### 统计代写|随机过程代写stochastic process代考|The distribution of X

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## 统计代写|随机过程代写stochastic process代考|The distribution of X

EXERCISE 7.1. (a) We plug $\lambda_{n}=n \lambda$ and $\mu_{n}=0$ into (7.1), and note that $n$ starts with 1 and not 0 . The Kolmogorov forward equations become $P_{1}^{\prime}(t)=-\lambda P_{1}(t)$ and $P_{n}^{\prime}(t)=(n-1) \lambda P_{n-1}(t)-$ $n \lambda P_{n}(t), n=2,3, \ldots$, with the initial condition $P_{1}(0)=1$.
(b) To show that $P_{n}(t)=e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-1}, n=1,2, \ldots$, solve the Kolmogorov equations, we write $P_{1}(t)=e^{-\lambda t}$, so $P_{1}^{\prime}(t)=-\lambda e^{-\lambda t}=-\lambda P_{1}(t)$. Also,
$P_{n}^{\prime}(t)=-\lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-1}+e^{-\lambda t}(n-1) \lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-2}=-\lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-1}+$ $e^{-\lambda t}(n-1) \lambda\left(e^{-\lambda t}-1+1\right)\left(1-e^{-\lambda t}\right)^{n-2}=-\lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-1}-(n-1) \lambda e^{-\lambda t}(1-$ $\left.e^{-\lambda t}\right)^{n-1}+(n-1) \lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-2}=(n-1) \lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-2}-n \lambda e^{-\lambda t}(1-$ $\left.e^{-\lambda t}\right)^{n-1}=(n-1) \lambda P_{n-1}(t)-n \lambda P_{n}(t) .$
(c) The distribution of $X(t)$ is geometric that models the number of trials until the first success where the probability of success is $p=e^{-\lambda t}$. Therefore, $E(X(t))=\frac{1}{p}=e^{\lambda t}$, and $\operatorname{Var}(X(t))=\frac{1-p}{p^{2}}=$ $\frac{1-e^{-\lambda t}}{e^{-2 \lambda t}}=e^{\lambda t}\left(e^{\lambda t}-1\right)$.
(d) If $\lambda=4$, the probability that there will be between 3 and 5 particles at week 1 is $P_{3}(1)+P_{4}(1)+$ $P_{5}(1)=e^{-4}\left(1-e^{-4}\right)^{3-1}+e^{-4}\left(1-e^{-4}\right)^{4-1}+e^{-4}\left(1-e^{-4}\right)^{5-1}=0.051989$. The mean at week 1 is $E(X(1))=e^{4}=54.59815$, and the standard deviation is $\sqrt{\operatorname{Var}(X(1))}=\sqrt{e^{4}\left(e^{4}-1\right)}=$ $54.09584$.
EXERCISE 7.2. (a) We plug $\lambda_{n}=n \lambda$ and $\mu_{n}=0$ into (7.1) and note that $n$ starts with $m$ and not 0 . The Kolmogorov forward equations become $P_{m}^{\prime}(t)=-m \lambda P_{m}(t)$ and $P_{n}^{\prime}(t)=(n-1) \lambda P_{n-1}(t)-$ $n \lambda P_{n}(t), n=2,3, \ldots$, with the initial condition $P_{m}(0)=1$.
(b) To verify that $P_{n}(t)=\left(\begin{array}{c}n-1 \ n-m\end{array}\right) e^{-m \lambda t}\left(1-e^{-\lambda t}\right)^{n-m}, n=m, m+1_{, \ldots}$, solve the Kolmogorov equations, we write $P_{m}(t)=e^{-m \lambda t}$, so $P_{m}^{\prime}(t)=-m \lambda e^{-m \lambda t}=-m \lambda P_{m}(t)$. Further, $P_{n}^{\prime}(t)=$ $-m \lambda\left(\begin{array}{c}n-1 \ n-m\end{array}\right) e^{-m \lambda t}\left(1-e^{-\lambda t}\right)^{n-m}+\left(\begin{array}{c}n-1 \ n-m\end{array}\right) e^{-m \lambda t}(n-m) \lambda e^{-\lambda t}\left(1-e^{-\lambda t}\right)^{n-m-1}–m \lambda P_{n}(t)+$ $\left(\begin{array}{l}n-1 \ n-m\end{array}\right) e^{-m \lambda t}(n-m) \lambda\left(e^{-\lambda t}-1+1\right)\left(1-e^{-\lambda t}\right)^{n-m-1}=-m \lambda P_{n}(t)-(n-m) \lambda P_{n}(t)+$ $(n-m)\left(\begin{array}{c}n-1 \ n=m\end{array}\right) \lambda e^{-m \lambda t}\left(1-e^{-\lambda t}\right)^{n-m-1}=-n \lambda P_{n}(t)+(n-1) \lambda\left(\begin{array}{c}n-2 \ n=m=1\end{array}\right) e^{-m \lambda t}(1-$ $\left.e^{-\lambda t}\right)^{n-m-1}=(n-1) \lambda P_{n-1}(t)-n \lambda P_{n}(t)$.
(c) The distribution of $X(t)$ is a negative binomial that models the number of trials until the $m$ th success, where the probability of success is $p=e^{-\lambda t}$. Therefore, the mean and the variance are $E(X(t))=\frac{m}{p}=m e^{\lambda t}$, and $\operatorname{Var}(X(t))=\frac{m(1-p)}{p^{2}}=m e^{\lambda t}\left(e^{-\lambda t}-1\right) .$
(d) $P_{12}(2)=\left(\begin{array}{c}12-1 \ 12-5\end{array}\right) e^{-(5)(0.2)(2)}\left(1-e^{-(0.2)(2)}\right)^{12-5}=0.0189$. The mean and standard deviations are $E(X(2))=(5) e^{(0.2)(2)}=7.459123$, and $\sqrt{\operatorname{Var}(X(2))}=\sqrt{(5) e^{0.4}\left(e^{0.4}-1\right)}=1.915354 .$

