### 数学代写|随机过程统计代写Stochastic process statistics代考|STAT7004

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• Statistical Inference 统计推断
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• Longitudinal Data Analysis 纵向数据分析
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## 数学代写|随机过程统计代写Stochastic process statistics代考|Waiting time for a gain

Let $\left{X_{i}\right}$ be sequence of i.i.d r.v.s with common distribution $P\left(X_{k}=1\right)=p, p\left(X_{k}=-1\right)=q, p+q=1$ and $S_{n}=X_{1}+\ldots+X_{n}, S_{0}=0$
In gambling terminology $S_{n}, n \geq 1$ is the Peter’s accumulated gain at the end of the $n^{\text {th }}$ trial if Peter and Paul play for unit stakes. Now consider the event $A_{n}=\left{S_{1} \leq 0, S_{2} \leq 0, \ldots, S_{n-1} \leq 0, S_{n}=1\right}$.

Thus, the $n^{\text {th }}$ trial is the first to render Peter’s accumulated gain positive. The event $A_{n}$ is called first visit to $+1$ or the index $n$ is the passage time through $l$ in random walk terminology.
Let $\phi_{n}=P\left(A_{n}\right)$. Define $\phi_{0}=0, \phi_{1}=p$.
If the event holds for $n>1$, then $S_{1}=-1$ and there exists a smallest integer $v$ $<n$ such that $S_{v}=0$. The outcome of the first $n$ trials may be described as follows: (1) At the first trial Peter looses an unit amount. (2) It takes $v-1$ further trials for Peter to reestablish the initial situation. (3) It takes exactly $n-v$ further trials for Peter to attain a positive net gain. These events depend on non-overlapping blocks of trials and are therefore mutually independent, $(2)$ and (3) have probabilities $\phi_{t-1}$ and $\phi_{n-v}$.

Now the event $A_{n}$ occurs iff the events (1) to (3) occcur for some $v<n$. Summing over all possible $1 \leq v \leq n-1$, we get
$$\phi_{n}=q\left(\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}\right)=q \sum_{v=2}^{n-1} \phi_{v-1} \phi_{n-v}$$
Multiplying both sides by $s^{n}$ and summing $n=2,3, \ldots$, we get
\begin{aligned} \sum_{n=2}^{\infty} \phi_{n} s^{n} &=q s\left{\sum_{n=2}^{\infty} \phi_{n-1} s^{n-1}\right}^{2} \ &=q s\left{\sum_{n=0}^{\infty} \phi_{n} s^{n}\right}^{2}=q s \Phi^{2}(s) \end{aligned}
$\left(\right.$ since $\left.\phi_{0}=0\right)$
Also
$$\sum_{n=2}^{\infty} \phi_{n} s^{n}=\Phi(s)-\phi_{0}-\phi_{1} s=\Phi(s)-p s$$
where $\Phi(s)$ is the G.F. of $\left{\phi_{n}\right}$.
Equation (3.3) follows from the fact that
$$\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}=(n-1) \text { th term of }\left{\phi_{n}\right}^{*}\left{\phi_{n}\right} .$$
and
\begin{aligned} & \sum_{n=2}^{\infty} q\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}\right} s^{n} \ =& q s \sum_{n=2}^{\infty}\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}\right} s^{n-1} \end{aligned}
Hence from (3.3) and (3.4), $\Phi(s)-p s=q s \Phi^{2}(s)$.

## 数学代写|随机过程统计代写Stochastic process statistics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.
$$f_{2 n}^{+}=P\left(B_{n} \cap\left(X_{1}=1\right)\right) \text { and } f_{2 n}^{-}=P\left(B_{n} \cap\left(X_{1}=-1\right)\right) \text {. }$$
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}
Hence probability that a return to equilibrium occurs sooner or later will be \begin{aligned} F(1) &=\sum_{n=1}^{\infty} f_{2 n} \ &=1-(1-4 p q)^{\frac{1}{2}}=1-|p-q| \ &= \begin{cases}2 q & \text { if } p>q \ 2 p & \text { if } p<q \ 1 & \text { if } p=q=\frac{1}{2}\end{cases} \end{aligned}
Hence, if $p=q=\frac{1}{2}$ a return to equilibrium is certain.

## 数学代写|随机过程统计代写Stochastic process statistics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.
\text { Let } \begin{aligned} 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} \end{aligned}

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 $n$,
$$\sum_{n=m}^{\infty} n^{k} P[N=n] \leq \sum_{n=m}^{\infty} e^{i n} P[N \geq n] \leq A \sum_{n=m}^{\infty}\left(\delta e^{t}\right)^{n}<\infty \text { if } \delta e^{t}<1$$
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}^{\infty} 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 $S D\left(X_{1}, \ldots, X_{n-1}\right)\left(X_{1}, \ldots, X_{n-1}\right.$, need not be i.i.d. r.v.s).

## 数学代写|随机过程统计代写Stochastic process statistics代考|Waiting time for a gain

φn=q(φ1φn−2+…+φn−2φ1)=q∑在=2n−1φ在−1φn−在

\begin{对齐} \sum_{n=2}^{\infty} \phi_{n} s^{n} &=q s\left{\sum_{n=2}^{\infty} \phi_{n- 1} s^{n-1}\right}^{2} \ &=q s\left{\sum_{n=0}^{\infty} \phi_{n} s^{n}\right}^{ 2}=q s \Phi^{2}(s) \end{对齐}\begin{对齐} \sum_{n=2}^{\infty} \phi_{n} s^{n} &=q s\left{\sum_{n=2}^{\infty} \phi_{n- 1} s^{n-1}\right}^{2} \ &=q s\left{\sum_{n=0}^{\infty} \phi_{n} s^{n}\right}^{ 2}=q s \Phi^{2}(s) \end{对齐}
(自从φ0=0)

∑n=2∞φnsn=披(s)−φ0−φ1s=披(s)−ps

\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}=(n-1) \text { }\left{\phi_{n}\right }^{*}\left{\phi_{n}\right} 。\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}=(n-1) \text { }\left{\phi_{n}\right }^{*}\left{\phi_{n}\right} 。

\begin{aligned} & \sum_{n=2}^{\infty} q\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}\右} s^{n} \ =& q s \sum_{n=2}^{\infty}\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_ {1}\right} s^{n-1} \end{对齐}\begin{aligned} & \sum_{n=2}^{\infty} q\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_{1}\右} s^{n} \ =& q s \sum_{n=2}^{\infty}\left{\phi_{1} \phi_{n-2}+\ldots+\phi_{n-2} \phi_ {1}\right} s^{n-1} \end{对齐}

## 数学代写|随机过程统计代写Stochastic process statistics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 )

F(1)=∑n=1∞F2n =1−(1−4pq)12=1−|p−q| ={2q 如果 p>q 2p 如果 p<q 1 如果 p=q=12

## 数学代写|随机过程统计代写Stochastic process statistics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(一个,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,一个).

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<∞对所有人ķ=1,2,…证明 (i) 我们将证明，更具体地说，存在一个>0和

0<d<1 独立于 n 和 磷[ñ≥n]≤一个dn

让 小号1=X1+…+Xr,小号2=Xr+1+Xr+2+…+X2r,…, 小号ķ∗=X(ķ−1)r+1+…+Xķr

∑n=米∞nķ磷[ñ=n]≤∑n=米∞和一世n磷[ñ≥n]≤一个∑n=米∞(d和吨)n<∞ 如果 d和吨<1

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