lecture-18.tex

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\documentclass[letterpaper,10pt,english]{article}
\input{header}
\title{Lecture-18: Tractable Random Processes}
\author{}

\begin{document}
\maketitle

\section{Examples of Tractable Stochastic Processes}
In general, it is very difficult to characterize a stochastic process completely in terms of its finite dimensional distribution. 
However, we have listed few analytically tractable examples below, where we can completely characterize the stochastic process. 
\subsection{Independent and identically distributed processes}
Let $\{X_t: t \in T\}$ be an independent and identically distributed (\emph{iid}) random process, with a common distribution $F(x)$. 
Then, the finite dimensional distribution for this process for any finite $S \subseteq T$ can be written as 
\begin{align*}
F_S(x) = P\left(\{X_s(\omega) \leq x_s, s \in S\}\right) = \prod_{s \in S}F(x_s).
\end{align*}
It's easy to verify that the first and the second moments are independent of time indices. 
Since $X_t = X_0$ in distribution, 
\meq{3}{
&m_X = \E X_0 , && R_X = \E X_0^2, && C_X = \text{Var}(X_0).
}

%\subsubsection{Independent increments process}
\subsection{Stationary processes}
A stochastic process $X_t$ is \textbf{stationary} if all finite dimensional distributions are shift invariant, that is for finite $S \subseteq T$ and $t > 0$, we have 
\begin{align*}
F_S(x_S) = P(\{X_s(\omega) \leq x_s, s \in S\}) = P(\{X_{s+t}(\omega) \leq x_s, s \in S\}) = F_{t+S}(x_S).
\end{align*}
In particular, all the moments are shift invariant. 
Since $X_t = X_0$ and $(X_t,X_s) = (X_{t-s}, X_{0})$ in distribution, we have 
\begin{xalignat*}{5}
&m_X = \E X_0 , && R_X(t-s,0) = \E X_{t-s}X_0, && C_X(t-s,0) = R_X(t-s,0) - m_X^2.
\end{xalignat*}

\subsection{Markov processes}
A stochastic process $X_t$ is \textbf{Markov} if conditioned on the present state, future is independent of the past.  
That is, for any ordered index set $T$ containing any two indices $u > t$, we have  %such that $t < u$, % \in T$, 
\eq{
P(\{X_u \leq x_u\}| \sF_t) = P(\{X_{u} \leq x_u\}| \sigma(X_t)).
}
%For a stationary Markov process, we would have 
We will study this process in detail in coming lectures. 

\subsection{L\'evy processes}
A right continuous with left limits stochastic process $X = (X_t \in \R: t \in T \subseteq \R_+)$ with $X_0 = 0$ almost surely, is a \textbf{L\'evy process} if the following conditions hold. 
\begin{enumerate}[(L1)]
%\item $X_{0}=0$ almost surely.
\item The increments are independent. For any $ 0\leq t_{1}<t_{2}<\cdots <t_{n}<\infty$, $X_{t_{2}}-X_{t_{1}},X_{t_{3}}-X_{t_{2}},\ldots ,X_{t_{n}}-X_{t_{n-1}}$ are independent.
\item The increments are stationary. For any $s<t$,  $X_{t}-X_{s}$, is equal in distribution to $X_{t-s}$.
\item Continuous in probability. For any $\epsilon > 0$ and $t\geq 0$ it holds that $\lim _{h\rightarrow 0}P(|X_{t+h}-X_{t}|>\epsilon )=0$. 
\end{enumerate}
\begin{shaded*}
\begin{exmp}
Two examples of L\'evy processes are Poisson process and Wiener process. 
The distribution of Poisson process at time $t$ is Poisson with rate $\lambda t$ and the distribution of Wiener process at time $t$ is zero mean Gaussian with variance $t$. 
\end{exmp}
\end{shaded*}
\begin{thm} A L\'evy process has infinite divisibility. 
That is, for all $n \in \N$ %the following equality holds 
\begin{align*}
\E e^{\theta X_t} = \left(\E e^{\theta X_{{t}/{n}}}\right)^n.
\end{align*}
Further, if the process has finite moments $\mu_n(t) = \E X_t^n$ then the following Binomial identity holds
\begin{align*}
\mu_n(t+s) = \sum_{k=0}^n\binom{n}{k}\mu_k(t)\mu_{n-k}(s).
\end{align*}
\end{thm}
\begin{proof}
The first equality follows from the independent and stationary increment property of the process, 
and the fact that we can write 
\begin{align*}
X_t = \sum_{k=1}^nX_{\frac{kt}{n}}-X_{\frac{(k-1)t}{n}}.
\end{align*}
Second property also follows from the the independent and stationary increment property of the process, 
and the fact that we can write 
\begin{align*}
X_{t+s}^n = (X_{t} + X_{t+s}-X_{t})^n = \sum_{k=0}^n\binom{n}{k}X_t^k(X_{t+s}-X_t)^{n-k}.
\end{align*}

