Poisson Process
Counting Process
 $N(t)\geq 0$
 $N(t)\in\mathbb{N}$
 $s<t,N(s)\leq N(t)$
 $s<t,N(t)N(s)$ equals the number of events that have occurred in the interval $(s,t]$
Definition 1 of Poisson Process

$N(0)=0$

independent increments

$\forall s,t\geq 0$
$$P(N(t+s)N(s)=n)=e^{\lambda t}\frac{(\lambda t)^n}{n!},n=0,1$$
Definition 2 of Poisson Process
 $N(0)=0$
 The process has stationary and independent increments
 $P(N(t+h)N(t)=1)=\lambda h+o(h)$
 $P(N(t+h)N(t)\geq 2)=o(h)$
Sequence of interarrival times

${X_n,n\geq 1}$: $X_n$ denotes the time of the $n$th event
$$P(X_1>t)=P(N(t)=0)=e^{\lambda t}$$

$X_n$ are independent identically distributed exponential random variables having mean $\frac{1}{\lambda}$
Arrival time of $n$th event

$S_n=\sum_{i=1}^nX_i,n\geq 1$: Gamma Distribution

order statistics $Y_{(i)}$ corresponding to $Y_i$: $Y_{(i)}$ is the $k$th minimum of $Y_{i}$
$$ f(y_1,y_2,\cdots,y_n)=n!\prod_{i=1}^nf(y_i) $$

$Y_i$ are uniformly distributed over $(0,t)$, then the joint density funciton of order statistics is
$$ f(y_1,y_2,\cdots,y_n)=\frac{n!}{t^n} $$

Given that $N(t)=n$, the $n$ arrival times $S_1,\cdots,S_n$ have the same distribution as the order statistics corresponding to $n$ independent random variables uniformly distributed on the interval $(0,t)$
Two Type Poisson random variables
 event occurs at time $s$, then, indepentently of all else, it is classified as being a typeI event with probability $P(s)$ and a typeII event with probability $1P(s)$
 $N_1(t)$ and $N_2(t)$ are independent Poisson random variables having respective means $\lambda tp$ and $\lambda t(1p)$
$$ p=\frac{1}{t}\int^t_0P(s)ds $$
M/G/1 Busy Period
 M/G/1 queueing system
 customers arrive in accordance with a Poisson process with rate $\lambda$
 upon arrival, either enter service if the server is free or else join the queue
 successive service times are independent and identically distributed according to $G$: $Y_i$ denotes the sequence of service times
 busy period begins: an arrival finds the server free
 busy period ends: no longer any customers in the system
 busy period last a time $t$ and consits of $n$ services (Probability $B(t,n)$) iff
 $S_k\leq Y_1+\cdots Y_k$
 $Y_1+\cdots Y_n=t$
 There are $n1$ arrivals in $(0,t)$
$$B(t,n)=\int_0^te^{\lambda t}\frac{(\lambda t)^{n1}}{n!}dG_n(t)$$
$$B(t)=\sum_{n=1}^\infty B(t,n)$$