### 统计代写|随机过程作业代写stochastic process代考| Returns to equilibrium

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## 统计代写|随机过程作业代写stochastic process代考|Returns to equilibrium

Let $A_{k}$ be the event of equalization of the accumulated number of successes and failures occurs at the $k$ th trial if $S_{k}=0$. Let $u_{k}=P\left(S_{k}=0\right)$. The number of trials is necessarily even and the probability of a return to the origin at the $2 n$th trial is given by
$$U_{2 n}=\left(\begin{array}{c} 2 n \ n \end{array}\right) p^{n} q^{n}=(-1)^{n}\left(\begin{array}{c} -\frac{1}{2} \ n \end{array}\right)(4 p q)^{n}$$
The G.F. of $\left{U_{2 n}\right}$ is $U(s)=\sum_{n=0}^{\infty} U_{2 n} s^{2 n}$
$$=\sum_{n=0}^{\infty}(-1)^{n}\left(\begin{array}{c} -\frac{1}{2} \ n \end{array}\right)\left(4 p q s^{2}\right)^{n}=\left(1-4 p q s^{2}\right)^{-\frac{1}{2}}$$
$$B_{2 n}=\left[S_{k} \neq 0, \text { for } k=1, \ldots, 2 n-1, S_{2 n}=0\right] .$$
Let $P\left(B_{2 n}\right)=f_{2 n}$.
Consider two sub-events with $X_{1}=1, X_{1}=-1$ and denote their probabilities by $f_{2 n}^{+}$and $f_{2 n}^{-}$, i.e.
$$f_{2 n}^{+}=P\left(B_{n} \cap\left(X_{1}=1\right)\right) \text { and } f_{2 n}^{-}=P\left(B_{n} \cap\left(X_{1}=-1\right)\right) .$$
Now $f_{2 n}^{-}=q \phi_{2 n-1}$ (because first $2 n-2$ partial sums $X_{2}+X_{3}+\ldots+X_{n} \leq 0$, but the next one is positive)
As before let $\phi_{n}=P\left[S_{1} \leq 0, S_{2} \leq 0, \ldots, S_{n}=1\right]$
Then the G.F. of $\left{f_{2 n}^{-}\right}$is
\begin{aligned} F^{-}(s) &=\sum_{n=1}^{\infty} f_{2 n}^{-} s^{2 n}=s q \sum_{n=1}^{\infty} \phi_{2 n-1} s^{2 n-1} \ &=q s \Phi(s)=q s \frac{1-\left(1-4 p q s^{2}\right)^{\frac{1}{2}}}{2 q s} \end{aligned}
By symmetry, $F^{+}(s)=F^{-}(s)$ and hence
\begin{aligned} \sum_{n=1}^{\infty} f_{2 n} s^{2 n} &=F(s)=F^{+}(s)=F^{-}(s)=1-\left(1-4 p q s^{2}\right)^{\frac{1}{2}} \ (&\left.=1-\frac{1}{U(S)} \text { in general }\right) \end{aligned}

## 统计代写|随机过程作业代写stochastic process代考|Sequential Analysis

An important problem arising in Wald’s sequential analysis is concerned with the random variable $N=N(a, b)$, where $N=\min \left{n \mid S_{n} \leq-b\right.$ or $\left.S_{n} \geq a\right}$ is the first exist time from the interval $(-b, a)$.
We ignore the trivial case $P\left(X_{i}=0\right)=1$.
Let $X_{i}$ are i.i.d. r.v.s and $S_{n}=X_{1}+\ldots+X_{n}$.
Theorem $3.1$ (C. Stein 1947)
$N$ is a proper random variable with finite moments of all order, i.e.
(i) $P(N<\infty)=1$ and (ii) $E(N)^{k}<\infty$ for all $k=1,2, \ldots$ Proof (i) We shall show, more specifically that there exists $A>0$ and
$$0<\delta<1 \text { independent of } n \text { and } P[N \geq n] \leq A \delta^{n}$$
Let $C=a+b$ and $r$ be a positive integer.
Let $S_{1}^{}=X_{1}+\ldots+X_{r}, S_{2}^{}=X_{r+1}+X_{r+2}+\ldots+X_{2 r}, \ldots$,
$$S_{k}^{*}=X_{(k-1) r+1}+\ldots+X_{k r}$$

