty.
%\end{align*}
%\end{proof}
\begin{prop}[Equivalence] Blackwell's theorem and key renewal theorem are equivalent.
\end{prop}
\begin{proof} Let's assume key renewal theorem is true. We select $h$ as a simple function with value unity on interval $[0, a]$ and zero elsewhere. That is,
\begin{equation*}
h(x) = 1_{\{ x \in [0,a]\}}.
\end{equation*}
It is easy to see that this function is directly Riemann integrable. %With this selection of $h$, Blackwell's theorem follows.

To see how we can prove the key renewal theorem from Blackwell's theorem, observe from Blackwell's theorem that,
\begin{align*}
\lim_{t \to \infty}\frac{dm(t)}{dt} \stackrel{(a)}= \lim_{a \to 0}\lim_{t \to \infty} \frac{m(t
+a)-m(t)}{a}=\frac{1}{\mu}.
\end{align*}     
where in $(a)$ we can exchange the order of limits under certain regularity conditions. 
We defer the formal proof for a later stage.
\end{proof}
%\begin{rem} 
Key renewal theorem is very useful in computing the limiting value of some function $g(t)$, probability or expectation of an event at an arbitrary time $t$, for a renewal process. This value is computed by conditioning on the time of last renewal prior to time $t$.
%\end{rem}

%\section{Key Renewal Theorem and Applications}
%
%\subsection{Key Renewal Theorem}
%\begin{thm}[Key Renewal Theorem] Let $N(t)$ be a renewal process having mean $m(t)$, and \textit{iid} inter-arrival times with mean $\mu$ and distribution function $F$. If $F$ is non-lattice, and if a function $h(t)$ is directly Riemann integrable, then
%\begin{equation}
%\label{eqn:Key Renewal Theorem}
%\lim_{t \to \infty} \int_{0}^{\infty}h(t-x)dm(x)=\frac{1}{\mu}\int_{0}^{\infty}h(t)dt,
%\end{equation}
%where 
%\begin{xalignat*}{3}
%&m(t) = \sum_{n \in \N}F_n(t),&& \mu= \int_{0}^{\infty}\bar{F}(t).
%\end{xalignat*}
%\end{thm}
%%\textbf{Remark:} and we can deduce one theorem from the other. 

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Alternating renewal processes}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Alternating renewal processes form an important class of renewal processes, and model many interesting applications. 
We find one natural application of  key renewal theorem  in this section. 
%\begin{defn}[Alternating Renewal Process] 

Let $\{(Z_n,Y_n),~n \in \N\}$ be an \textit{iid} random process, where $Y_n$ and $Z_n$ are not necessarily independent. 
A renewal process where each inter-renewal time $T_n$ consist of \textbf{on} time $Z_n$ followed by \textbf{off} time $Y_n$, is called an \textbf{alternating renewal process}. 
We denote the distributions for on, off, and renewal periods by $H, G$, and $F$, respectively. Let 
\begin{equation*}
P(t)=\Pr\{\text{on at time}~ t\}.
\end{equation*}
%\end{defn}
%Let $Y_n \sim G$, $Z_n \sim H, F \sim Z_n+Y_n \triangleq X_n$. The random variable $Z_n$ denotes the ON time of a system and $Y_n$ denotes the OFF time of the system. Let $P(t)=P(\text{ON at time}~ t)$. 
%\begin{rem}
To see that the alternating renewal process is indeed a renewal process, it needs to be established that $\{T_n:n\in\N\}$ is an \textit{iid} sequence. 
However, this follows trivially from the fact that $\{f(Y_n,Z_n):n\in\N\}$ is an \textit{iid} sequence whenever $\{(Z_n,Y_n),~n \in \N\}$ is an \textit{iid} sequence. 
Let $f(a,b) = a+b$ to see that $\{T_n = Y_n + Z_n: n\in\N\}$ is an \textit{iid} sequence.
%\end{rem}