\end{proof}
%In particular, all the moments are shift invariant. 
%Since $X_t = X_0$ and $(X_t,X_s) = (X_{t-s}, X_{0})$ in distribution, we have 
%\begin{xalignat*}{5}
%&m_X = \E X_0 , && R_X(t-s,0) = \E X_{t-s}X_0, && C_X(t-s,0) = R_X(t-s,0) - m_X^2.
%\end{xalignat*}
\end{document}

\section{Examples of Stochastic Processes}
\begin{exmp}[Queues] Queues are complex stochastic processes and consist of two stochastic processes arrival and service, coupled through a buffer. 
Number of arrivals and arrival instants could be discrete or continuous random variable. 
For a discrete arrival case, arrival process can be characterized by the time epochs of discrete arrivals, denoted $\{A_n: n \in \N \}$. 
Similarly, service requirement of each incoming arrival can also be a discrete or continuous random variable.
Service of each discrete arrival can be considered to be a random amount of time, $\{S_n: n \in \N\}$. 
Queue can have a finite or infinite waiting area, and can be served by single or multiple servers. 
Important performance metrics for queues are mean waiting time of arrivals and mean queue length. 
These metrics are affected by the service policy that determines how to serve incoming arrivals. 
Few important service policies are first come first out, last in first out, processor sharing etc.
%If a queue can only be served at a fixed rate, then arrivals would have to wait in the buffer till their service time. 
Queues have applications in operations research, industrial engineering, telecommunications networks, among others. 
\end{exmp}
\begin{exmp}[Gambler's ruin] One can model many gambling games with random walks, where wins or losses on each bet can be thought of as a random step. If a gambler starts with certain capital, and he wants to quit gambling after he makes a certain amount of money, one is interested in probability of a gambler getting bankrupt before it can quit gambling. These questions are related to hitting times of a random walk. Random walks have deep relations to Brownian motion.
\end{exmp}
\begin{exmp}[Urn Models] In these models, one is interested in ball and urns. One is interested in distribution of balls in urns, when one can randomly throw balls into urns. Balls can be of multiple colors and may or may not be replaced after putting into urns. These models have applications in influence maximization and epidemic control.
\end{exmp}
\begin{exmp}[Branching Processes] This is used by biologists to model population. In this model, one assumes that every individual in a population has a probability distribution over number of children it can have. Each children can be assumed to have identical and independent distribution for their progenies in next generation. These type of models can answer questions related to survival of species.
\end{exmp}
\begin{exmp}[Random Graphs] A typical graph $G$ consists of vertex set $V$ and edge set $E \subseteq V \times V$. Both of these can be random in general. In classical settings, usually $V = [n]$ and $E$ is selected randomly from set of all possible edges $[n] \times [n]$, without self-loops. These models are exploited in study of various type of networks, and can be used to answer questions regarding network properties.
\end{exmp}
\end{document}
\section{Poisson processes}
\begin{defn}[Point Process] A stochastic process $\{N(t), t\geqslant 0\}$ is a \textbf{point process} if
\begin{enumerate}
  \item $N(0) = 0$.
  \item $t\mapsto N(t) (\omega)$ is non-decreasing, integer valued, right continuous and at points of discontinuity (wherever it has jumps) $(N(t)- N(t^{-}))\leqslant 1, \hspace{0.1cm} \sForall \quad \omega \in \Omega$. 
\end{enumerate}
\end{defn} 
\begin{defn}[Simple Point Process] A \textbf{simple point process} is a point process of jump size 1.
\end{defn}

