11.1.2 Basic Concepts of the Poisson Process
 $$ the number of car accidents at a site or in an area;
 $$ the location of users in a wireless network;
 $$ the requests for individual documents on a web server;
 $$ the outbreak of wars;
 $$ photons landing on a photodiode.
Poisson random variable: Here, we briefly review some properties of the Poisson random variable that we have discussed in the previous chapters. Remember that a discrete random variable $X$ is said to be a Poisson random variable with parameter $\mu$, shown as $X \sim Poisson(\mu)$, if its range is $R_X=\{0,1,2,3,...\}$, and its PMF is given by \begin{equation} \nonumber P_X(k) = \left\{ \begin{array}{l l} \frac{e^{\mu} \mu^k}{k!}& \quad \text{for } k \in R_X\\ 0 & \quad \text{ otherwise} \end{array} \right. \end{equation} Here are some useful facts that we have seen before:
 If $X \sim Poisson(\mu)$, then $EX=\mu$, and $\textrm{Var}(X)=\mu$.
 If $X_i \sim Poisson(\mu_i)$, for $i=1,2,\cdots, n$, and the $X_i$'s are independent, then \begin{align*} X_1+X_2+\cdots+X_n \sim Poisson(\mu_1+\mu_2+\cdots+\mu_n). \end{align*}
 The Poisson distribution can be viewed as the limit of binomial distribution.
Theorem
Let $Y_n \sim Binomial\big(n,p=p(n)\big)$. Let $\mu>0$ be a fixed real number, and $\lim_{n \rightarrow \infty} np=\mu$. Then, the PMF of $Y_n$ converges to a $Poisson(\mu)$ PMF, as $n \rightarrow \infty$. That is, for any $k \in \{0,1,2,...\}$, we have \begin{equation}%\label{} \nonumber \lim_{n \rightarrow \infty} P_{Y_n}(k)=\frac{e^{\mu} \mu^k}{k!}. \end{equation}
Poisson Process as the Limit of a Bernoulli Process:
Suppose that we would like to model the arrival of events that happen completely at random at a rate $\lambda$ per unit time. Here is one way to do this. At time $t=0$, we have no arrivals yet, so $N(0)=0$. We now divide the halfline $[0, \infty)$ to tiny subintervals of length $\delta$ as shown in Figure 11.2.Thus, by Theorem 11.1, as $\delta \rightarrow 0$, the PMF of $N(t)$ converges to a Poisson distribution with rate $\lambda t$. More generally, we can argue that the number of arrivals in any interval of length $\tau$ follows a $Poisson(\lambda \tau)$ distribution as $\delta \rightarrow 0$.
Consider several nonoverlapping intervals. The number of arrivals in each interval is determined by the results of the coin flips for that interval. Since different coin flips are independent, we conclude that the above counting process has independent increments.
Definition of the Poisson Process:
The above construction can be made mathematically rigorous. The resulting random process is called a Poisson process with rate (or intensity) $\lambda$. Here is a formal definition of the Poisson process.Let $\lambda>0$ be fixed. The counting process $\{N(t), t \in [0, \infty)\}$ is called a Poisson process with rates $\lambda$ if all the following conditions hold:
 $N(0)=0$;
 $N(t)$ has independent increments;
 the number of arrivals in any interval of length $\tau>0$ has $Poisson(\lambda \tau)$ distribution.
Example
The number of customers arriving at a grocery store can be modeled by a Poisson process with intensity $\lambda=10$ customers per hour.
 Find the probability that there are $2$ customers between 10:00 and 10:20.
 Find the probability that there are $3$ customers between 10:00 and 10:20 and $7$ customers between 10:20 and 11.
 Solution

 Here, $\lambda=10$ and the interval between 10:00 and 10:20 has length $\tau=\frac{1}{3}$ hours. Thus, if $X$ is the number of arrivals in that interval, we can write $X \sim Poisson(10/3)$. Therefore, \begin{align*} P(X=2)&=\frac{e^{\frac{10}{3}} \left(\frac{10}{3}\right)^2}{2!}\\ &\approx 0.2 \end{align*}

Here, we have two nonoverlapping intervals $I_1 =$(10:00 a.m., 10:20 a.m.] and $I_2=$ (10:20 a.m., 11 a.m.]. Thus, we can write
\begin{align*} P\bigg(\textrm{$3$ arrivals in $I_1$ $\;$ and $\;$ $7$ arrivals in $I_2$}\bigg)&=\\ & P\bigg(\textrm{$3$ arrivals in $I_1$}\bigg) \cdot P\bigg(\textrm{$7$ arrivals in $I_2$}\bigg). \end{align*} Since the lengths of the intervals are $\tau_1=1/3$ and $\tau_2=2/3$ respectively, we obtain $\lambda \tau_1=10/3$ and $\lambda \tau_2=20/3$. Thus, we have \begin{align*} P\bigg(\textrm{$3$ arrivals in $I_1$ $\;$ and $\;$ $7$ arrivals in $I_2$}\bigg)&=\frac{e^{\frac{10}{3}} \left(\frac{10}{3}\right)^3}{3!} \cdot \frac{e^{\frac{20}{3}} \left(\frac{20}{3}\right)^7}{7!} \\ &\approx 0.0325 \end{align*}