For the renewal process with $n$th inter-renewal time $T_n$ for each $n \in \N$, the $n$th renewal instant is denoted by $S_n = \sum_{i=1}^nT_n$. 
We can define a stochastic process $\{W(t) \in \{0,1\}, t \geqslant 0\} $ that takes values $1$  and $0$, when the renewal process is in on and off state respectively. 
In particular, we can write 
\eq{
W(t) &= 1\{A(t) < Y_{N(t)+1}\}. 
}
It is easy to see that $W$ is a regenerative process with regenerative sequence $S$. 
%We are interested in finding the probability
%\eq{
%P(t) &= P\{W(t) = 1\}.
%}
\begin{thm}[on probability] 
\label{Thm:OnProbability}
Let $m$ be the renewal function associated with the renewal process $\{S_n: n \in \N\}$ with a non-lattice inter-renewal duration distribution $F$. 
If $\E[Z_n+Y_n]< \infty $, then
\begin{equation*}
P(t) = \bar{H}(t)+\int_{0}^{t}\bar{H}(t-y)dm(y).
\end{equation*}
\end{thm} 
\begin{proof} 
We recall that $P(t) = P\{W(t)=1\}$ and compute the kernel 
\eq{
P\{W(t) = 1, T_1 > t\} = P\{Z_1 > t, T_1 > t\} = P\{Z_1 > t\} = \bar{H}(t). 
} 
Hence, we can write the renewal equation for $P(t)$ as 
\eq{
P(t) &= \bar{H}(t) + (F\ast P)(t).
}
Result follows from the solution of renewal equation. % with renewal function $m$ for the renewal process $\{S_n: n\in \N\}$. 
%To find time dependent probability $P(t)$, we can partition the event of system being ON at time $t$ on value of last renewal time $S_{N(t)}$. That is, we can write
%\begin{equation*}
%\{\text{ON at time}~ t\} =\bigcup_{y \in [0, t)}\{\text{ON at time } t, S_{N(t)} = y\}.% = \bigcup_{y \in [0, t)}\{Z_1 > t- y, S_{N(t)} = y\}.
%\end{equation*}
%Since any ON time is possibly only dependent on the corresponding OFF time and no past renewal times, conditioned on $\{S_{N(t)} = y \}$, the system stays ON at time $t$ \textit{iff} ON time is longer than $t-y$ conditioned on renewal time being larger than $t-y$. That is, 
%\begin{equation*}
%\{\text{ON at time } t| S_{N(t)} = y\} =\{Z_1 > t - y| Z_1 + Y_1 > t-y\}.
%\end{equation*}
%Since, for $y>0$, we have $\Pr\{Z_1 > t - y| Z_1 + Y_1 > t-y\} = \frac{\bar{H}(t-y)}{\bar{F}(t-y)}$, it follows that
%\begin{equation*}
%P(t) = \bar{H}(t)+ \int_{0}^t {\bar{H}(t-y)}{\bar{F}(t-y)}dF_{S_{N(t)}}(y)
%\end{equation*}
%In view of the density of $S_{N(t)}$ from Remark~\ref{Remark:DensityLastRenewal}, the result follows.
\end{proof}

