### 统计代写|随机过程代写stochastic process代考|The R script below simulates

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## 统计代写|随机过程代写stochastic process代考|The R script below simulates

EXERCISE 2.4. (a) The R script below simulates the trajectories and terminates them if the barrier is hit. Otherwise, trajectories continue for 1,000 steps. The total number of trajectories that hit the barrier is counted. We also record the number of steps (for part (b)) and the $y$-coordinate (for part (c)).

nhits<- 0

set.seed (50118)

###### defining walk as matrix

walk<- matrix (NA, nrow=1001, ncol=2)
nsteps<-c()
ycoord<- c()

###### defining random steps

Istep<- matrix $(c(1,0,-1,0,0,1,0,-1)$, nrow=4, ncol=2, byrow=TRUE)

###### simulating trajectories

for (j in 1:10000)
1
walk $[1,]<-c(0,0$, #setting initial value to the origin
for (i in 2:1001)
\&
walk $[i,]<-$, walk $[i-1,]+$, rstep $[\operatorname{sample}(1: 4, \operatorname{size}=1),$,
if $($ walk $[i, 1]==30) \quad$ {
nhits=nhits+1
break ?
1
nsteps [j]<- i
ycoord $[j]<-$ ifelse $(i==1001,99999$, walk $[i, 2])$
1
print (nhits)
1764
So, of the 10,000 trajectories, 1,764 hit the vertical barrier. Thus, the estimated probability to hit the barrier is $0.1764$.

(b) The average number of steps it takes a trajectory to hit the barrier, provided it did hit the barrier within the 1,000 steps, is estimated as
mean (nsteps [nsteps ! =1001])
$623.2053$
It took on average $623.2$ steps to hit the barrier for the $17.64 \%$ of the trajectories that terminated at the barrier.
(a) Estimate the expected value of the $y$-coordinate at the time when the random walk hits the barriet What should this value be from the theoretical point of view? Hint: deduce from a symmetry argument.
mean $($ ycoord $[$ ycoord $!=99999]))$
$0.1066364$
The estimated average $y$-coordinate for $17.64 \%$ of the trajectories that hit the barrier was $0.1066$. From the theoretical viewpoint, using the symmetry of the random walk, we can argue that the $y$ coordinate should be equal to 0 .

## 统计代写|随机过程代写stochastic process代考|By running the following script

EXERCISE 2.5. By running the following script, we simulate trajectories and calculate the number of those that hit the barrier. The plot is given below.
Nhits<- c()

###### specifying seed

set.seed $(96770)$

###### defining walk as matrix

walk<- matrix (NA, nrow=1001, ncol=2)

###### defining random steps

rstep<- matrix $(c(1,0,-1,0,0,1,0,-1)$, nrow=4, ncol=2, byrow=TRUE $)$

###### varying the barrier value

for (barrier in $1: 50$ ) \&
nhits<- 0
Hyimulating trajectordes
for (j in 1:100)
1
walk $[1,]<-c(0,0$, #setting initial value to the origin
for (i in 2:1001)
{
walk $[1,]<-\operatorname{walk}[i-1,]+\operatorname{rstep}[\operatorname{sample}(1: 4, \operatorname{size}=1),$,
if (walk[i,1]==barrier) |
nhits=nhits+1
break ?
)
1
Nhits [barrier] =nhits
1

print (Nhits)
$$\begin{array}{rrrrrrrrrrrrr} {[1]} & 93 & 94 & 93 & 94 & 87 & 85 & 72 & 66 & 66 & 66 & 61 & 61 \ {[13]} & 66 & 43 & 49 & 49 & 44 & 46 & 37 & 38 & 39 & 31 & 33 & 29 \ {[25]} & 30 & 28 & 19 & 24 & 17 & 18 & 23 & 13 & 22 & 12 & 8 & 9 \ {[37]} & 8 & 10 & 8 & 11 & 6 & 8 & 4 & 7 & 4 & 6 & 3 & 4 \ {[49]} & 4 & 1 & & & & & & & & & & \end{array}$$
plot $(1: 50$, Nhits/100, col=”blue”, xlab=”Position of barrier”, ylab=”Probability of hitting barrier”, panel.first=grid())

