# 数学代写|概率论代写Probability theory代考|MATHS7103

## 数学代写|概率论代写Probability theory代考|UNIFORM DISTRIBUTION

The $\operatorname{pdf} \mathrm{f}(\mathrm{x})$ of the uniform distribution is given below.
$$\mathrm{f}(\mathrm{x})=\frac{1}{b-a},-\infty<a \leq x \leq b<\infty$$
The uniform distribution with the above pdf is denoted by $\mathrm{U}(\mathrm{a}, \mathrm{b})$. The pdf of $\mathrm{U}(0,5)$ is given in Figure 4.18.
Mean $\frac{a+b}{2}$
Variance $\frac{(b-a)^2}{1}$
$$E\left(X^{\mathrm{n}}\right) \frac{1}{n+1} \sum_{i=0}^n a^i b^{n-i}$$

Moment generating function $\mathrm{M}(\mathrm{t})=\frac{e^{b t}-e^{a t}}{t(b-a)}$
Characteristic function $\varphi(t)=\frac{e^{i b t}-e i^{a t}}{i t(b-a)}$

## 数学代写|概率论代写Probability theory代考|DEFINITIONS AND EXAMPLES

Up to these times we basically dealt with random objects representing independent random variables. For example, for working with the sequence of independent variables $\xi_1, \xi_2, \ldots$ it is often enough to know the set of the distribution functions or the corresponding characteristic functions. Our reader remembers that in this case work with sums of independent summands is not essentially complicated: it is enough to remember that this sum corresponds to a product of characteristic functions, and if a product of such independent variables is given, then we can remember that the expected value of such a product is expressed simply in terms of products of individual mean values (if they exist) of the random factors. It becomes essentially more unpleasant to deal with dependent random variables, even if this dependence is not so complicated. In this case, for example, already for finding the dispersion of a sum of several dependent summands not only variances of the individual summands should be known but also all the mixed moments as covariances of these random variables. So, prima facie the situation becomes rather wonderful when dealing with a set of independent variables we consciously go to regarding some random variable transformations leading to wittingly dependent random objects. Order statistics ärc an cxample of such objcets.

1) Let there be a batch of some details (for example, incandescent light bulbs), that must satisfy some standards (for example, the mean time of work should be not less than some value $t$ ). The question is dealt about dealing this batch from a seller to a customer. How can the seller persuade the customer in the compliance of the production to the standard? The simplest thing is to display all the details to the test bench, to get all the data amount their longevity, to count the average longevity and to say that this concrete batch that became broken satisfied all the necessary quantities and qualities. Here two problems arise. One problem: there is nothing to be bought. Another problem: the divergence of the lifetimes of the bulbs can be very great and the customer may not expect the end of the experiment. What is to be done in this situation? First, it is known that methods of mathematical statistics allow us to test only some representatives of the production, such as the number $n$. We can obtain the exact lifetimes $t_l, \ldots, t_n$ of these $n$ details and to use the corresponding statistical methods to verify the hypothesis about the needed quality of all the production. However, a “temporary” problem also arises here. Some of the lamps may soon be broken, while others may be very longevously. But the customer already has booked the means of delivering production into his enterprise. It is good that some bulbs (but not all, for example, $m$ of $n$ bulbs) have already burned out. Let $t_{l, n}<t_{2, n}<\ldots<t_{m, n}$ be the lifetimes of the bulbs.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|UNIFORM DISTRIBUTION

$$\mathrm{f}(\mathrm{x})=\frac{1}{b-a},-\infty<a \leq x \leq b<\infty$$

$$E\left(X^{\mathrm{n}}\right) \frac{1}{n+1} \sum_{i=0}^n a^i b^{n-i}$$

## 数学代写|概率论代写Probability theory代考|DEFINITIONS AND EXAMPLES

1）假设有一批一些细节（例如白炽灯泡），必须满足一些标准（例如，平均工作时间应该不小于某个值吨）。问题是关于将这批从卖方到客户的交易。卖方如何说服客户生产符合标准？最简单的事情是在测试台上显示所有细节，获取所有数据来衡量它们的寿命，计算平均寿命，然后说这批破碎的混凝土满足了所有必要的数量和质量。这里出现了两个问题。一个问题：没有什么可买的。另一个问题：灯泡寿命的差异可能非常大，客户可能不会期望实验结束。在这种情况下该怎么办？首先，众所周知，数理统计方法只允许我们测试一些代表产品，例如数字n. 我们可以得到准确的生命周期吨l,…,吨n这些n详细信息并使用相应的统计方法来验证有关所有产品所需质量的假设。然而，这里也出现了一个“暂时”的问题。有些灯可能很快就坏了，而另一些可能会很长。但是客户已经预订了将生产交付到他的企业的方式。一些灯泡（但不是全部，例如，米的n灯泡）已经烧坏了。让吨l,n<吨2,n<…<吨米,n是灯泡的寿命。

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: