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# 数学代写|随机过程Stochastic Porcesses代考|STAT507 Renewal Theory

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## 数学代写|随机过程Stochastic Porcesses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
$$F^{(0)}(t)=\left{\begin{array}{l} 0 \text { if } t<0 \\ 1 \text { if } t \geq 0 . \end{array}\right.$$ Now \begin{aligned} 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] \text { (by nonnegativeness of } X_1 \text { ) } \end{aligned}

## 数学代写|随机过程Stochastic Porcesses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 $H(t)$.
Proof (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 \ & =\sum_{n=1}^{\infty} F^{(n)}(t) \text { provided the series is convergent. } \end{aligned}
(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
\begin{aligned} \mathscr{X}(s) & =\int_0^{\infty} e^{-s t} d H(t)=\mathscr{F}(s)+\mathscr{X}(s) \mathscr{F}(s) \ \mathscr{F}(s) & =\frac{\mathscr{X}(s)}{1+\mathscr{X}(s)}, \text { where } \mathscr{F}(s)=\int_0^{\infty} e^{-s t} d F(t) \end{aligned}
and $\mathscr{L}(s)=\frac{\mathscr{F}(s)}{1-\mathscr{F}(s)}(\operatorname{Re}(s)>0)$. 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.

## 数学代写|随机过程Stochastic Porcesses代考|Renewal Equation

(一个) $P[N(t)=n]=F^{(n)}(t)-F^{(n+1)}(t)$
(二) $H(t)=\sum_{n=1}^{\infty} F^{(n)}(t)$
(C) $H(t)=F(t)+\int_0^t H(t-u) d F(u)$ ，所閆积分方程电新理论 (Renewal equation) 。
（四) $N(t), t \in[0, \infty)$ 完全由 $H(t)$.

$$H(t)=\sum_{n=0}^{\infty} n P[N(t)=n] \quad=P[N(t)=1]+2 P[N(t)=2]+\ldots=F^{(1)}(t)-F^{(2)}(t)+2 F^{(2)}(t)-2 F^{(3)}(t)+\ldots \quad=F^{(1)}(t)+F^{(2)}(t)+F^{(3)}($$
(级数收敛将在练习 $4.5$ 中证明)
\begin{aligned} & \text { (c) } H(t)=\sum_{n=1}^{\infty} F^{(n)}(t)=F^{(1)}(t)+\sum_{n=2}^{\infty} F^{(n)}(t) \ & =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) \quad=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( \end{aligned}
(四) $H(t)=F(t)+\int_0^t H(t-u) d F(u)=F(t)+H^* F(u)$ 在郆哩 $*$ 是卷积算子。

$$\mathrm{X}(s)=\int_0^{\infty} e^{-s t} d H(t)=\mathrm{F}(s)+\mathrm{X}(s) \mathrm{F}(s) \mathrm{F}(s) \quad=\frac{\mathrm{X}(s)}{1+\mathrm{X}(s)} \text {, where } \mathrm{F}(s)=\int_0^{\infty} e^{-s t} d F(t)$$

df) 函数。 因此 $N(t)$ 完全由 $H(t)$.

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