### 统计代写|随机过程作业代写stochastic process代考| Wald’s fundamental identity

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## 统计代写|随机过程作业代写stochastic process代考|Wald’s fundamental identity

Let $X_{1}, X_{2}, \ldots$ are i.i.d. r.v.s with $S_{n}=X_{1}+X_{2}+\ldots+X_{n}$ and $N$ is a stopping rule.

Let $F_{n}(x)=P\left[S_{n} \leq x\right], F_{1}(x)=F(x)=P\left[X_{1} \leq x\right]$ and m.g.f. of $X_{1}$ is given by $\phi(\theta)=\int_{-\infty}^{\infty} e^{\theta x} d F(x)<\infty$ if $\phi(\sigma)<\infty$, where $\sigma=\operatorname{Re}(\theta)$ We also assume that $$\phi(\sigma)<\infty \text { for all } \sigma,-\beta<\sigma<\alpha<\infty, \alpha, \beta>0$$
Under these conditions, $P\left[e^{X}<1-\delta\right]>0$ and $P\left[e^{X}>1+\delta\right]>0, \delta>0$. $\phi(\theta)$ has a minimum at $\theta=\theta_{0} \neq 0$, where $\theta_{0}$ is the root of the equation $\phi(\theta)=1 .$
Wald’s Sequential Analysis presented the so-called Wald’s identify
$$E\left(e^{\theta S_{N}} /[\phi(\theta)]^{N}\right)=1 \text { for } \phi(\theta)<\infty \text { and }|\phi(\theta)| \geq 1 \text {. }$$
Actually we shall give the proof of a more general theorem in Random walk due to Miller and Kemperman (1961).

Define $F_{n}(x)=P\left[S_{n} \leq x ; N \geq n\right], N=\min \left{n \mid S_{n} \notin(-b, a), 0<a, b<\infty\right}$ and the series $F(z, \theta)=\sum_{n=0}^{\infty} z^{n} \int_{-b}^{a} e^{\theta x} d F_{n}(x)$.
Then
$$E\left(e^{\theta S_{N}} z^{N}\right)=1+[z \phi(\theta)-1] F(z, \theta) \text { for all } \theta$$
which is known as Miller and Kemperman’s Identity.

If $\phi(q)=1 / z$ we get Wald’s Identify.
Proof Let $F_{0}(x)=\left{\begin{array}{lll}0 & \text { if } & x \leq 0 \ 1 & \text { if } & x \geq 0\end{array}\right.$
and $\quad F_{n}(x)=P\left[S_{n} \leq x ; N \geq n\right], n \geq 1$
$$=P\left[-ba \text { ) }$$
is the joint probability that the time $N$ for absorption is $n$ and that the position reached when absorption occurs between $x$ and $x+d x$. Hence if we take Laplace transform with respect to $n$ and with respect to $x$ over absorbing states we have
\begin{aligned} E\left(e^{\theta S_{N}} z^{N}\right) &=\sum_{n=1}^{\infty} z^{n}\left(\int_{-\infty}^{-b} e^{\theta x} d F_{n}(x)+\int_{a}^{\infty} e^{\theta x} d F_{n}(x)\right) \ &=\sum_{n=1}^{\infty} z^{n}\left(\int_{-\infty}^{\infty}-\int_{-b}^{a}\right) e^{\theta x} d F_{n}(x) \ &=\sum_{n=1}^{\infty} z^{n} \int_{-\infty}^{\infty} e^{\theta x} d F_{n}(x)-F(z, \theta)+1 \end{aligned}
where $F(z, \theta)=\sum_{n=0}^{\infty} z^{n} \int_{-h}^{a} e^{\theta x} d F_{n}(x)$.

## 统计代写|随机过程作业代写stochastic process代考|Fluctuation Theory

In this section $X_{1}, X_{2}, \ldots, X_{n}, \ldots$ are i.i.d. r.v.s.
Theorem $3.3$ If $E\left|X_{i}\right|<\infty$, then \begin{aligned} P[N(b)&<\infty]=1 \text { if } E X_{i} \leq 0 \ &<1 \text { if } E X_{i}>0 \end{aligned}
For Proof see Chung and Fuchs (1951) and Chung and Ornstein (1962), Memoirs of American Math. Society.

Definition $3.2$ If $S$ is uncountable, and $S_{n}=X_{1}+\ldots+X_{n}$ are Markov, $X_{i}$ ‘s being independent, then $x$ is called a possible value of the state space $S$ of the Markoy chain if there exits an $n$ such that
$P\left[\left|S_{n}-x\right|<\delta\right]>0$ for all $\delta>0$. A state $x$ is called recurrent if $P\left[\left|S_{n}-X\right|<\delta\right.$ i.o. $]=1$ i.e. $S_{n} \varepsilon(x-\delta, x+\delta)$ i.o. with probability one.
We shall conclude this section by stating two very important and famous theorems whose proofs are beyond the scope of this book.
Theorem 3.4 (Chung and Fuchs)
Either every state is recurrent or no state is recurrent. (ref. Spitzer-Random Walk (1962)).
Theorem $3.5$ (Chung and Ornstein)
If $E\left|X_{i}\right|<\infty$, then recurrent values exist iff $E\left(X_{i}\right)=0$.

## 统计代写|随机过程作业代写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
$$(q / p)^{k} \frac{\lambda_{1}^{u-k}(s)-\lambda_{2}^{a-k}(s)}{\lambda_{1}^{a}(s)-\lambda_{2}^{a}(s)}$$

$$\lambda_{1}(s)=\frac{1+\left(1-4 p q s^{2}\right)^{1 / 2}}{2 p s}, \lambda_{2}(s)=\frac{1-\left(1-4 p q s^{2}\right)^{1 / 2}}{2 p s} .$$
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
$$F_{j a}(s)=\frac{\left[\lambda_{1}(s)\right]^{j+b}-\left[\lambda_{2}(s)\right]^{j+b}}{\left[\lambda_{1}(s)\right]^{a+b}-\left[\lambda_{2}(s)\right]^{a+b}}$$
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|随机过程作业代写stochastic process代考|Wald’s fundamental identity

Wald’s Sequential Analysis 提出了所谓的 Wald 标识

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