We have, $P[N \geq k r] \leq P\left[\left|S_{1}^{}\right|}\right|}\right|}\right|0$.
If $p=1$, then $E\left(S_{k}^{}\right)^{2}=r E X_{i}^{2}+r(r-1)\left(E X_{i}\right)^{2}$ (since $X_{i}$ ‘s are i.i.d.) Since $E\left(X_{i}^{2}\right)>0, E\left(S_{k}^{}\right)^{2}>C^{2}$ by choosing $r$ large enough. But $p=1 \Rightarrow$ $E\left(S_{k}^{*}\right)^{2} \leq C^{2}$, which is a contradiction. Therefore $p \neq 1$ and $P(N<\infty)=1$. (ii) For $t>0$ and positive integer $k, n^{k}<e^{t n}$ for large $n$,
$$\sum_{n=m}^{\infty} n^{k} P[N=n] \leq \sum_{n=m}^{\infty} e^{t n} P[N \geq n] \leq A \sum_{n=m}^{\infty}\left(\delta e^{t}\right)^{n}<\infty \text { if } \delta e^{t}<1 .$$
Hence
\begin{aligned} E\left(N^{k}\right) &=\sum_{n=1}^{\infty} n^{k} P[N=n] \ &=\sum_{n=1}^{m-1} n^{k} P[N=n]+\sum_{n=m}^{m} n^{k} P[N=n] \end{aligned}
Definition $3.1 \quad N$ is called a stopping rule if $N$ is a non-negative integer-valued random variable and the event $[N \geq n]$ depends on $X_{1}, X_{2}, \ldots X_{n-1}$ only, i.e. $\lfloor N=n]$ is measurable with respect to $5\left(X_{1}, \ldots, X_{n-1}\right)\left(X_{1}, \ldots, X_{n-1}\right.$, need not be i.i.d. r.v.s).

## 统计代写|随机过程作业代写stochastic process代考|Wald’s Equation and Wald’s Identity

Theorem $3.2$ (Wald’s equation) Let $\left{X_{i}\right}$ be a sequence of i.i.d. r.v.s with $E(N)<\infty$. If $E\left|X_{1}\right|<\infty$ then $E\left(S_{N}\right)=\left(E X_{1}\right) E N$.
If moreover, $\sigma^{2}=\operatorname{var}\left(X_{1}\right)<\infty$, then $E\left(S_{N}-N \mu\right)^{2}=\sigma^{2} E(N)$, where $\mu=E\left(X_{1}\right)$.
Proof $E\left(S_{N}\right)=\sum_{n=1}^{\infty} E\left(S_{N} \mid N=n\right) P[N=n]$
$$=\sum_{n=1}^{\infty} \sum_{i=1}^{n} P[N=n] E\left(X_{i} \mid N=n\right)$$
$$=\sum_{i=1}^{\infty} \sum_{n=i}^{\infty} P[N=n] E\left(X_{i} \mid N=n\right)$$
(interchanging the order of summation)
$$\left|\sum_{i=1}^{\infty} \sum_{n=i}^{\infty} E\left(X_{i} \mid N=n\right) P(N=n)\right| \leq \sum_{i=1}^{\infty} \sum_{n=i}^{\infty} E\left(\left|X_{i}\right| \mid N=n\right) P(N=n)$$
$$=E\left|X_{t}\right| E(N)<\infty$$
(Fubini condition is satisfied)
Therefore
\begin{aligned} E\left(S_{N}\right) &=\sum_{i=1}^{\infty} P[N \geq i] E\left(X_{i} \mid N \geq i\right)\left(\text { since } N \geq i \text { depends on } X_{1}, \ldots, X_{i=1}\right. \text { only) }\ &=\sum_{i=1}^{\infty} P[N \geq i] E\left(X_{i}\right)=E\left(X_{i}\right) E(N) \end{aligned}