Second Definition of the Poisson Process:
Let $N(t)$ be a Poisson process with rate $\lambda$. Consider a very short interval of length $\Delta$. Then, the number of arrivals in this interval has the same distribution as $N(\Delta)$. In particular, we can write \begin{align*} P(N(\Delta)=0) &=e^{\lambda \Delta} \\ &=1\lambda \Delta+ \frac{\lambda^2}{2} \Delta^2\cdots \; (\textrm{Taylor Series}). \end{align*} Note that if $\Delta$ is small, the terms that include second or higher powers of $\Delta$ are negligible compared to $\Delta$. We write this as \begin{align} P(N(\Delta)=0) =1\lambda \Delta+ o(\Delta) \hspace{30pt} (11.1) \end{align} Here $o(\Delta)$ shows a function that is negligible compared to $\Delta$, as $\Delta \rightarrow 0$. More precisely, $g(\Delta)=o(\Delta)$ means that \begin{align*} \lim_{\Delta \rightarrow 0} \frac{g(\Delta)}{\Delta}=0. \end{align*} Now, let us look at the probability of having one arrival in an interval of length $\Delta$. \begin{align*} P(N(\Delta)=1) &=e^{\lambda \Delta} \lambda \Delta \\ &=\lambda \Delta \left(1\lambda \Delta+ \frac{\lambda^2}{2} \Delta^2\cdots \right) \; (\textrm{Taylor Series})\\ &=\lambda \Delta+\left(\lambda^2 \Delta^2+ \frac{\lambda^3}{2} \Delta^3\cdots \right)\\ &=\lambda \Delta+o(\Delta). \end{align*} We conclude that \begin{align} P(N(\Delta)=1)=\lambda \Delta+o(\Delta) \hspace{30pt} (11.2) \end{align} Similarly, we can show that \begin{align} P(N(\Delta) \geq 2)=o(\Delta) \hspace{30pt} (11.3) \end{align} In fact, equations 11.1, 11.2, and 11.3 give us another way to define a Poisson process.Let $\lambda>0$ be fixed. The counting process $\{N(t), t \in [0, \infty)\}$ is called a Poisson process with rate $\lambda$ if all the following conditions hold:
 $N(0)=0$;
 $N(t)$ has independent and stationary increments
 we have \begin{align*} &P(N(\Delta)=0) =1\lambda \Delta+ o(\Delta),\\ &P(N(\Delta)=1)=\lambda \Delta+o(\Delta),\\ &P(N(\Delta) \geq 2)=o(\Delta). \end{align*}
Arrival and Interarrival Times:
Let $N(t)$ be a Poisson process with rate $\lambda$. Let $X_1$ be the time of the first arrival. Then, \begin{align*} P(X_1>t)&=P\big(\textrm{no arrival in }(0,t] \big) \\ &=e^{\lambda t}. \end{align*} We conclude that \begin{align} \nonumber F_{X_1}(t) = \left\{ \begin{array}{l l} 1e^{\lambda t} & \quad t>0 \\ & \quad \\ 0 & \quad \text{otherwise} \end{array} \right. \end{align} Therefore, $X_1 \sim Exponential(\lambda)$. Let $X_2$ be the time elapsed between the first and the second arrival (Figure 11.4).If $N(t)$ is a Poisson process with rate $\lambda$, then the interarrival times $X_1$, $X_2$, $\cdots$ are independent and \begin{align*} X_i \sim Exponential(\lambda), \; \textrm{ for }i=1,2,3, \cdots. \end{align*}
Example
Let $N(t)$ be a Poisson process with intensity $\lambda=2$, and let $X_1$, $X_2$, $\cdots$ be the corresponding interarrival times.
 Find the probability that the first arrival occurs after $t=0.5$, i.e., $P(X_1>0.5)$.
 Given that we have had no arrivals before $t=1$, find $P(X_1>3)$.
 Given that the third arrival occurred at time $t=2$, find the probability that the fourth arrival occurs after $t=4$.
 I start watching the process at time $t=10$. Let $T$ be the time of the first arrival that I see. In other words, $T$ is the first arrival after $t=10$. Find $ET$ and $\textrm{Var}(T)$.
 I start watching the process at time $t=10$. Let $T$ be the time of the first arrival that I see. Find the conditional expectation and the conditional variance of $T$ given that I am informed that the last arrival occurred at time $t=9$.
 Solution

