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# 数学代写|随机过程Stochastic Porcesses代考|STATS217 The Poisson process

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## 数学代写|随机过程Stochastic Porcesses代考|The Poisson process

We already mentioned the Poisson process in Chapters 2 and 3 . It is a particular continuous-time Markov chain. In Chapter 4, we asserted that the Wiener process and the Poisson process are the two most important stochastic processes for applications. The Poisson process is notably used in the basic queueing models.

The Poisson process, which will be denoted by ${N(t), t \geq 0}$, is also a pure birth process (see Subsection 3.3.4). That is, $N(t)$ designates the number of births (or of events, in general) that occurred from 0 up to time $t$. A process of this type is called a counting process.

Definition 5.1.1. Let $N(t)$ be the number of events that occurred in the interval $[0, t]$. The stochastic process ${N(t), t \geq 0}$ is called a counting process.
Counting processes have the following properties, which are deduced directly from their definition.

Properties. i) $N(t)$ is a random variable whose possible values are $0,1, \ldots$.
ii) The function $N(t)$ is nondecreasing: $N\left(t_2\right)-N\left(t_1\right) \geq 0$ if $t_2>t_1 \geq 0$. Moreover, $N\left(t_2\right)-N\left(t_1\right)$ is the number of events that occurred in the interval $\left(t_1, t_2\right]$

## 数学代写|随机过程Stochastic Porcesses代考|The telegraph signal

As we did with the Brownian motion and in Examples 5.1.4 and 5.1.5, we can define stochastic processes from a Poisson process. An interesting particular transformation of the Poisson process is the telegraph signal ${X(t), t \geq 0}$, defined as follows:
X(t)=(-1)^{N(t)}=\left{\begin{aligned} 1 & \text { if } N(t)=0,2,4, \ldots \ -1 & \text { if } N(t)=1,3,5, \ldots \end{aligned}\right.
An example of a trajectory of a telegraph signal is shown in Fig. 5.2.
Remark . Note that $X(0)=1$, because $N(0)=0$. Thus, the initial value of the process is deterministic. To make the starting point of the process random, we can simply multiply $X(t)$ by a random variable $Z$ that is independent of $X(t)$, for all $t$, and that takes on the value 1 or $-1$ with probability $1 / 2$. It is as if we tossed a fair coin at time $t \geq 0$ to determine whether $X(t)=1$ or $-1$.
The process ${Y(t), t \geq 0}$, where $Y(t):=Z \cdot X(t)$, for all $t \geq 0$, is called a random telegraph signal. We may write that $Z=Y(0)$. Moreover, to be precise, we then use the expression semirandom telegraph signal to designate the process ${X(t), t \geq 0}$. We already encountered the random telegraph signal in Example 2.3.2.

To obtain the distribution of the random variable $X(t)$, it suffices to calculate
\begin{aligned} P[X(t)=1] &=\sum_{k=0}^{\infty} P[N(t)=2 k]=\sum_{k=0}^{\infty} e^{-\lambda t} \frac{(\lambda t)^{2 k}}{(2 k) !} \ &=e^{-\lambda t} \frac{e^{\lambda t}+e^{-\lambda t}}{2}=\frac{1+e^{-2 \lambda t}}{2} \quad \forall t \geq 0 \end{aligned}
because
$$\cosh \lambda t:=\frac{e^{\lambda t}+e^{-\lambda t}}{2}=\sum_{k=0}^{\infty} \frac{(\lambda t)^{2 k}}{(2 k) !}$$

## 数学代与写随机过程STOCHASTIC PORCESSES代考|THE POISSON PROCESS

ii) 功能 $N(t)$ 是非递减的: $N\left(t_2\right)-N\left(t_1\right) \geq 0$ 如果 $t_2>t_1 \geq 0$. 而且， $N\left(t_2\right)-N\left(t_1\right)$ 是区间内发生的事件数 $\left(t_1, t_2\right]$

## 数学代寻|随机过程STOCHASTIC PORCESSES代考|THE TELEGRAPH SIGNAL

$$P[X(t)=1]=\sum_{k=0}^{\infty} P[N(t)=2 k]=\sum_{k=0}^{\infty} e^{-\lambda t} \frac{(\lambda t)^{2 k}}{(2 k) !} \quad=e^{-\lambda t} \frac{e^{\lambda t}+e^{-\lambda t}}{2}=\frac{1+e^{-2 \lambda t}}{2} \quad \forall t \geq 0$$

$$\cosh \lambda t:=\frac{e^{\lambda t}+e^{-\lambda t}}{2}=\sum_{k=0}^{\infty} \frac{(\lambda t)^{2 k}}{(2 k) !}$$

## Matlab代写

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