\begin{cor}[limiting on probability]
\label{cor:LimitingOnProbability}
If $\E[Z_n+Y_n]< \infty $ and $F$ is non-lattice, then 
\begin{equation*}
\lim_{t \to \infty}P(t)=\frac{\E[Z_n]}{\E[Y_n]+\E[Z_n]}.
\end{equation*}
\end{cor}
\begin{proof} Since $H$ is the distribution function of the non-negative random variable $Z_n$, it follows that 
\begin{align*}
\lim_{t \to \infty}\bar{H}(t) = 0, \text{ and } \int_0^\infty \bar{H}(t)dt = \E Z_n.
\end{align*}
Applying key renewal theorem to Theorem~\ref{Thm:OnProbability}, we get the result.
%\begin{align*}
%P(t)&= \Pr\{\text{ON at time t}, S_{N(t)}=0\}+\Pr\{\text{ON at time t}, S_{N(t)}>0\}\\
%&=\Pr\{\text{ON at time t}, S_{N(t)}=0\}+\int_{y=0}^{t}\Pr\{\text{ON at time t}| S_{N(t)}=y\}dF_{S_{N(t)}}(y)\\
%&=\Pr\{Z_1>t)+\int_{y=0}^{t}\Pr\{Z>t-y| Z+Y > t-y\}dF_{S_{N(t)}}(y)\\
%&\stackrel{(a)}{=}\bar{H}(t)+\int_{y=0}^{t}\frac{\bar{H}(t-y)}{\bar{F}(t-y)}\bar{F}(t-y)dm(y)\\
%&= \bar{H}(t)+\int_{y=0}^{t}\bar{H}(t-y))dm(y),\\
%\end{align*}
%where $(a)$ follows from the remark following Theorem 3.Now apply key renewal theorem to obtain the required result. Since $\bar{H}(t) \to 0$ as $t \to \infty$, we get
%\begin{flalign}
%P(t) \to \frac{\int_{0}^{\infty}\bar{H}(t)dt}{\mu}=\frac{\E[Z_n]}{\E[Y_n]+\E[Z_n]}.
%\end{flalign}
\end{proof}
%Many processes of practical interest can be modeled by an alternate renewal process. 