Let $N_{n}=\min (N, n)$. Now let $N_{n} \rightarrow N$ monotonically, it follows from the Monotone convergence theorem that
$$E N_{n} \rightarrow E(N) \text { as } n \rightarrow \infty$$
Since $\left.\left{\left(S_{n}-n \mu\right)^{2}-n \sigma^{2}, g_{n}\right), n \geq 1\right}$ is a martingale (prove it).
We can apply optional sampling theorem to obtain (see Appendix iv)
$$E\left(S_{N_{n}}-n \mu\right)^{2}=\sigma^{2} E N_{n}$$
Now let $m \geq n$. Since martingales have orthogonal increments we have, by (3.7) and (3.8),
\begin{aligned} E\left(S_{N_{m}}-\mu N_{m}-\right.&\left.\left(S_{N_{n}}-\mu N_{n}\right)\right)^{2}=E\left(S_{N_{m}}-\mu N_{m}\right)^{2}-E\left(S_{N_{n}}-\mu N_{n}\right)^{2} \ =& \sigma^{2}\left(E N_{m}-E N_{n}\right) \rightarrow 0 \text { as } n, m \rightarrow \infty \end{aligned}
that is $S_{N_{n}}-\mu N_{n}$ converges in $L_{2}$ as $n \rightarrow \infty$.
However, since we already know that $S_{N_{n}}-\mu N_{n} \rightarrow S_{N}-\mu N$ as $n \rightarrow \infty$, it follows that
$$E\left(S_{N_{n}}-\mu N_{n}\right)^{2} \rightarrow E\left(S_{N}-\mu N\right)^{2} \text { as } n \rightarrow \infty,$$
which together with (3.7) and (3.8), completes the proof.

## 统计代写|随机过程作业代写stochastic process代考|Returns to equilibrium

=∑n=0∞(−1)n(−12 n)(4pqs2)n=(1−4pqs2)−12

F2n+=磷(乙n∩(X1=1)) 和 F2n−=磷(乙n∩(X1=−1)).

F−(s)=∑n=1∞F2n−s2n=sq∑n=1∞φ2n−1s2n−1 =qs披(s)=qs1−(1−4pqs2)122qs

∑n=1∞F2ns2n=F(s)=F+(s)=F−(s)=1−(1−4pqs2)12 (=1−1在(小号) 一般来说 )

## 统计代写|随机过程作业代写stochastic process代考|Sequential Analysis

Wald 序列分析中出现的一个重要问题与随机变量有关ñ=ñ(一种,b)， 在哪里N=\min \left{n \mid S_{n} \leq-b\right.$或$\left.S_{n} \geq a\right}N=\min \left{n \mid S_{n} \leq-b\right.$或$\left.S_{n} \geq a\right}是间隔中的第一个存在时间(−b,一种).

ñ是具有所有阶的有限矩的适当随机变量，即
(i)磷(ñ<∞)=1(ii)和(ñ)ķ<∞对全部ķ=1,2,…证明 (i) 我们将证明，更具体地说，存在一种>0和
0<d<1 独立于 n 和 磷[ñ≥n]≤一种dn

∑n=米∞nķ磷[ñ=n]≤∑n=米∞和吨n磷[ñ≥n]≤一种∑n=米∞(d和吨)n<∞ 如果 d和吨<1.

## 统计代写|随机过程作业代写stochastic process代考|Wald’s Equation and Wald’s Identity

=∑n=1∞∑一世=1n磷[ñ=n]和(X一世∣ñ=n)
=∑一世=1∞∑n=一世∞磷[ñ=n]和(X一世∣ñ=n)
（交换求和顺序）
|∑一世=1∞∑n=一世∞和(X一世∣ñ=n)磷(ñ=n)|≤∑一世=1∞∑n=一世∞和(|X一世|∣ñ=n)磷(ñ=n)
=和|X吨|和(ñ)<∞
（满足 Fubini 条件）

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