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

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• (Generalized) Linear Models 广义线性模型
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• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

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

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 $X_{1}$ )

$=P\left[S_{n} \leq t, S_{n}+X_{n+1}>t\right]$
$=P\left[t-X_{n+1}<S_{n} \leq t\right]$
$=\int_{0}^{\infty} P\left[t-X_{n+1}<S_{n} \leq t \mid u<X_{n+1} \leq u+d u\right] d F(u)$
$=\int_{0}^{\infty} P\left[t-u<S_{n} \leq t \mid u<X_{n+1} \leq u+d u\right] d F(u)$
$=\int_{0}^{\infty} P\left[t-u<S_{n} \leq t\right] d F(u)\left(\right.$ since $S_{n}$ is independent of $\left.X_{n+1}\right)$
$=\int_{0}^{\infty}\left{F^{(n)}(t)-F^{(n)}(t-u)\right} d F(u)$
$=\int_{0}^{\infty} F^{(n)}(t) d F(u)-\int_{0}^{\infty} F^{(n)}(t-u) d F(u)$
$=F^{(n)}(t)-\int_{0}^{t} F^{(n)}(t-u) d F(u)$

## 统计代写|随机过程作业代写stochastic process代考|Renewal Equation

Theorem 4.1
(a) $P[N(t)=n]=F^{(n)}(t)-F^{(n+1)}(t)$
(b) $H(t)=\sum_{n=1}^{\infty} F^{(n)}(t)$
(c) $H(t)=F(t)+\int_{0}^{t} H(t-u) d F(u)$, the so-called integral equation of Renewal Theory (Renewal equation).
(d) ${N(t), t \in[0, \infty)}$ is completely determined by $H(t)$.
$\operatorname{Proof}(\mathrm{b})$
\begin{aligned} H(t) &=\sum_{n=0}^{\infty} n P[N(t)=n] \ &=P[N(t)=1]+2 P[N(t)=2]+\ldots \ &=F^{(1)}(t)-F^{(2)}(t)+2 F^{(2)}(t)-2 F^{(3)}(t)+\ldots \ &=F^{(1)}(t)+F^{(2)}(t)+F^{(3)}(t)+\ldots \end{aligned}
$=\sum_{n=1}^{\infty} F^{(n)}(t)$ provided the series is convergent.
(convergence of the series will be proved in Exercise 4.5)
(c) $H(t)=\sum_{n=1}^{\infty} F^{(n)}(t)=F^{(1)}(t)+\sum_{n=2}^{\infty} F^{(n)}(t)$

\begin{aligned} &=F(t)+\sum_{n=1}^{\infty} F^{(n+1)}(t)=F(t)+\sum_{n=1}^{\infty} \int_{0}^{t} F^{(n)}(t-u) d F(u) \ &=F(t)+\int_{0}^{t} \sum_{n=1}^{\infty} F^{(n)}(t-u) d F(u) \text { (by Fubini Theorem) } \ &=F(t)+\int_{0}^{t} H(t-u) d F(u) \end{aligned}
(d) $H(t)=F(t)+\int_{0}^{t} H(t-u) d F(u)=F(t)+H^{} F(u)$ where $$is the convolution operator. Taking Laplace transform on both sides$$
\mathscr{L}(s)=\int_{0}^{\infty} e^{-s t} d H(t)=\mathscr{F}(s)+\mathscr{2}(s) \mathscr{F}(s)
$$or \mathscr{F}(s)=\frac{\mathscr{L}(s)}{1+\mathscr{L}(s)}, where \mathscr{F}(s)=\int_{0}^{\infty} e^{-s t} d F(t) and \mathscr{L}^{\prime}(s)=\frac{\mathscr{F}(s)}{1-G^{\top}(s)}(\operatorname{Re}(s)>0). This shows that H(t) and F(x) can be determined uniquely one from the other, since Laplace transform determines a non-decreasing (specially a d.f.) function uniquely. Hence N(t) is completely determined by H(t). Now N(t)=\max \left{n \mid S_{n} \leq t\right} and E N(t)=\sum_{n=1}^{\infty} F^{(n)}(t) if E N(t)<\infty. The next theorem will prove that all moments of N(t) is finite. ## 统计代写|随机过程作业代写stochastic process代考|Renewal Theorems 1. Elementary Renewal Theorem (Feller 1941)$$
\lim _{t \rightarrow \infty} \frac{H(t)}{t}= \begin{cases}1 / \mu & \text { if } 0<\mu=E(x)<\infty \ 0 & \text { if } \mu=\infty\end{cases}

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