\subsection{Age and excess times}
%\begin{exmp}[Age and Excess Time] 
Consider a renewal process with renewal instants $\{S_n: n \in \N\}$, and \textit{iid} inter-renewal times $\{X_n: n \in \N\}$ with the common non-lattice distribution $F$. 
%We are interested in finding the limiting distribution of the age $A(t)$ and excess time $Y(t)$. 
At time $t$, the last renewal occurred at time $S_{N(t)}$, and the next renewal will occur at time $S_{N(t)+1}$. 
Recall that the age $A(t)$ is the time since the last renewal and the excess time $Y(t)$ is the time till the next renewal. 
That is,
\begin{xalignat*}{3}
&A(t) = t-S_{N(t)},&&Y(t) = S_{N(t)+1}-t.
\end{xalignat*}   
We are interested in finding the limiting distribution of age and excess times. 
That, is for a fixed $x$, we wish to compute
\begin{xalignat*}{3}
&\lim_{t \to \infty}\Pr\{A(t) \leq x\}, && \lim_{t \to \infty}\Pr\{Y(t) \leq x\} .
\end{xalignat*}
%To compute age we consider the renewal process $\{S_n: n \in \N\}$ with \textit{iid} inter-renewal times $\{X_n: n \in \N\}$ with the common distribution $F$. 
%\eq{
%\lim_{t\to \infty} \Pr\{A(t)\leq x\} &= \frac{1}{\mu}\int_{0}^x \bar{F}(u)du
%}
%\begin{shaded*}
%\end{shaded*}
%\end{exmp}
\begin{prop} 
Limiting age distribution for a renewal process with non-lattice distribution $F$ is 
\eq{
\lim_{t \to \infty} \Pr\{A(t) \leq x\} &= \frac{1}{\mu}\int_{0}^{x}\bar{F}(t)dt.
}
\end{prop}
\begin{proof}
We will call this renewal process to be on, when the age is less than $x$. 
That is, we consider an alternative renewal process $W$ such that
\eq{
W(t) &= 1\{A(t) \leq x\}.% = 1\{S_{N(t)} \geq t-x\}
}
This is an alternating renewal process with finite probability of off times being zero. 
Further, we can write the $n$th on time $Z_n$ for this renewal process as
\eq{
Z_n &= \min\{x, X_n\}. 
}
%Now, observe that $\Pr\{A(t) \leq x\}=\E[1_{\{A(t) \leq x\}}]$ which is the mean time when the ``age at $t$" is less than $x$ which is equal to $\E[\min\{x,X\}]$. Hence, we get\
From limiting on probability of alternating renewal process, we get 
\eq{
\lim_{t \to \infty} \Pr\{A(t) \leq x\} &= \lim_{t \to \infty}P\{W(t) = 1\} = \frac{\E \min\{x,X\}}{\E X} = \frac{1}{\mu}\int_{0}^{x}\bar{F}(t)dt.
}
\end{proof}
\begin{proof}[Alternative proof]
Another way of evaluating $\lim_{t\to \infty} \Pr\{A(t)\leq x\}$ is to note that $\{A(t)\leq x\} = \{S_{N(t)}\geq t-x\}$. 
From the distribution of $S_{N(t)}$ and the fact that the support of renewal function $m(t)$ is positive real life, %we get
\begin{align*}
\Pr\{A(t)\leq x\} &= \Pr\{S_{N(t)}\geq t-x\} %\\
%& = \int_{-\infty}^{\infty} \Pr\{S_{N(t)} \geq t-x | S_{N(t)}=y\}dF_{S_{N(t)}}(y) \\
%= \int_{t-x}^\infty dF_{S_{N(t)}}(y) 
= \int_{t-x}^{\infty}\bar{F}(t-y)dm(y) = \int_{-\infty}^{x}\bar{F}(y)dm(t-y)= \int_{0}^x \bar{F}(u)dm(t-u). 
\end{align*}
%where $(a)$ follows from a change of variable $u = t-y$. In the limit, $dm(u) \to \frac{du}{\mu}$, as $t \to \infty$, and hence 
Applying key renewal theorem, we get the result.
\end{proof}
%\begin{shaded*}
We see that the limiting distribution of age and excess times are identical. %$\Pr\{Y(t)\leq x\}$ also yield the same limit as $t \to \infty$. 
This can be observed by noting that if we consider the reversed processes (an identically distributed renewal process), $Y(t)$,  
the ``excess life time" at $t$ is same as the age at $t$, $A(t)$ of the original process.
%\end{shaded*}
\begin{prop} 
Limiting excess time distribution for a renewal process with non-lattice distribution $F$ is 
\eq{
\lim_{t \to \infty} \Pr\{Y(t) \leq x\} &= \frac{1}{\mu}\int_{0}^{x}\bar{F}(t)dt.
}
\end{prop}
\begin{proof}
We can repeat the same proof for limiting age distribution, 
where we obtain an alternative renewal process by defining on times when the excess time is less than $x$. 
That is, we consider an alternative renewal process $W$ such that
\eq{
W(t) &= 1\{Y(t) \leq x\}.% = 1\{S_{N(t)} \geq t-x\}
}
%This is an alternating renewal process with finite probability of off times being zero. 
%Further, we can write the $n$th on time $Z_n$ for this renewal process as
%\eq{
%Z_n &= \min\{x, X_n\}. 
%}
%%Now, observe that $\Pr\{A(t) \leq x\}=\E[1_{\{A(t) \leq x\}}]$ which is the mean time when the ``age at $t$" is less than $x$ which is equal to $\E[\min\{x,X\}]$. Hence, we get\
%From limiting on probability of alternating renewal process, we get 
%\eq{
%\lim_{t \to \infty} \Pr\{Y(t) \leq x\} &= \lim_{t \to \infty}P\{W(t) = 1\} = \frac{\E \min\{x,X\}}{\E X} = \frac{1}{\mu}\int_{0}^{x}\bar{F}(t)dt.
%}
\end{proof}
%\section{Limiting Mean Excess Time}
\begin{cor}
Limiting mean excess time for a renewal process with \textit{iid} inter-renewal times $\{X_n: n \in \N\}$ having non-lattice distribution $F$ and mean $\mu$ is 
\eq{
\lim_{t \to \infty}\E Y(t) &= \frac{\E[X^2]}{2\mu}.
}
\end{cor}
\begin{proof} 
One can get the limiting mean from the limiting distribution by integrating its complement. 
This involves exchanging limit and integration, which can be justified using monotone convergence theorem. 
Hence, exchanging integrals using Fubini's theorem and integrating by parts, we get 
\eq{
\lim_{t\to\infty}\E Y(t) &= \int_{0}^{\infty}\lim_{t\to\infty}P\{Y(t) > x\}dx = \frac{1}{\mu}\int_{0}^{\infty}\int_{x}^{\infty}\bar{F}(t)dt = \frac{1}{2\mu}\int_{0}^{\infty}\bar{F}(t)dt^2 =  \frac{1}{2\mu}\int_{0}^{\infty}t^2dF(t).
}
Alternatively, one can derive it directly from the regenerative process theory. 
\end{proof}
%Consider a non-lattice renewal process and we are interested in computing the mean excess time of the process. We start by writing the renewal equation of mean excess life time, $\E[Y(t)]$.
%\begin{align*}
%\E[Y(t)]&= \E[Y(t)|S_{N(t)}=0]F^c(t)+ \int_{0}^{t} \E[Y(t)|S_{N(t)}=y]F^c(t-y)dm(y)\\
%&=\E[X_1-t |X_1>t]F^c(t)+ \int_{0}^{t} \E[X-(t-y)|X>t-y]F^c(t-y)dm(y).
%\end{align*}
%From Key Renewal theorem, we have 
%
%\begin{align*}
%\lim_{t \to \infty}\E[Y(t)]&=\frac{1}{\mu} \int_{0}^{\infty} \E[X-t|X-t >0]F^c(t) dt\\
%&= \frac{1}{\mu} \int_{t=0}^{\infty} \int_{x=t}^{\infty}(x-t) dF(x) dt\\
%&= \frac{1}{\mu} \int_{x=0}^{\infty} \int_{t=0}^{x}(x-t) dF(x) dt\\
%&= \frac{\E[X^2]}{2\mu}.
%\end{align*}

