### 统计代写|随机过程代写stochastic process代考|the independence and stationarity

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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|随机过程代写stochastic process代考|the independence and stationarity

EXERCISE 3.1. We use the independence and stationarity of increments of a Poisson process to derive the expression for the joint probability distribution. We write
$$\begin{gathered} P(N(s)=m, N(t)=n)=P(N(t)-N(s)=n-m, N(s)=m) \ =P(N(t)-N(s)=n-m) P(N(s)=m)=P(N(t-s)=n-m) P(N(s)=m) \ =\frac{(\lambda(t-s))^{n-m}}{(n-m) !} e^{-\lambda(t-s)} \cdot \frac{(\lambda s)^{m}}{m !} e^{-\lambda s}=\frac{(t-s)^{n-m} s^{m}}{(n-m) ! m !} \lambda^{n} e^{-\lambda t} \ =\left(\begin{array}{c} n \ m \end{array}\right)\left(\frac{s}{t}\right)^{m}\left(1-\frac{s}{t}\right)^{n-m} \frac{\lambda^{n}}{n !} e^{-\lambda t} . \end{gathered}$$
EXERCISE 3.2. Assume that $s<t$. We compute the covariance function, using the independence and stationarity of the increments. We have
\begin{aligned} &\operatorname{Cov}(N(s), N(t))=E[N(s) N(t)]-E[N(s)] E[N(t)] \ &=E[(N(t)-N(s)+N(s))(N(s))]-E[N(s)] E[N(t)] \ &=E[(N(t)-N(s)) N(s)]+E[N(s)]^{2}-E[N(s)] E[N(t)] \ &=E[N(t)-N(s)] E[N(s)]+\operatorname{Var}[N(s)]+[E(N(s))]^{2}-E[N(s)] E[N(t)] \ &=E[N(t-s)] E[N(s)]+\operatorname{Var}[N(s)]+[E(N(s))]^{2}-E[N(s)] E[N(t)] \ &=\lambda(t-s) \lambda s+\lambda s+(\lambda s)^{2}-\lambda s \lambda t=\lambda s . \end{aligned}
EXERCISE 3.3. (a) $P(N(5)=16 \mid N(1)=2, N(2)-N(1)=3$ )
$$\begin{gathered} =\frac{P(N(5)-N(2)=11, N(2)-N(1)=3, N(1)=2)}{P(N(2)-N(1)=3, N(1)=2)} \ =\frac{P(N(3)=11) P(N(1)=3) P(N(1)=2)}{P(N(1)=3) P(N(1)=2)}=P(N(3)=11)=\frac{((5)(3))^{11}}{11 !} e^{-(5)(3)}=0.0663 . \end{gathered}$$
(b) Since $E\left(S_{100}\right)-(100)\left(\frac{1}{5}\right)-20$, the 100th claim is expected to be seen on the 20 th business day, that is, on January 27th.

## 统计代写|随机过程代写stochastic process代考|Phone calls that result

EXERCISE 3.4. (a) Phone calls that result in sales occur with rate $(0.15)\left(\frac{60}{5}\right)=1.8$ per hour.
Therefore, in the next two hours, there will be, on average $(2)(1.8)=3.6$ successful sales.
(b) The total number of phone calls is a Poisson process with a rate of $60 / 5=12$ per hour. Phone calls that result in a sale and those that don’t form independent Poisson processes with rates $1.8$ and $10.2$ per hour, respectively. Therefore,
$$\begin{gathered} P\left(N(1)=15, N_{\text {sale }}(1)=5\right)=P\left(N_{\text {sale }}(1)=5, N_{\text {no sale }}(1)=10\right) \ =P\left(N_{\text {sale }}(1)=5\right) P\left(N_{\text {no sale }}(1)=10\right)=\frac{(1.8)^{5}}{5 !} e^{-1.8} \frac{(10.2)^{10}}{10 !} e^{-10.2}=0.00325 . \end{gathered}$$
(c) $P(N(4)=10 \mid N(1)=3)=P(N(4)-N(1)=7)=P(N(3)=7)=\frac{((1.8)(3))^{7}}{7 !} e^{-(1.8)(3)}=0.119987$.
EXERCISE 3.5. (a) $N_{1}(t)$ and $N_{2}(t)$ are splitted Poisson processes with the means
\begin{aligned} E\left(N_{1}(t)\right)=\lambda \int_{0}^{t} P(\text { disease is contracted at time } s \text {, symptoms show }\ &=\lambda \int_{0}^{t} F(t-s) d s={u=t-s}=\lambda \int_{0}^{t} F(u) d u, \end{aligned}
and
\begin{aligned} E\left(N_{2}(t)\right)=& \lambda \int_{0}^{t} P(\text { disease is contracted at time } s, \text { no symptoms show by }\ &=\lambda \int_{0}^{t}(1-F(t-s)) d s={u=t-s}=\lambda \int_{0}^{t}(1-F(u)) d u . \end{aligned}
(b) Suppose by a fixed time $t, \hat{E}\left(N_{1}(t)\right)$ individuals are observed who show symptoms of a disease. From here, we can estimate the rate of contracting the disease as $\hat{\lambda}=\frac{\hat{E}\left(N_{1}(t)\right)}{\int_{0}^{t} F(u) d u}$. Plugging this into the expression for the expected value of $N_{2}(t)$, we can calculate the estimated number of individuals infected but not yet showing symptoms by time $t$ as
$$\hat{E}\left(N_{2}(t)\right)-\frac{\hat{E}\left(N_{1}(t)\right) \int_{0}^{t}(1-F(u)) d u}{\int_{0}^{t} F(u) d u} .$$
(c) Suppose the incubation period until symptoms show is an exponentially distributed random variable with a mean of 2 days. Thus, $F(u)=1-e^{-u / 2}, u \geq 0$. Given that $\hat{E}\left(N_{1}(10)\right)=1000$, we estimate the number of individuals who are infected but haven’t shown the symptoms yet as
$$\hat{E}\left(N_{2}(10)\right)=\frac{1000 \int_{0}^{10} e^{-\frac{u}{2}} d u}{\int_{0}^{10}\left(1-e^{-\frac{u}{2}}\right) d u}=\frac{(1000)(2)\left(1-e^{-\frac{10}{2}}\right)}{10-(2)\left(1-e^{-\frac{10}{2}}\right)}=247.8979,$$