\begin{defn}[Stationary Increment Point Process] A point process $\{N(t), t\geqslant 0\}$ is called \textbf{stationary increment point process }, if for any collection of $0\leqslant t_{1}<t_{2}...,<t_{n}$, we have $(N(t_{n})-N(t_{n-1}),N(t_{n-1})-N(t_{n-2}),...,N(t_{1}))$ having the same joint distribution as $(N(t_{n}+t)-N(t_{n-1}+t),...,N(t_{1}+t)), ~ \sForall t \geqslant 0$.
\end{defn}
\begin{defn}[Stationary Independent Increment Point Process]   A point process $\{N(t), t\geqslant 0\}$ is called \textbf{stationary independent increment process}, if it has stationary increments and the increments are independent random variables.
\end{defn}

\begin{figure}[hhhh]
\center
	\input{Figures/Poisson}
	\caption{Sample path of a Poisson process.}
	\label{Fig:Poisson}
\end{figure}
\noindent The points of discontinuity corresponds to the arrival instants of the point process. Let $X_{n}$  denote the inter arrival time between $(n-1)^{th}$ and $n^{th}$ arrival. Further, let, $S_{0}=0,~ S_{n}= \sum^{n}_{k=1}X_{k}$. $S_{n}$ is the arrival instant of of $n^{th}$ point. The arrival at time zero is not counted.

\begin{defn}[Poisson Process]
A simple point process $\{N(t),~ t\geqslant0\} $ is called a \textbf{Poisson process} with rate $0< \lambda< \infty$, if inter-arrival times $\{X_{n},~n\geqslant 1\}$ are \emph{iid} $\exp(\lambda)$ random variables, i.e.
 \begin{equation*}
  P\{X_{1}\leqslant x\} = 
	\begin{cases}
		1-e^{-\lambda x}, & x\geqslant 0   \\
		0,  & \text{ else}.
	\end{cases}
\end{equation*}
\end{defn}

\textbf{Remarks:} Observe that 
\begin{align*}
\{S_{n}\leqslant t\} =  \{N(t)\geqslant n \},\\
\{S_{n}\leqslant t, S_{n+1}> t \} = \{N(t)= n\},\quad\mathrm{and} \\
P\{X_{n} = 0\} = P\{X_n\leqslant 0\} = 0.
\end{align*}
Also, by Strong Law of Large Numbers (SLLN), 
\begin{equation*}
\lim_{n \to \infty} \frac{S_{n}}{n} = E[X_{1}] = \frac{1}{\lambda}\quad\mathrm{a.s.} 
\end{equation*}
Therefore, we have $S_n \rightarrow \infty$, a.s. This implies $P\{\omega: N(t)(\omega) < \infty\} =1$. To see this, let's pick an $\omega \in \Omega$ such that $ N(t)(\omega) = \infty$, then $S_{n}(\omega)\leqslant t,\quad \sForall n$. This implies $S_{\infty}(\omega)\leqslant t$  and $\omega \not\in \{\omega: S_{n}(\omega) \rightarrow \infty \}.$ Hence, probability measure for such $omega$'s is zero and the claim follows. 