\begin{lem}
Limiting empirical time average of excess time for a renewal process with \textit{iid} inter-renewal times $\{X_n: n \in \N\}$ having non-lattice distribution $F$ and mean $\mu$ is almost surely 
\eq{
\lim_{t\to \infty}\frac{1}{t}\int_0^t Y(u) du &= \frac{\E X^2}{2\mu}.
}
\end{lem}
\begin{proof}
Recall that excess times are linearly decreasing in each renewal duration, with value $X_n$ to $0$ in $n$th renewal duration of length $X_n$. 
Conditioned on $N(t)$, one can write 
\eq{
\int_0^t Y(u) du &= \frac{1}{2}\sum_{n=1}^{N(t)}X_n^2 + \int_{S_{N(t)}}^t(S_{N(t)+1}-u)du.
}
In particular, one can write the following
\eq{
\frac{\sum_{n=1}^{N(t)}X_n^2}{2N(t)}\left(\frac{N(t)}{t}\right) &\leq \frac{1}{t}\int_0^t Y(u) du \leq \frac{\sum_{n=1}^{N(t)+1}X_n^2}{2(N(t)+1)}\left(\frac{N(t)+1}{t}\right).
}
Result follows from strong law of large number by taking limits on both sides. 
%It can also be established that  since, for each sample path $\{Y(t)= y(t)\}$ and sample values $\{x_i,i\in\N\}$, we have $\int_0^ty(\tau)d\tau = \frac{1}{2}\sum_{i=1}^{n(t)}x_i^2+\int_{\tau=s_{n(t)}}^t y(\tau)d\tau$.
\end{proof}