## 统计代写|随机过程代写stochastic process代考|Let 𝑁𝑁(𝑡𝑡) denote

EXERCISE 3.6. (a) Let $N(t)$ denote the number of high road surface distress areas on a $t$-mile stretch of the road. It is a Poisson process with a rate $\lambda=2.8 . \mathrm{So}, E(N(10))=(2.8)(10)=28$.
(b) The code below simulates 30 distances between distressed surface areas. These distances are independent and exponentially distributed with mean $\frac{1}{2.8}=0.357143$ miles.

###### specifying parameters

lambda $<-2.8$
Nareas<- 30

###### defining states

$\mathrm{N}<-0$ : Nareas

dist<- c()

###### setting initial value for distance

dist $[1]<-0$

###### specifying seed

set.seed (777754)
for (i in 2:(Nareast1))
dist $[i]<-$ dist $[i-1]+\operatorname{round}((-1 / \operatorname{lambda}) * \log (\operatorname{runif}(1)), 2)$

###### plotting trajectory

plot(dist, $N$, type=” $n^{n}$, xlab=”Number of miles”, $y l a b=”$ Number of distressed areas”,
panel.first $=$ grid ())
segments (dist [-length (dist) ], $N$ [-length (dist) ], dist $[-1]-0.07, \mathbb{N}[-1$ ength(dist) ],
$l w d=2, \quad$ col= 4 )
points (dist, N, pch=20, col=4)
points (dist $[-1], N[-$ length (dist) $]$, pch=1, col=4)

## 统计代写|随机过程代写stochastic process代考|the independence and stationarity

=磷(ñ(5)−ñ(2)=11,ñ(2)−ñ(1)=3,ñ(1)=2)磷(ñ(2)−ñ(1)=3,ñ(1)=2) =磷(ñ(3)=11)磷(ñ(1)=3)磷(ñ(1)=2)磷(ñ(1)=3)磷(ñ(1)=2)=磷(ñ(3)=11)=((5)(3))1111!和−(5)(3)=0.0663.
(b) 由于和(小号100)−(100)(15)−20，预计第 100 次索赔将在第 20 个工作日，即 1 月 27 日看到。

## 统计代写|随机过程代写stochastic process代考|Phone calls that result

(b) 电话总数是一个泊松过程，比率为60/5=12每小时。导致销售的电话和那些不形成具有费率的独立泊松过程的电话1.8和10.2每小时，分别。所以，

（C）磷(ñ(4)=10∣ñ(1)=3)=磷(ñ(4)−ñ(1)=7)=磷(ñ(3)=7)=((1.8)(3))77!和−(1.8)(3)=0.119987.

(b) 假设按固定时间吨,和^(ñ1(吨))观察到表现出疾病症状的个体。从这里，我们可以估计感染疾病的比率为λ^=和^(ñ1(吨))∫0吨F(在)d在. 将其代入表达式以获得预期值ñ2(吨)，我们可以按时间计算估计感染但尚未出现症状的人数吨作为

(c) 假设直到出现症状的潜伏期是一个指数分布的随机变量，平均为 2 天。因此，F(在)=1−和−在/2,在≥0. 鉴于和^(ñ1(10))=1000，我们估计感染但尚未出现症状的人数为

## 统计代写|随机过程代写stochastic process代考|Let 𝑁𝑁(𝑡𝑡) denote

(b) 下面的代码模拟了受损表面区域之间的 30 距离。这些距离是独立的并且呈指数分布，均值12.8=0.357143英里。

ñ<−0: 纳瑞斯

###### 指定种子

set.seed (777754)
for (i in 2:(Nareast1))
dist[一世]<−距离[一世−1]+圆形的⁡((−1/拉姆达)∗日志⁡(鲁尼夫⁡(1)),2)

###### 绘制轨迹

panel.first=grid ())

l在d=2,col= 4 )

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