## 统计代写|随机过程代写stochastic process代考|In the Kolmogorov forward equations

EXERCISE 7.3. (a) In the Kolmogorov forward equations (7.1), we use $\lambda_{n}=0$, and $\mu_{n}=n \mu$, and the fact that the initial population size is $N$. We write $P_{N}^{\prime}(t)=-N \mu P_{N}(t)$ and $P_{n}^{\prime}(t)=$ $(n+1) \mu P_{n+1}(t)-n \mu P_{n}(t), n=0,1, \ldots, N-1$, with the initial condition $P_{N}(0)=1$.
(b) The probabilities $P_{n}(t)=\left(\begin{array}{l}N \ n\end{array}\right) e^{-n \mu t}\left(1-e^{-\mu t}\right)^{N-n}, n=0, \ldots, N$, solve the Kolmogorov forward equations since $P_{N}(t)=e^{-N \mu t}$ and so, $P_{N}^{\prime}(t)=-N \mu e^{-N \mu t}=-N \mu P_{N}(t)$. Also,
$$\begin{gathered} P_{n}^{\prime}(t)=-n \mu\left(\begin{array}{l} N \ n \end{array}\right) e^{-n \mu t}\left(1-e^{-\mu t}\right)^{N-n}+\left(\begin{array}{l} N \ n \end{array}\right) e^{-n \mu t}(N-n) \mu e^{-\mu t}\left(1-e^{-\mu t}\right)^{N-n-1} \ =-n \mu P_{n}(t)+(n+1) \mu\left(\begin{array}{c} N \ n+1 \end{array}\right) e^{-(n+1) \mu t}\left(1-e^{-\mu t}\right)^{N-(n+1)}=(n+1) \mu P_{n+1}(t)-n \mu P_{n}(t) . \end{gathered}$$
(c) The distribution of $X(t)$ is binomial with parameters $N$ and $p=e^{-\mu t}$. Therefore, $E(X(t))=N p=$ $N e^{-\mu t}$, and $\operatorname{Var}(X(t))=N p(1-p)=N e^{-\mu t}\left(1-e^{-\mu t}\right)$.
(d) $P_{12}(3)=\left(\begin{array}{c}15 \ 12\end{array}\right) e^{-(12)(0.02)(3)}\left(1-e^{-(0.02)(3)}\right)^{15-12}=0.0437$. The mean and standard deviation are $E(X(3))=15 e^{-(0.02)(3)}=14.12647$, and $\sqrt{\operatorname{Var}(X(3))}=\sqrt{15 e^{-(0.02)(3)}\left(1-e^{-(0.02)(3)}\right)}=$ $0.907007 .$
EXERCISE 7.4. (a) We are given that $\lambda=1.3$ and $\mu=0.2$. We need to compute
$$P_{4}(2)=\left(1-P_{0}\right)\left(1-\frac{\lambda}{\mu} P_{0}\right)\left(\frac{\lambda}{\mu} P_{0}\right)^{n-1}=\left(1-P_{0}\right)\left(1-\frac{1.3}{0.2} P_{0}\right)\left(\frac{1.3}{0.2} P_{0}\right)^{4-1}$$
where
$$P_{0}=\frac{\mu e^{(\lambda-\mu) t}-\mu}{\lambda e^{(\lambda-\mu) t}-\mu}=\frac{0.2 e^{(1.3-0.2)(2)}-0.2}{1.3 e^{(1.3-0.2)(2)}-0.2}=0.139172 .$$
Thus, $P_{4}(2)=0.060783$. The mean and variance are $E(X(2))=e^{(\lambda-\mu) t}=e^{(1.3-0.2)(2)}=9.025013$ and $\operatorname{Var}(X(2))=\frac{\lambda+\mu}{\lambda-\mu} e^{(\lambda-\mu) t}\left(e^{(\lambda-\mu) t}-1\right)=\frac{1.3+0.2}{1.3-0.2} e^{(1.3-0.2)(2)}\left(e^{(1.3-0.2)(2)}-1\right)=98.76253$.
(b) Below we simulate a 50 -step trajectory of the process that starts in state 1 and has parameters $\lambda=$ $1.3$ and $\mu=0.2$.