%\newpage

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%5  
\subsection{The Inspection Paradox}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Define $X_{N(t)+1}=A(t)+Y(t)$ as the length of the renewal interval containing $t$, in other words, the length of current renewal interval. 
\begin{thm}[inspection paradox]
For any $x$, the length of the current renewal interval to be greater than $x$ is more likely than that for an ordinary renewal interval with distribution function $F$.  
That is, %for a renewal process with inter-renewal duration distribution $F$ 
\eq{
P\{X_{N(t)+1} >x\} \geq \bar{F}(x). 
}
\end{thm}
%\begin{proof} 
%Clearly, $X_{N(t)+1}$ is a regenerative process with the regenerative sequence being the renewal instants. 
%We define kernel function for $x < t$ as 
%\eq{
%K(t) &= P\{X_{N(t)+1} > x, S_1 > t\} = P\{S_1 > x\vee t\} = P\{S_1 > t\} = \bar{F}(t).
%}
%Then, in terms of renewal function $m$ and the kernel function $K(t)$, 
%one can write the complementary distribution of $X_{N(t)+1}$ using the solution to the renewal equation as
%\eq{
%P\{X_{N(t)+1} >x\} &= \bar{F}(t) + \int_{0}^{t}\bar{F}(t-y)dm(y) = (\bar{F} + \bar{F}\ast m)(t) = 1+ m(t) - (F + F \ast m)(t).
%}
%Recall, $F +  F \ast m = m$, hence we have
%\eq{
%P\{X_{N(t)+1} >x\} &= 1 + m(t) - \int_{0}^{t}\bar{F}(t-y)dm(y).
%}
%\end{proof}
\begin{proof}
Conditioning on the joint distribution of last renewal instant and number of renewals, we can write 
\begin{align*}
\Pr\{X_{N(t)+1}>x\}&= \int_{0}^t\Pr\{X_{N(t)+1} > x | S_{N(t)} = y, N(t)=n\}dF_{(S_{N(t)}, N(t))}.
%&= P(X_2 >x)\\
%&= \bar{F}(x).
\end{align*}
Now we have,
\begin{align*}
\Pr\{X_{N(t)+1}>x | S_{N(t)}=y, N(t)=n\} %& = \Pr\{X_{N(t)+1}>x | X_1+\cdots+X_n=y, X_{n+1}>t-y\} \\
& = \Pr\{X_{n+1}>x | X_{n+1}>t-y\} = \frac{\Pr\{X_{n+1}>\text{max}(x,t-y)\}}{\Pr\{X_{n+1}>t-y\}}.% \\
%& \geq \bar{F}(x). 
\end{align*}
Applying Chebyshev's sum inequality to increasing positive functions $f(z) = 1\{t > x\}$ and $g(z) = 1\{z > t-y\}$, we get 
\eq{
\E f(X_{n+1})g(X_{n+1})/\E g(X_{n+1}) \geq \E f(X_{n+1}) = \bar{F}(x).
}
We get the result by integrating over the joint distribution.  
%\begin{align*}
%\Pr\{X_{N(t)+1}>x\}\geq \Pr\{X_{1}>x\}.
%\end{align*}
\end{proof}
We also have a weaker version of inspection paradox involving the limiting distribution of $X_{N(t)+1}$.
\begin{lem}
For any $x$, the limiting probability of length of the current renewal interval being greater than $x$ is larger than the corresponding probability of an ordinary renewal interval with distribution function $F$. 
That is, %for a renewal process with inter-renewal duration distribution $F$ 
\eq{
\lim_{t \to \infty}P\{X_{N(t)+1} >x\} \geq \bar{F}(x). 
}
\end{lem}
\begin{proof}
Consider an alternating renewal process $W$ for which the on time is the renewal duration if greater than $x$, and zero otherwise. 
That is, 
\eq{
W(t) &= 1\{X_{N(t)+1} > x\}.
}
Hence, each renewal duration consists of either on or off intervals, depending if the renewal duration length is greater than $x$ or not.
%during a cycle (if the renewal interval is greater than $x$), or totally OFF otherwise. Formally, 
We can denote $n$th on and off times by $Z_n$ and $Y_n$ respectively, where
\begin{xalignat*}{3}
&Z_n = X_n 1\{X_n>x\}, && Y_n =  X_n 1\{X_n\leq x\}.
\end{xalignat*}
From the definition, we have
\eq{
\E W(t) &= P\{X_{N(t)+1} > x\} = P\{\text{ on at time } t\}.
}
%Now we have,
%\begin{align*}
%\Pr \{X_{N(t)+1}>x\} &= \Pr\{\text{length of the interval containing } t>x\}\\
%&= \Pr\{ \text{on at time } t \}.
%\end{align*}
From the alternating renewal process theorem, we conclude that 
\begin{align*}
\lim_{t\to \infty}\Pr\{X_{N(t)+1}>x\} %&= \frac{\E Z_n}{\mu} \\
&= \frac{\E X1\{X>x\}}{\E X}. %= \frac{\int_{x}^\infty y dF(y)}{\mu}. %&\geq \Pr[X_1\geq x]. 
\end{align*}
The result follows from Chebyshev's sum inequality applied to positive increasing function $f(z) = z$ and $g(z) = 1\{z > x\}$.
\end{proof}
%\begin{rem}
The inspection paradox states, in essence, that if we pick a point $t$, it is more likely that an inter-renewal interval with larger length will contain $t$ than the smaller ones. 
For instance, if $X_i$ were equally likely to be $\epsilon$ or $1-\epsilon$, we see that the mean of any inter arrival length is $1$ for any value of $\epsilon\in(0,1)$. 
However, for small values $\epsilon$, it is more likely that a given $t$ will be in an interval of length $1-\epsilon$ than in an interval of length $\epsilon$.
%\end{rem}


