# 统计代写|随机过程代写stochastic process代考|STAT3021

## 统计代写|随机过程代写stochastic process代考|Exercises and Complements

Exercise 3.1 In a simple random walk with two absorbing barriers at 0 and a let the position $X_n$ at the $n$th step be given by $X_n=X_{n-1}+Z_n$ where $Z_n$ ‘s are i.i.d. r.vs. taking values 1 and $-1$ with corresponding probabilities $p$ and $q=1-p$. Let $\pi_k(n)$, be the probability of absorption at 0 of the random walk in $n$-steps starting from position $k$.
Show that the generating function $G_k(s)=\sum_{n=0}^{\infty} \pi_k(n) s^n,|s|>1$ is given by
$$(q / p)^k \frac{\lambda_1^{a-k}(s)-\lambda_2^{a-k}(s)}{\lambda_1^a(s)-\lambda_2^a(s)} \text {, }$$

$$\lambda_1(s)=\frac{1+\left(1-4 p q s^2\right)^{1 / 2}}{2 p s}, \lambda_2(s)=\frac{1-\left(1-4 p q s^2\right)^{1 / 2}}{2 p s} .$$
Also show that
$$\pi_k(n)=2^n p^{(n-k) / 2} q^{(n+k) / 2} \int_0^1 \cos ^{n-1}(\pi x) \sin (\pi x) \sin (k \pi x) d x .$$
What will be the value of $\pi_k(n)$ in case of simple absorbing barrier at 0 when playing against an infinitely rich opponent?

Exercise 3.2 In a random walk with two absorbing barriers at $-n$ and $a$, let the position $X_n$ at the $n$th step be given by $X_n=X_{n-1}+Z_n$. where $Z_n$ ‘s are i.i.d. r.v.s taking values 1 ,.
$-1.0$ with corresponding probabilities $p . q .1-p-q$.
If $f_{j a}^{(n)}=P\left(-b<X_1, X_2, \ldots, X_{n-1}<a, X_n=a \mid X_0=j\right)$,
Show that the generating function of $\left{f_{j a}^{(n)}\right}$ is given by
$$F_{s a}(s)=\frac{\left.\mid \lambda_1(s)\right]^{j+b}-\left[\lambda_2(s)\right]^{j+b}}{\left[\lambda_1(s)\right]^{u+b}-\left[\lambda_2(s)\right]^{a+b}}$$
where $\lambda_1(s)$ and $\lambda_2(s)$ are the roots of the equation
$$p s \lambda^2-\lambda[1-s(1-p q)]+q s=0 .$$
If the random walk starts from the origin, what will be the expression of the generating function.

## 统计代写|随机过程代写stochastic process代考|Renewal Theory

Let $X_n, n=1,2, \ldots$, be the nonnegative i.i.d r.v.s with $S_n=X_1+\ldots+X_n, n \geq 1$, $S_0=0$. $F$ is the common d.f. of $X$ and assume $P\left(X_n=0\right)<1$. Define $N(t)=$ sup $\left{n \mid S_n \leq t\right}$. The process ${N(t), t \geq 0}$ is called the Renewal Process.

To fix our ideas $X_i$ can be taken to represent the life time of the machines being replaced. The first machine is installed at time $t=0$ and is replaced instantaneously at time $t=X_1$. The replaced machine is again replaced at time $t=X_1+X_2$, and so on. If we write $S_n=X_1+\ldots+X_n$, the partial sum $S_n$ can be interpreted to be the time at which the nth replacement is made. $N(t)$ is the largest value of $n$ for which $S_n \leq t$. In other words $N(t)$ is the number of renewals that would have occurred at time $t$. The Renewal Theory, in a sense, is a special case of a Random Walk with absorbing barrier. We are sampling the $X_i$ until $S_n$ shoots the barrier at time $t$ and $N(t)+1$ is the sample size when we stop. Hence the Renewal Theory is also linked with Sequential Analysis in statistics.
${N(t), t \in(0, \infty)}$ is called the Renewal Counting Process. We can also write $N(t)=\max \left{n \mid S_n \leq t\right}$

We want to find $P[N(t)=n]$ given $F$. To compute this we proceed as follows:
\begin{aligned} P\left[S_2 \leq t\right] &=\int_0^{\infty} F(t-u) d F(u) \ &=\int_0^t F(t-u) d F(u) \ &=F^* F(t)=F^{(2)}(t), \ldots \ P\left[S_n \leq t\right] &=F^{(n)}(t)=\int_0^t F^{(n-1)}(t-u) d F(u), n \geq 1 \end{aligned}
Define $\quad F^{(0)}(t)=\left{\begin{array}{l}0 \text { if } t<0 \\ 1 \text { if } t \geq 0 .\end{array}\right.$ Now $P[N(t)=n]=P\left[S_1 \leq t, S_2 \leq t, \ldots, S_n \leq t, S_{n+1}>t\right]$
$=P\left[S_n \leq t, S_{n+1}>t\right]$ (by nonnegativeness of $X_1$ )

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Exercises and Complements

$$\begin{gathered} (q / p)^k \frac{\lambda_1^{a-k}(s)-\lambda_2^{a-k}(s)}{\lambda_1^a(s)-\lambda_2^a(s)}, \ \lambda_1(s)=\frac{1+\left(1-4 p q s^2\right)^{1 / 2}}{2 p s}, \lambda_2(s)=\frac{1-\left(1-4 p q s^2\right)^{1 / 2}}{2 p s} . \end{gathered}$$

$$\pi_k(n)=2^n p^{(n-k) / 2} q^{(n+k) / 2} \int_0^1 \cos ^{n-1}(\pi x) \sin (\pi x) \sin (k \pi x) d x .$$

$-1.0$ 有相应的概率 $p . q \cdot 1-p-q$.

$$F_{s a}(s)=\frac{\left.\mid \lambda_1(s)\right]^{j+b}-\left[\lambda_2(s)\right]^{j+b}}{\left[\lambda_1(s)\right]^{u+b}-\left[\lambda_2(s)\right]^{a+b}}$$

$$p s \lambda^2-\lambda[1-s(1-p q)]+q s=0 .$$

## 统计代写|随机过程代写stochastic process代考|Renewal Theory

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