\subsection{Moment Generating Function and Density Function of $S_n$}
We know that time of $n^{\mathrm{th}}$ event $S_n$ is sum of $n$ consecutive \emph{iid} inter-arrival times $X_k$, i.e. $S_n = \sum^{n}_{k=1}X_{k}$. Therefore, moment generating function $\mathbb{E} [ e^{\alpha S_{n}} ]$ of $S_n$ is given by 
 \begin{equation*}
  \mathbb{E} [ e^{\alpha S_{n}} ] = \left(\mathbb{E}[e^{\alpha X_{1}}]\right)^{n}. 
 \end{equation*} 
Since each $X_k$ is \emph{iid} exponential with rate $\lambda$, it is easy to see that moment generating function of intr-arrival time $X_1$ is 
 \begin{equation*}
  \mathbb{E} [ e^{\alpha X_1} ] = 
		\begin{cases}
		\frac{\lambda}{\lambda-\alpha}, & \alpha < \lambda \\
		\infty, & \alpha \geqslant \lambda.
		\end{cases} 
 \end{equation*} 
   %\begin{eqnarray*}
%\mathbb{ E}[e^{\alpha X_{1}}]  =& \int^{\infty}_{0}\lambda e^{-\lambda t} e^{\alpha t} dt \\
  %=& \lambda \int^{\infty}_{0} e ^{-(\lambda-\alpha)t} dt\\
  %\end{eqnarray*}
   %If $\alpha \geqslant \lambda,~\mathbb{E} [e^{\alpha X_{1}}] = \infty$. Else, 
  %\begin{eqnarray*}
  %\mathbb{E} [e^{\alpha X_{1}}]=&\lambda \left[ \frac{e^{-(\lambda-\alpha)t}}{-(\lambda-\alpha)}\right]^{\infty}_{0} \\
   %=& \lambda\left[0-(-\frac{1}{\lambda-\alpha})\right] = \frac{\lambda}{\lambda-\alpha}. 
   %\end{eqnarray*}
Substituting the moment generating function of inter-arrival time $X_1$ in moment generating function of $n^{\mathrm{th}}$ event time $S_n$, we obtain
\begin{equation*}
   \mathbb{E}[e^{\alpha S_{n}}] = 
	\begin{cases} 
	\left(\frac{\lambda}{\lambda-\alpha}\right)^{n}, & \alpha < \lambda, \\
   \infty, &\text{else}.
	\end{cases}
\end{equation*}

\begin{thm}[Arrival Time] Density function of $S_n$ is Gamma distributed with parameters $n$ and $\lambda$. That is,
\begin{equation*}
f_{S_n}(s) =\frac{\lambda (\lambda s)^{n-1}} {(n-1)!} e^{-\lambda s}.
\end{equation*}
\end{thm}
\begin{proof} Notice that $X_i$'s are \emph{iid} and $S_1 = X_1$. In addition, we know that $S_n = X_n + S_{n-1}$. Since, $X_n$ is independent of $S_{n-1}$, we know that distribution of $S_n$ would be convolution of distribution of $S_{n-1}$ and $X_1$. Since $X_n$ and $S_1$ have identical distribution, we have $f_{S_{n}}=f_{S_{n-1}}*f_{S_1}$. The result follows from straightforward induction.
\end{proof}

Process $N(t)$ is of real interest, and we can compute the distribution of $N(t)$ for each $t$ from the distribution of $S_n$ in the following.
%\begin{figure}[hhhh]
%\center
	%\include{Figures/Poisson}
 %% \caption{}\label{}
%\end{figure}
\begin{thm}[Poisson process] Process $N(t)$ is Poisson distributed with parameter $\lambda$ for each $t$. That is,
	\begin{equation*}
	P\{N(t)=n)\}= e^{-\lambda t}\frac{(\lambda t)^{n}}{n!}.
	\end{equation*}
\end{thm}
\begin{proof}

\begin{eqnarray*}
   P\{N(t) =n\}=&  P\{S_{n}\leqslant t, S_{n+1} >t)\\
   =&  \int^{t}_{0} P\left\{ {S_{n+1}>t}|{S_{n}=s}\right\}f_{S_n}(s)  ds\\
   &\stackrel{(a)}{=}& \int^{t}_{0} P\{X_{n+1}>t-s\} f_{S_n}(s) ds\\
   =&  \int^{t}_{0}e^{-\lambda(t- s)} \frac{\lambda^{n}s^{n-1}}{(n-1)!}e^{-\lambda s}  ds\\
   =&\frac{e^{-\lambda t} (\lambda t)^{n}}{n !}.
\end{eqnarray*}
 where (a) follows from the memoryless property of exponential distribution. %($P\{X_{n+1}>s+t|X_{k+1}>t)=P\{X_{k+1}>s)$).\\
\end{proof}