 Since $X_1 \sim Exponential(2)$, we can write \begin{align*} P(X_1>0.5) &=e^{(2 \times 0.5)} \\ &\approx 0.37 \end{align*} Another way to solve this is to note that \begin{align*} P(X_1>0.5) &=P(\textrm{no arrivals in }(0,0.5])=e^{(2 \times 0.5)}\approx 0.37 \end{align*}
 We can write \begin{align*} P(X_1>3X_1>1) &=P(X_1>2) \; (\textrm{memoryless property})\\ &=e^{2 \times 2}\\ &\approx 0.0183 \end{align*} Another way to solve this is to note that the number of arrivals in $(1,3]$ is independent of the arrivals before $t=1$. Thus, \begin{align*} P(X_1>3X_1>1) &=P\big(\textrm{no arrivals in }(1,3] \;  \; \textrm{no arrivals in }(0,1]\big)\\ &=P\big(\textrm{no arrivals in }(1,3]\big)\; (\textrm{independent increments})\\ &=e^{2 \times 2}\\ &\approx 0.0183 \end{align*}
 The time between the third and the fourth arrival is $X_4 \sim Exponential(2)$. Thus, the desired conditional probability is equal to \begin{align*} P(X_4>2X_1+X_2+X_3=2)&=P(X_4>2) \; (\textrm{independence of the $X_i$'s})\\ &=e^{2 \times 2}\\ &\approx 0.0183 \end{align*}
 When I start watching the process at time $t=10$, I will see a Poisson process. Thus, the time of the first arrival from $t=10$ is $Exponential(2)$. In other words, we can write \begin{align*} T=10+X, \end{align*} where $X \sim Exponential(2)$. Thus, \begin{align*} ET&=10+EX\\ &=10+\frac{1}{2}=\frac{21}{2}, \end{align*} \begin{align*} \textrm{Var}(T)&=\textrm{Var}(X)\\ &=\frac{1}{4}. \end{align*}
 Arrivals before $t=10$ are independent of arrivals after $t=10$. Thus, knowing that the last arrival occurred at time $t=9$ does not impact the distribution of the first arrival after $t=10$. Thus, if $A$ is the event that the last arrival occurred at $t=9$, we can write \begin{align*} E[TA]&=E[T]\\ &=\frac{21}{2}, \end{align*} \begin{align*} \textrm{Var}(TA)&=\textrm{Var}(T)\\ &=\frac{1}{4}. \end{align*}

Now that we know the distribution of the interarrival times, we can find the distribution of arrival times \begin{align*} &T_1=X_1,\\ &T_2=X_1+X_2,\\ &T_3=X_1+X_2+X_3,\\ & \hspace{30pt} \vdots \end{align*} More specifically, $T_n$ is the sum of $n$ independent $Exponential (\lambda)$ random variables. In previous chapters we have seen that if $T_n=X_1+X_2+\cdots+X_n$, where the $X_i$'s are independent $Exponential(\lambda)$ random variables, then $T_n \sim Gamma(n, \lambda)$. This has been shown using MGFs. Note that here $n \in \mathbb{N}$. The $Gamma(n, \lambda)$ is also called Erlang distribution, i.e, we can write \begin{align*} T_n \sim Erlang(n,\lambda)=Gamma(n, \lambda), \; \textrm{ for }n=1,2,3, \cdots. \end{align*} The PDF of $T_n$, for $n=1,2,3,\cdots$, is given by \begin{align*} f_{T_n}(t)=\frac{\lambda^n t^{n1}e^{\lambda t}}{(n1)!}, \; \textrm{ for }t>0. \end{align*} Remember that if $X \sim Exponential(\lambda)$, then \begin{align*} &E[X]=\frac{1}{\lambda},\\ &\textrm{Var}(X)=\frac{1}{\lambda^2}. \end{align*} Since $T_n=X_1+X_2+\cdots+X_n$, we conclude that \begin{align*} &E[T_n]=n EX_1=\frac{n}{\lambda},\\ &\textrm{Var}(T_n)=n \textrm{Var}(X_n)=\frac{n}{\lambda^2}. \end{align*} Note that the arrival times are not independent. In particular, we must have $T_1 \leq T_2 \leq T_3 \leq \cdots$.
If $N(t)$ is a Poisson process with rate $\lambda$, then the arrival times $T_1$, $T_2$, $\cdots$ have $Gamma(n, \lambda)$ distribution. In particular, for $n=1,2,3,\cdots$, we have \begin{align*} E[T_n]=\frac{n}{\lambda}, \; \textrm{and} \; \textrm{Var}(T_n)=\frac{n}{\lambda^2}. \end{align*}