\begin{prop}
If the inter arrival time is non-lattice and $\E[X^2] < \infty$, we have 
\begin{align*}
\lim_{t \to \infty} \left(m(t)-\frac{t}{\mu}\right) = \frac{\E[X^2]}{2\mu^2}-1.
\end{align*} 
\begin{proof}
From definition of excess time $Y(t)$ and Wald's lemma for stopping time $N(t)+1$ for renewal processes, it follows  
\eq{
\E S_{N(t)+1} = \E[\sum_{i=1}^{N(t)+1}X_n] = \mu (m(t)+1) = t + \E[Y(t)].
}
\end{proof}

\end{prop}

\appendix

\section{Chebyshev's sum inequality}
\begin{lem}
Let $f:\R \to \R_{+}$ and $g : \R \to \R_{+}$ be arbitrary functions with the same monotonicity.  
For any random variable $X$, functions $f(X)$ and $g(X)$ are positive and
 \eq{
 \E[f(X)g(X)] \geq \E[f(X)]\E[g(X)].
 }
\end{lem}
\begin{proof} 
Let $Y$ be a random variable independent of $X$ and with the same distribution. 
Then, 
\eq{
(f(X) - f(Y))(g(X) - g(Y)) \geq 0.
}
Taking expectation on both sides the result follows. 
\end{proof} 
%In particular, this inequality implies
%\eq{
%\E[X1\{X\geq x\}] &\geq \E[X]P\{X\geq x\}.
% }

\end{document}


\subsection{Example:}
 Suppose the number of commodities desired by a customer at a store follows a distribution $G$. The ordering policy of the store is as follows: For some fixed $s,~S$, if the inventory level after serving a customer is $x$, then the amount ordered is
 
 

     \begin{displaymath}
        \left\{
         \begin{array}{lr}
           S-x & \text{if } x <s\\
           0 & \text{if } x \geq s
         \end{array}
       \right.
    \end{displaymath} 

Let $L(t)$ denote the inventory level at time $t$. We are interested in finding $\lim_{t \to \infty}\mathbb{P}(L(t) \geq y)$. 
Let $X_n$ denote inter-restocking times. Let $\{L(t)\geq y\}$ denote ON period. $X_n$ forms an 
alternating renewal process with the above mentioned ON time. 
From alternating renewal process theorem, we have 

\begin{align*}
\lim_{t \to \infty}\mathbb{P}(L(t) \geq y) &= \frac{\E[\text{ON time}]}{\E[X_1]}\\
&=\frac{\E[\sum_{i=1}^{N_y}X_i]}{\E[\sum_{i=1}^{N_s}X_i]}=\frac{\E[N_x]}{\E[N_s]}.
\end{align*}

where $N_y= \min\{n \in \N: \sum_{i=1}^{n}D_i > S-y\}$  and $D_1,D_2 \hdots$ denote the successive customer demands. Since $D_i$ are iid, we can interpret $N_y-1$ as the number of renewals till time $S-y$. $D_i$ is the inter arrival time of the process. Thus   

\begin{align*}
\lim_{t \to \infty}\mathbb{P}(L(t) \geq y) =\frac{m_G(S-x)+1}{m_G(S-s)+1}, s \leq x \leq S.
\end{align*}