\textbf{Remark:} The Poisson process is not a stationary process. That is, the finite dimensional distributions (fdd) are not shift invariant. In the following section, we show that the Poisson process is a \textit{stationary,  independent increment} process. To this end, we will use an important property of exponential distribution- namely memoryless property. Memoryless property of exponential distribution will facilitate the computation of fdd of the Poisson process via one dimensional marginal distribution. 

\begin{prop}[Memoryless Distribution] Exponential distribution with  continuous support is the only distribution satisfying memoryless property.
\end{prop}

%\begin{eqnarray*}    
%            P\{N(t1}& = &n_{1}),..., N(tk}=n_{k}\hspace{1.0cm} fdd
%             \end{eqnarray*}
%
%We will S.t Poisson process has another very important property- Stationary indep. increment process\\
%
%$ 0 \leqslant t_{1} \leqslant  t_{2} \hspace{0.5cm} P[N(t1}=n_{1}, N(t2}=n_{2}]=0$ \hspace{1.0cm} if $n_{2}< n_{1}$\\
%$n_{2}\geqslant n_{1} \hspace{1.5cm} P[N(t1}=n, N(t2}-N(t1}=n_{2}-n_{1}]$
%
%
%If $N(t)$ is independent increment process.
%\begin{eqnarray*}
%=&P [N(t1}=n_{1}]\\
%P (N(t2}-N(t1} =& n_{2}-n_{1}) n2-n1\\
% P (N(t2}-N(t1} =&\frac{[\lambda (t2-t1)]}{(n2-n1)!} e^{-\lambda (t2-t1)!}\\
%\end{eqnarray*}
%Need\\
%\begin{eqnarray*}
% N(t1}=& n1\overrightarrow(n2-n1) n2\\
% arrivals =& \phi \frac{(\lambda t1)^{n1}}{(n1)!}e^{-\lambda t1}\\
%\end{eqnarray*}
%What it means all fdd can be computed from $1-d$ marginal in a stationary indep. incr. property.\\
%Its not stationary $ \rightarrow N(t1}$ and $N(t2}$ not same distr.\\
%$N(t2}$ in some sense $> N(t1}$ so not same distr. intution\\
%To s.t ${N(t), t\geqslant 0}$ stationery ind. increment process.\\
%First we show indep. increment.\\
%We need an important property of exponential property of memorylessness.\\
%$X\sim exp (\lambda$\\
%$S>0, t> 0$\\
%\begin{eqnarray*}
%% \nonumber to remove numbering (before each equation)
%  P\{X>S +t / X>t) =& P\{X>S) \\
%  In \hspace{0.2cm}reliability\hspace{0.2cm} theory & x& life time\\
%  P\{X>t+S/ X>t) &\leqslant& P\{X>S)
%\end{eqnarray*}
%Class of distr. satisfying ``New better than used distr''\\
%\begin{eqnarray*}
%X>t+s>S\\
%P\{X>s+t)/ X>t)\\
%=& \frac{P\{X>S+t, X>t)}{P\{X>t)}\\
%=& \frac{P\{X>S+t)}{P\{X>t)}\\
%=& e^{-\lambda S}\\
%=& P\{X>S)
%\end{eqnarray*}
\begin{proof}
Let $X$ be a random variable with a distribution function $F$ with memoryless property defined on $\mathbb{R}^{+}$. Let $g(t) \triangleq P\{X > t\} = 1 - F(t)$. Due to memoryless property of $F$, we notice that
\begin{eqnarray*}
  P\{X>s\} =& P\{ X > t+s| X>t\} \\
  P\{X>s\} =& \frac{ P\{ X>t+s, X>t\}}{P\{X>t\}}.
\end{eqnarray*}
Since $\{X > t + s\} =\{ X>t+s, X>t\}$, we have $g(t+s) = g(t)g(s)$ and hence $g(0) = g^2(0)$. Therefore, $g(0)$ is either unity or zero. Note, that $g$ is a right continuous (RC) function and is non-negative. 

