# 统计代写|随机过程代写stochastic process代考|MATH3801

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

Consider an ordinary renewal process generated by the sequence $\left[X_n, n \geq 1\right}$ of interarrival times. Suppose that each time a renewal occurs, a reward is received or a cost is incurred (e.g. for replacement of a failed component). Denote the reward or cost associated with $n$th renewal by $Y_n, n \geq 1$. Now $Y_n$ will usually depend on $X_n$, the duration of the interarrival time between $(n-1)^{\text {th }}$ and $n$th renewal. In such a case we assume that the sequence of pairs $\left[X_n, Y_n, n \geq 1\right}$ are i.i.d. The sequence of pairs of i.i.d. random variables $\left{X_n, Y_n\right}$ is said to generate a Renewal Reward process. Now $Y(t)=\sum_{n=1}^{N(t)} Y_n$ gives the total reward carried (or cost incurred) by time $t$ and ${Y(t), t \geq 0}$ is called the renewal reward process generated by $\left{X_n, Y_n\right}$.
Theorem 4.7 Suppose $E(X)<\infty, E|Y|<\infty$, then
(a) $Y(t) / t \rightarrow E(Y) / E(X)$ as $t \rightarrow \infty$ a.s.
(b) $E(Y(t) / t) \rightarrow E(Y) / E(X)$ as $t \rightarrow \infty$.

(a) $\frac{Y(t)}{t}=\frac{Y(t)}{N(t)} \cdot \frac{N(t)}{t}=\frac{\sum_{n=1}^{N(t)} Y_n}{N(t)} \cdot \frac{N(t)}{t}$
By Kolmogorov’s SILN for i.i.d. r.v.s $S_n / n \rightarrow E(X)$ a.s. as $n \rightarrow \infty$.
Let $M(t)=\inf \left{n: S_n>t\right}=N(t)+1$.
Since $N(t) / t$ or $\frac{M(t)}{t}=(N(t)+1) / t \rightarrow 1 / E(X)>0$ a.s. as $t \rightarrow \infty$ $M(t)$ or $N(t) \rightarrow \infty$ as $t \rightarrow \infty$ if $0<E(X)<\infty$.
Let $S_n^=\sum_{i=1}^n Y_i, A=\left{\omega: S_n^(\omega) / n \rightarrow E(Y)\right}, B={\omega: M(t, \omega) \rightarrow \infty}$ and $C=$ $\left{\omega: S_{M(t)}^*(\omega) / M(t) \rightarrow E(Y)\right}$

## 统计代写|随机过程代写stochastic process代考|Age and Block Replacement Policies

The usual replacement policy implies replacement of a component as and when it fails, by a similar new one. The most important replacement policies general use are age and block replacement policies.

Definition 4.5 In age replacement policy a unit is replaced upon failure or at age $T$, which ever comes first.

Definition 4.6 In a block replacement policy the unit in operation is replaced upon failure and at times $T, 2 T, 3 T, \ldots$

The age replacement policy is administratively more difficult to implement, as it requires the age of the unit to be recorded. Whereas block replacement policy, though simpler to administer as the age of the unit need not be recorded, leads to more frequent replacements of relatively new items. The word “block” is used to imply that in practice a block or set of components will be replaced at times $T, 2 T, 3 T, \ldots$ regardless of which have failed. Suppose that the successive life times of the units are r.v.s with a common d.f. $F(x)$ having mean $\mu$.
$\mathrm{T}$ is called the replacement interval under these policies.
Let $N(t)=$ Number of renewals in $[0, t]$ for ordinary renewal process,
$N_A(t, T)=$ Number of renewals in $[0, t]$ under age replacement policy,
$N_D(t, T)=$ Number of renewals in $[0, t]$ under block replacement policy,
Denote by
$H(t)=E N(t), H_A(t, T)=E N_A(t, T), H_D(t, T)=E N_B(t, T)$ the corresponding renewal functions.

# 随机过程代考

## 统计代写|随机过程代写stochastic process代考|Age and Block Replacement Policies

$\mathrm{T}$ 在这些政策下称为更换间隔。

$N_A(t, T)=$ 续订次数 $[0, t]$ 根据年龄更替政策，
$N_D(t, T)=$ 续订次数 $[0, t]$ 根据块替换策略，

$H(t)=E N(t), H_A(t, T)=E N_A(t, T), H_D(t, T)=E N_B(t, T)$ 相应的更新功能。

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