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% Optimal Foraging Theory Revisited: Appendix A. Limits of Markov
%                                    Renewal Processes
%
% (c) Copyright 2007 by Theodore P. Pavlic
%

\chapter{Limits of Markov Renewal Processes}
\label{app:markov_renewal_sto_limits}\aimention{Andrei A.
Markov}%
\index{stochasticity!Markov renewal process|(}%
\index{stochasticity!Markov renewal process!limits|(}

\index{Markov renewal process|see{stochasticity, Markov renewal
process}}\index{renewal process|see{stochasticity, Markov renewal
process}}\index{Poisson process|see{stochasticity, Poisson process}}Take
a \index{stochasticity!probability space}probability space
$(\set{U},\Sigma,\Pr)$. Let $(M(t_s): t_s \in \R_{\geq0})$ be a
\index{stochasticity!Poisson process}\aimention{Sim\'{e}on-Denis
Poisson}Poisson process from the space with rate $\lambda \in \R_{>0}$
and interevent time process $(\Upsilon_M)$. Take $(\tau_M)$ to be a
sequence of non-negative random variables from this space where $\tau_M$
is independent of $\Upsilon_N$ for all $M,N \in \N$. From these, define
the random processes $(T_N)$ and $(T^N)$ with
%
\begin{equation*}
        T_N \triangleq \Upsilon_N + \tau_N
        \quad \text{ and } \quad
        T^N 
        \triangleq \sum\limits_{i=1}^N T_N 
        = \sum\limits_{i=1}^N \Upsilon_i + \tau_i
\end{equation*}
%
for all $N \in \N$ and $\zeta \in \set{U}$. Define the
\index{stochasticity!Markov renewal process}\aimention{Andrei A.
Markov}Markov renewal process $(N(t): t \in \R_{\geq0})$ by
%
\begin{equation*}
        N(t)
        \triangleq
        \sup
        \left\{ 
        N \in \N : \sum\limits_{i=1}^N T_i \leq t 
        \right\}
\end{equation*}
%
For all outcomes $\zeta \in \set{U}$,
%
\begin{equation*}
        T^2 
        = T_1 + T_2
        = \Upsilon_1 + \Upsilon_2 + \tau_1 + \tau_2
        = A^2 + B^2
\end{equation*}
%
where random variables $A^2: \set{U} \mapsto \extR$ and $B^2: \set{U}
\mapsto \extR$ are defined by $A^2 \triangleq \Upsilon_1 + \Upsilon_2$
and $B^2 \triangleq \tau_1 + \tau_2$ for all $\zeta \in \set{U}$. For
brevity, we assume that the probability measures associated with these
two random variables are absolutely continuous with respect to the
\aimention{Henri L. Lebesgue}Lebesgue measure\footnote{By the
\aimention{Sim\'{e}on-Denis Poisson}Poisson assumption, this must be
true for $A^2$.}. The main result of this appendix holds for the general
case as well. Because $T^2$ is the sum of two independent random
variables, for all $x \in \extR$, $f_{T^2}(x) = f_{A^2}(x) * f_{B^2}(x)$
where the $*$ operator denotes \index{mathematics!functions!convolution
($*$)}convolution \citep{PapoulisPillai02}. Therefore,
%
\begin{align*}
        \E\left( \frac{1}{T^2} \right)
        &= 
        \int_{-\infty}^\infty 
        \frac{1}{x} \left(f_{A^2}(x) * f_{B^2}(x)\right) \total x\\
        %&= 
        %\int_{-\infty}^\infty 
        %\left(\frac{1}{x} f_{A^2}(x)\right) * f_{B^2}(x) \total x\\
        &= 
        \int_{-\infty}^\infty 
        \int_{-\infty}^\infty 
        \frac{1}{x} f_{A^2}(x-t) f_{B^2}(t) \total t \total x\\
        &= 
        \int_{-\infty}^\infty 
        \int_{-\infty}^\infty 
        \frac{1}{x} f_{A^2}(x-t) f_{B^2}(t) \total x \total t\\
        %&= 
        %\int_{-\infty}^\infty 
        %f_{B^2}(t)
        %\int_{-\infty}^\infty 
        %\frac{1}{x} f_{A^2}(x-t) \total x \total t\\
        &= 
        \int_{-\infty}^\infty 
        f_{B^2}(t)
        \int_{-\infty}^\infty 
        \frac{1}{y+t} f_{A^2}(y) \total y \total t
\end{align*}
%
However, $f_{B^2}$ is a probability density where $f_{B^2}(t)=0$ for all
$t < 0$, and so
%
\begin{equation*}
        \E\left( \frac{1}{T^2} \right)
        \leq
        \int_{-\infty}^\infty \frac{1}{y} f_{A^2}(y) \total y
\end{equation*}
%
Because $(\Upsilon_M)$ is the interevent time process for
\index{stochasticity!Poisson process}\aimention{Sim\'{e}on-Denis
Poisson}Poisson process $(M(t_s):t_s \in \R_{\geq0})$, $A^2$ is
\index{Erlang-2|see{stochasticity, distributions, Erlang-2}}%
\index{stochasticity!distributions!Erlang-2}Erlang-2 distributed%
\footnote{\index{stochasticity!distributions!Erlang-2|(indexdef}The
Erlang-2 distribution is characterized by density $f_{A^2}(y) =
\lambda^2 y \exp(-\lambda y)$ for $y >
0$.\index{stochasticity!distributions!Erlang-2|)indexdef}} with
parameter $\lambda$. Thus, $\E( 1/T^2 ) \leq \lambda$. So, there exists
an $N \in \N$ such that $\E( 1 / T^N ) < \infty$. By results of
\citet{JohnsMiller63}, for all $K \in \N$,
%
\begin{equation*}
        \aslim\limits_{t \to \infty} \frac{ N(t) }{t} = 
        \lim\limits_{t \to \infty} \frac{ \E( N(t) ) }{t} = 
        \aslim\limits_{N \to \infty} \frac{ N }{ T^N } =
        \lim\limits_{N \to \infty} 
        \E\left( \frac{ N }{ T^N } \right) =
        \frac{K}{ \E(T^K) } =
        \frac{1}{ \E(T_1) }
\end{equation*}
%
where
%
\begin{equation*}
        \frac{1}{ \E(T_1) }
        =
        \frac{1}{ \E( \Upsilon_1 ) + \E( \tau_1 ) }
        =
        \frac{1}{ \frac{1}{\lambda} + \E( \tau_1 ) }
\end{equation*}
%
This could be called the \emph{long-term encounter rate} of $(N(t):t \in
\R_{\geq0})$. Now define the process $(T(t): t \in \R_{\geq0})$ by
%
\begin{equation*}
        T(t)
        \triangleq
        T^{N(t)}
        =
        \sum\limits_{i=1}^{N(t)} T_1
\end{equation*}
%
for all $t \in \R_{\geq0}$. It is similarly the case that
%
\begin{equation*}
        \aslim\limits_{t \to \infty} \frac{ T(t) }{t} = 
        \lim\limits_{t \to \infty} \frac{ \E( T(t) ) }{t} = 1
\end{equation*}
\index{stochasticity!Markov renewal process!limits|)}%
\index{stochasticity!Markov renewal process|)}