## 统计代写|随机过程代写stochastic process代考|Conditioning on the outcome

EXERCISE 2.7. (a) Conditioning on the outcome of the first step, we see that the probability $\boldsymbol{P}{i}$ solves the recurrence relation $\boldsymbol{P}{i}=p \boldsymbol{P}{i+1}+q \boldsymbol{P}{i-1}$ with the border constraints $\boldsymbol{P}{A}=0$ and $\boldsymbol{P}{B}=1$. Assuming first that $\frac{q}{p} \neq 1$, we look for the solution in the form $\boldsymbol{P}{i}=c\left(\frac{q}{p}\right)^{i}+d$ where $c$ and $d$ are some constants that can be found from the boundary conditions: $\boldsymbol{P}{A}=0=c\left(\frac{q}{p}\right)^{A}+d$ and $\boldsymbol{P}{B}=1=$ $c\left(\frac{q}{p}\right)^{B}+d$. From here, $c=-\frac{1}{\left(\frac{q}{p}\right)^{\Lambda}-\left(\frac{q}{p}\right)^{B}}$ and $d=\frac{\left(\frac{q}{p}\right)^{A}}{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{B}}$, and thus, $P{i}=\frac{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{i}}{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{B}}$.
Now assume $\frac{q}{p}=1$. We look for the solution of the recurrence relation in the form $\boldsymbol{P}{\ell}=c i+d$. Again, from the boundary conditions, $\boldsymbol{P}{A}=0=c A+d$ and $\boldsymbol{P}{B}=1=c B+d$. Hence, $c=\frac{1}{B-A}$ and $d=-\frac{A}{B-A}$, leading to $\boldsymbol{P}{i}=\frac{i-A}{B-A}$.
(b) By conditioning on the first step, we see right away that the expectation satisfies the recurrence relation $\boldsymbol{E}{i}=p \boldsymbol{E}{i+1}+q \boldsymbol{E}{i-1}+1$ with the boundary conditions $E{A}=E_{B}=0$. Because of the additive constant term, this equation is referred to as a non-homogeneous relation and the general solution is sought in the form $\boldsymbol{E}{i}=c\left(\frac{q}{p}\right)^{i}+d+\frac{i}{q-p}$, if $\frac{q}{p} \neq 1$, and $\boldsymbol{E}{i}=c i+d-i^{2}$, if $\frac{q}{p}=1$. The constants $c$ and $d$ are found from the boundary conditions. In the former case, they satisfy $\boldsymbol{E}{A}=0=c\left(\frac{q}{p}\right)^{A}+d+\frac{A}{q-p}$ and $\boldsymbol{E}{B}=0=c\left(\frac{q}{p}\right)^{B}+d+\frac{B}{q-p}$. Whence,
$$c=\frac{B-A}{q-p} \cdot \frac{1}{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{B}},$$
and
$$d=-\frac{B-A}{q-p} \cdot \frac{\left(\frac{q}{p}\right)^{B}}{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{B}}-\frac{B}{q-p}$$
resulting in