We will show that $g$ is an exponential function. That is, $g(t) = e^{\alpha t}$ for some $\alpha \geqslant 0$. We will prove this in stages. First, we show this is true for $t \in \mathbb{Z}^+$. Notice that we can obtain via induction
\begin{eqnarray*}
	g(2) =& g(1) g(1) = g^{2}(1), \mathrm{ and }\\
	g(m) =& [g(1)]^{m}.
\end{eqnarray*}
Since $g(1)$ is non negative, there exists a $\beta$ such that $g(1)=e^{\beta}$ and $g(m)= e^{m \beta}, m \in \mathbb{Z}_{+}$. Next we show that for any $n \in \mathbb{Z}_{+}$,        
\begin{equation*}
	g(1) =  g\left(\frac{1}{n}+..., +\frac{1}{n}\right) = \left[g\left(\frac{1}{n}\right)\right]^{n}.
\end{equation*}
Therefore, for same $\beta$ we used for $g(1)$, we have $g\left(\frac{1}{n}\right) = e^{\frac{\beta}{n}}$. Now, we show that $g$ is exponential for any $t \in \mathbb{Q}^+$. To this end, we see that for any $m, n \in \mathbb{Z}_{+}$, we have 
\begin{equation*}
	g\left(\frac{m}{n}\right) = \left[g\left(\frac{1}{n}\right)\right]^{m}= e^{\frac{m \beta}{n}}.
\end{equation*}
Now, we can show that $g$ is exponential for any real positive $t$ by taking a sequence of rational numbers $\{t_n\}$ decreasing to t. From right continuity of $g$, we obtain 
\begin{equation*}
	g(t) \stackrel{(a)}{=} \lim_{n\rightarrow \infty} g(t_n) =   \lim_{n\rightarrow \infty} e^{\beta t_{n}}= e^{\beta t}.
\end{equation*}
Since $P\{X > x\}$  is decreasing  with $x$, $\beta $ is negative.  
\end{proof}
\begin{figure}[hhhh]
\center
	\input{Figures/IndependentIncrements}
  \caption{Stationary independent increment property of Poisson process.}
	\label{Fig:IndependentIncrements}
\end{figure}
\begin{prop}[Stationary Independent Increment Property] Poisson process ${N(t), t\geqslant 0}$ has stationary independent increment property.
\end{prop}
\begin{proof}
To show that $N(t)$ has stationary independent increment property, it suffices to show that $N_t-N(t_{1}) \perp N(t_1)$ and $N(t) - N(t_1) \sim N(t-t_1)$. Since, we can use induction to show this stationary independence increment property for for any finite disjoint time-intervals. The memoryless property of exponential distribution is crucially used. And, we see that independent increment holds only if inter-arrival time is exponential. We can see in Figure~\ref{Fig:IndependentIncrements} that $t_1$ divides $X_{n+1}$ in two parts such that, $X_{n+1} = X_{n+1}^{'} + X_{n+1}^{''}$. Here,  $X_{n+1}^{''}$ is independent of $X_{n+1}^{'}$ and has same distribution as $X_{n+1}$. Therefore, 
\begin{align*}
\{ N(t_1) = n \} &\iff  \{ S_n = t_1 + X'_{n+1} \}, \\
\{ N(t) - N(t_1) \geqslant m \} &\iff \{ X''_{n+1} + \sum_{i=n+2}^{n+m} X_i \leqslant t - t_1 \}.
\end{align*}
Since, $\{X_i: i \geqslant n+2\}\cup\{X_{n+1}^{''}\}$ are independent of $\{X_i: i \leqslant n\}\cup{X_{n+1}^{'}}$, we have $N(t)-N(t_{1}) \perp N(t_1)$. Further, since $X_{n+1}^{''}$ has same distribution as $X_{n+1}$, we get $N(t) - N(t_1) \sim N(t-t_1)$. By induction we can extend this result to $(N(t_{n})-N(t_{n-1}),...,N(t_{1}))$. 
\end{proof}
\end{document}