###### specifying parameters

lambda<- $1.3$
$m u<-0.2$
njumps<- 50

###### defining state and time as vectors

$\mathbb{N}<-\mathrm{c}()$

###### specifying parameters

lambda<- $1.3$
mu<- $0.2$
njumps<- 50

N<- c()
time<-c()

N[1]<-
time[1]<- 0

set.seed(353332)
time<- col)

###### setting initial values

$\mathrm{N}[1]<-1$
time $[1]<-0$

###### specifying seed

set.seed (353332)

## 统计代写|随机过程代写stochastic process代考|the queue will accumulate faster

EXERCISE 7.5. (a) If $\lambda>\mu$, the queue will accumulate faster than customers go through the server, and so we expect an infinite number of customers in the system in the long run.
(b) For $\lambda=3$ and $\mu=5$, the long-run probability that there will be more than 2 customers in the system is $P(#$ of customers $>2)=1-P_{0}-P_{1}-P_{2}=1-\left(1-\frac{\lambda}{\mu}\right) \mu\left[1+\frac{\lambda}{\mu}+\left(\frac{\lambda}{\mu}\right)^{2}\right]=1-$ $\left(1-\frac{\lambda}{\mu}\right) \frac{1-\left(\frac{\lambda}{\mu}\right)^{3}}{1-\frac{\lambda}{\mu}}=\left(\frac{\lambda}{\mu}\right)^{3}=\left(\frac{3}{5}\right)^{3}=0.216 .$
(c) In the long run, the average number of customers in the system is
$$\lim _{t \rightarrow \infty} E(X(t))=\frac{\lambda}{\mu-\lambda}=\frac{3}{5-3}=1.5 .$$
(d) In the long run, the proportion of customers in the system who have to wait more than 1 minute is $P(T>1)=e^{-(\mu-\lambda) t}=e^{-(5-3)(1)}=0.135335$, or roughly $13.5 \%$.

EXERCISE 7.6. (a) Below we simulate a trajectory of a birth-and-death process with immigration and emigration, with parameters $\lambda=1, \mu=0.2, \alpha=0.3$, and $\beta=0.1$. The trajectory starts in state 10 and ends in state $25 .$

###### specifying parameters

lambda<- 1
$m<-0.2$
alpha<-0.3
beta<- $0.1$

###### defining state and time as vectors

$\mathrm{N}<-\mathrm{c}$ ()
time<- c()

###### setting initial values

$\mathrm{N}[1]<-10$
time $[1]<-0$

###### specifying seed

set.seed ( 93743765$)$

###### simulating trajectory

$1<-2$
repeat \&
time.birth<- $(-1 /(\mathrm{N}[i-1] * \operatorname{lambda}+\mathrm{l}$ pha) $) * \log (1-\operatorname{runif}(1))$
time deathc- $\left(-1 /\left(\mathbb{N}[1-1]^{\star}\right.\right.$ mutbeta $\left.)\right) \wedge \log (1-\operatorname{run} 1 f(1))$
if (time.birth < time. death | $N[i-1]==0$ ) (
time $[i]<-$ time $[i-1]+$ time.birth – $0.001$
$\mathbb{N}[i]<-\mathbb{N}[i-1]$
if $(N[i]==25)$ break
else।

## 统计代写|随机过程代写stochastic process代考|The distribution of X

(b) 表明磷n(吨)=和−λ吨(1−和−λ吨)n−1,n=1,2,…，求解 Kolmogorov 方程，我们写磷1(吨)=和−λ吨， 所以磷1′(吨)=−λ和−λ吨=−λ磷1(吨). 还，

(c) 分布X(吨)是几何的，它模拟试验次数，直到第一次成功，其中成功的概率是p=和−λ吨. 所以，和(X(吨))=1p=和λ吨， 和曾是⁡(X(吨))=1−pp2= 1−和−λ吨和−2λ吨=和λ吨(和λ吨−1).
(d) 如果λ=4，第 1 周有 3 到 5 个粒子的概率为磷3(1)+磷4(1)+ 磷5(1)=和−4(1−和−4)3−1+和−4(1−和−4)4−1+和−4(1−和−4)5−1=0.051989. 第 1 周的平均值是和(X(1))=和4=54.59815，标准差为曾是⁡(X(1))=和4(和4−1)= 54.09584.

