## 统计代写|随机过程代写stochastic process代考|Renewal Theorems

1. Elementary Renewal Theorem (Feller 1941)
$$\lim _{t \rightarrow \infty} \frac{H(t)}{t}= \begin{cases}1 / \mu & \text { if } 0<\mu-E(x)<\infty \ 0 & \text { if } \mu=\infty\end{cases}$$
2. Blackwell’s Renewal Theorem (1948)
$\lim _{t \rightarrow \infty} \frac{H(t+h)-H(t)}{h}=1 / \mu$ for fixed $h>0$ and $X_1$ is a continuous random variable. period of the lattice.
Definition 4.2 A random variable $X$ is said to have lattice distribution if $P[X=c+n d]>0$ where $c$ and $d(>0)$ are real constants and $n=\pm 1, \pm 2, \ldots$.
3. Key Renewal Theorem (W.L. Smith, 1953)
4. If $Q(t) \geq 0$ and non increasing and $\int_0^{\infty} Q(t) d t<\infty$ then $\lim {t \rightarrow \infty} \int_0^t Q(t-x) d H(x)=\frac{1}{\mu} \int_0^{\infty} Q(t) d t$ whenever $X$ is not arithmatic. A weaker version of the Elementary Renewal Theorem is the following due to J.L. Doob (1948) and is known as Doob’s Renewal Theorem $$\lim {t \rightarrow \infty} \frac{N(t)}{t}=\frac{1}{\mu} \text { a.s. }$$ Proof Assume that $X_1, X_2, \ldots$ and i.i.d. with $0<\mu<\infty$. and for all $0<\varepsilon<\mu,(\mu-\varepsilon) n \leq S_n \leq(\mu+\varepsilon) n$ for all large $n$. $$n t1. Replace t by n t to get 5.$$
6. n t \geq S_{N(t)} \geq(\mu-\varepsilon) N(t n) \text { for all large } n \text {. }
7. $$## 统计代写|随机过程代写stochastic process代考|Delayed and Equilibrium Renewal Processes There are two types of generalizations of ordinary Renewal Process. When the first interarrival time X_1 has a distribution function G different from the common d.f. F of X_2, X_3, \ldots, we call such a Renewal process a Modified or Delayed Renewal Process. A Modified Renewal Process in which$$
G(x)=\int_0^x \frac{1-F(t)}{\mu} d t,
$$is called the Equilibrium (Stationary) Renewal Process. Let H_D(t)=\sum_{n=1}^{\infty} G^* F^{(n-1)}(t) be the Renewal function of delayed Renewal process. Earlier we have seen for ordinary Renewal process$$
\frac{H(t)}{t} \rightarrow \frac{1}{\mu} \text { as } t \rightarrow \infty \text {, where } \mu=\int_0^{\infty} x d F(x) .
$$Similarly, \frac{H_D(t)}{t} \rightarrow \frac{1}{\mu} as t \rightarrow \infty (by Strong Law). Hence, the question is whether there is any Delayed Renewal Process for which H_D(t)=t / \mu ? Also we know that for Poisson process, where the underlying variables are exponential, H(t)=t / \mu. Let$$
F_e(x)=\int_0^x \frac{1-F(t)}{\mu} d t
$$The following proposition is an answer in affirmative to our last question. Proposition 4.1 There exists a modified Renewal process where the initial distribution G=F_e exists such that H_e(t)=t / \mu. Proof If H_e(t=t / \mu, then Laplace transform$$
H_e^(s)=\int_0^{\infty} e^{-s t}\left(\frac{t}{\mu}\right) d t=\frac{1}{\mu s^2} $$Also$$ H_e^(s)=\frac{f_e^(s)}{s\left[1-f^(s)\right]}
$$where f_e^(s) is the Laplace Transform of f_e and f^(s) is the Laplace Transform of f where F^{\prime}(x)=f(x). # 随机过程代考 ## 统计代写|随机过程代写stochastic process代考|Renewal Theorems 1. 基本更新定理 (Feller 1941)$$
\lim _{t \rightarrow \infty} \frac{H(t)}{t}={1 / \mu \quad \text { if } 0<\mu-E(x)<\infty 0 \quad \text { if } \mu=\infty
$$2. 布莱克威尔的更新定理 (1948) \lim _{t \rightarrow \infty} \frac{H(t+h)-H(t)}{h}=1 / \mu 对于固定 h>0 和 X_1 是一个连续随机变量。格的周期。 定义 4.2 随机变量 X 据说有格子分布，如果 P[X=c+n d]>0 在哪里 c 和 d(>0) 是实常数和 n=\pm 1, \pm 2, \ldots 3. 关键更新定理 (WL Smith, 1953) 4. 如果 Q(t) \geq 0 和不增加和 \int_0^{\infty} Q(t) d t<\infty 然后 \lim t \rightarrow \infty \int_0^t Q(t-x) d H(x)=\frac{1}{\mu} \int_0^{\infty} Q(t) d t 每 当 X 不是算术的。由于儿 Doob (1948)，基本更新定理的较弱版本如下，被称为 Doob 更新定理$$
\lim t \rightarrow \infty \frac{N(t)}{t}=\frac{1}{\mu} \text { a.s. }
$$证明假设 X_1, X_2, \ldots 和 \mathrm{iid} 0<\mu<\infty. 并为所有人 0<\varepsilon<\mu,(\mu-\varepsilon) n \leq S_n \leq(\mu+\varepsilon) n 对于所 有大 n. \ \ \mathrm{nt} 1. Replace 吨bynt \$$ 得到
$5 . \$ \$$5. nt \geq S_{ N(t)} \backslash g e q(\backslash m u-\backslash varepsilon) N( tn ) \backslash text { 对于所有大 } n \backslash text {0}} 6. \ \$$

## 统计代写|随机过程代写stochastic process代考|Delayed and Equilibrium Renewal Processes

$$G(x)=\int_0^x \frac{1-F(t)}{\mu} d t,$$

$$F_e(x)=\int_0^x \frac{1-F(t)}{\mu} d t$$

$$\left.H_e^{(} s\right)=\int_0^{\infty} e^{-s t}\left(\frac{t}{\mu}\right) d t=\frac{1}{\mu s^2}$$

$$\left.H_e^{(} s\right)=\frac{\left.f_e^{(} s\right)}{\left.s\left[1-f^{(} s\right)\right]}$$

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