$$\boldsymbol{E}{i}=\frac{B-A}{q-p} \cdot \frac{\left(\frac{q}{p}\right)^{i}-\left(\frac{q}{p}\right)^{B}}{\left(\frac{q}{p}\right)^{A}-\left(\frac{q}{p}\right)^{B}}-\frac{B-i}{q-p}$$ In the latter case, $c$ and $d$ solve $E{A}=0=c A+d-A^{2}$ and $E_{B}=0=c B+d-B^{2}$. From here, $c=A+B$ and $d=-A B$. Thus, $\boldsymbol{E}{i}=(A+B) i-A B-i^{2}=(B-i)(i-A)$. (c) We use the formulas derived above with $p=0.47, q=0.53, A=10, i=40$, and $B=80$. We obtain $$\boldsymbol{P}{40}=\frac{\left(\frac{0.53}{0.47}\right)^{10}-\left(\frac{0.53}{0.47}\right)^{40}}{\left(\frac{0.53}{0.47}\right)^{10}-\left(\frac{0.53}{0.47}\right)^{80}}=0.007962$$
and
$$\boldsymbol{E}_{40}=\frac{80-10}{0.53-0.47} \cdot \frac{\left(\frac{0.53}{0.47}\right)^{40}-\left(\frac{0.53}{0.47}\right)^{80}}{\left(\frac{0.53}{0.47}\right)^{10}-\left(\frac{0.53}{0.47}\right)^{80}}-\frac{80-40}{0.53-0.47}=490.7115$$
The probability of doubling the fortune in this rigged game is very small (about $0.008$ ), and the gambler will play, on average, about 491 games before he walks out of the casino.

## 统计代写|随机过程代写stochastic process代考|The R script below simulates

n命中<- 0

set.seed (50118)

###### 将walk定义为矩阵

walk<- 矩阵 (NA, nrow=1001, ncol=2)
nsteps<-c()
ycoord<- c()

###### 定义随机步骤

Istep<- 矩阵(C(1,0,−1,0,0,1,0,−1), nrow=4, ncol=2, byrow=TRUE)

###### 模拟轨迹

1

for (i in 2:1001)
\&
walk[一世,]<−， 走[一世−1,]+, r 步[样本⁡(1:4,尺寸=1),,

nhits=nhits+1

1
nsteps [j]<- i
ycoord[j]<−如果别的(一世==1001,99999， 走[一世,2])
1
print (nhits)
1764

(b) 轨迹撞到障碍物的平均步数，假设它确实在 1,000 步内撞到障碍物，估计为

623.2053

(a) 估计预期值是- 随机游走击中 barriet 时的坐标 从理论的角度来看，这个值应该是多少？提示：从对称论证中推导出来。

0.1066364

## 统计代写|随机过程代写stochastic process代考|By running the following script

Nhits<-c()

###### 将walk定义为矩阵

walk<- 矩阵（NA，nrow=1001，ncol=2）

###### 定义随机步骤

rstep<- 矩阵(C(1,0,−1,0,0,1,0,−1), nrow=4, ncol=2, byrow=TRUE)

###### 改变障碍值

nhits<- 0
Hyimulating 轨迹

1
walk[1,]<−C(0,0, #将初始值设置为原点
for (i in 2:1001)
{
walk[1,]<−走⁡[一世−1,]+rstep⁡[样本⁡(1:4,尺寸=1),,
if (walk[i,1]==barrier) |
nhits=nhits+1

)
1
Nhits [障碍] =nhits
1

[1]939493948785726666666161 [13]664349494446373839313329 [25]3028192417182313221289 [37]81081168474634 [49]41

## 统计代写|随机过程代写stochastic process代考|Conditioning on the outcome

(b) 通过以第一步为条件，我们立即看到期望满足递归关系和一世=p和一世+1+q和一世−1+1有边界条件和一种=和乙=0. 由于附加常数项，该方程被称为非齐次关系，并以形式寻求通解和一世=C(qp)一世+d+一世q−p， 如果qp≠1， 和和一世=C一世+d−一世2， 如果qp=1. 常数C和d由边界条件求得。在前一种情况下，它们满足和一种=0=C(qp)一种+d+一种q−p和和乙=0=C(qp)乙+d+乙q−p. 何处，
C=乙−一种q−p⋅1(qp)一种−(qp)乙,

d=−乙−一种q−p⋅(qp)乙(qp)一种−(qp)乙−乙q−p

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