(b) 核实磷n(吨)=(n−1 n−米)和−米λ吨(1−和−λ吨)n−米,n=米,米+1,…，求解 Kolmogorov 方程，我们写磷米(吨)=和−米λ吨， 所以磷米′(吨)=−米λ和−米λ吨=−米λ磷米(吨). 更远，磷n′(吨)= −米λ(n−1 n−米)和−米λ吨(1−和−λ吨)n−米+(n−1 n−米)和−米λ吨(n−米)λ和−λ吨(1−和−λ吨)n−米−1–米λ磷n(吨)+ (n−1 n−米)和−米λ吨(n−米)λ(和−λ吨−1+1)(1−和−λ吨)n−米−1=−米λ磷n(吨)−(n−米)λ磷n(吨)+ (n−米)(n−1 n=米)λ和−米λ吨(1−和−λ吨)n−米−1=−nλ磷n(吨)+(n−1)λ(n−2 n=米=1)和−米λ吨(1− 和−λ吨)n−米−1=(n−1)λ磷n−1(吨)−nλ磷n(吨).
(c) 分布X(吨)是一个负二项式，它对试验次数进行建模，直到米th 成功，其中成功的概率为p=和−λ吨. 因此，均值和方差为和(X(吨))=米p=米和λ吨， 和曾是⁡(X(吨))=米(1−p)p2=米和λ吨(和−λ吨−1).
(d)磷12(2)=(12−1 12−5)和−(5)(0.2)(2)(1−和−(0.2)(2))12−5=0.0189. 均值和标准差为和(X(2))=(5)和(0.2)(2)=7.459123， 和曾是⁡(X(2))=(5)和0.4(和0.4−1)=1.915354.

## 统计代写|随机过程代写stochastic process代考|In the Kolmogorov forward equations

(b) 概率磷n(吨)=(ñ n)和−nμ吨(1−和−μ吨)ñ−n,n=0,…,ñ，求解 Kolmogorov 正向方程，因为磷ñ(吨)=和−ñμ吨所以，磷ñ′(吨)=−ñμ和−ñμ吨=−ñμ磷ñ(吨). 还，

(c) 分布X(吨)是带参数的二项式ñ和p=和−μ吨. 所以，和(X(吨))=ñp= ñ和−μ吨， 和曾是⁡(X(吨))=ñp(1−p)=ñ和−μ吨(1−和−μ吨).
(d)磷12(3)=(15 12)和−(12)(0.02)(3)(1−和−(0.02)(3))15−12=0.0437. 均值和标准差是和(X(3))=15和−(0.02)(3)=14.12647， 和曾是⁡(X(3))=15和−(0.02)(3)(1−和−(0.02)(3))= 0.907007.

(b) 下面我们模拟从状态 1 开始并有参数的过程的 50 步轨迹λ= 1.3和μ=0.2.

λ<-1.3

ñ<−C()

λ<-1.3

N<- c()

N[1]<-

set.seed(353332)

ñ[1]<−1

###### 指定种子

set.seed (353332)

## 统计代写|随机过程代写stochastic process代考|the queue will accumulate faster

(b) 为λ=3和μ=5，系统中存在超过 2 个客户的长期概率为P(#P(#客户的>2)=1−磷0−磷1−磷2=1−(1−λμ)μ[1+λμ+(λμ)2]=1− (1−λμ)1−(λμ)31−λμ=(λμ)3=(35)3=0.216.
(c) 从长远来看，系统中的平均客户数为

(d) 从长远来看，系统中等待超过 1 分钟的客户比例为磷(吨>1)=和−(μ−λ)吨=和−(5−3)(1)=0.135335，或大致13.5%.

λ<- 1

ñ<−C()

ñ[1]<−10

###### 指定种子

set.seed ( 93743765)

###### 模拟轨迹

1<−2

time.birth<-(−1/(ñ[一世−1]∗拉姆达+l阶段）)∗日志⁡(1−鲁尼夫⁡(1))

ñ[一世]<−ñ[一世−1]

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