### 统计代写|统计推断代写Statistical inference代考|MAST90100

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

## 统计代写|统计推断代写Statistical inference代考|A more thorough treatment of random variables

In earlier sections of this chapter we refer rather vaguely to conditions on a set $B$ for $\mathrm{P}(X \in B)$ to be well defined and conditions on a function $g$ for $g(X)$ to be a random variable. We also suggest that we are not really interested in random variables as maps and that, for many situations, the notion of an underlying sample space is not particularly useful. In this section, we attempt to provide some justification for these assertions. The material here is technical and may be excluded without affecting understanding of other parts of the text. We start by providing an alternative definition of a random variable. This is equivalent to Definition 3.1.2 but uses more abstract concepts; key among them is the Borel $\sigma$-algebra.
Definition 3.8.1 (Borel $\sigma$-algebra on $\mathbb{R}$ )
Let $C$ be the collection of all open intervals of $\mathbb{R}$. The Borel $\sigma$-algebra on $\mathbb{R}$ is the (unique) smallest $\sigma$-algebra that contains $C$. We denote the Borel $\sigma$-algebra on $\mathbb{R}$ by $\mathcal{B}$. An element of $\mathcal{B}$ is referred to as a Borel set.

From the definition it is clear that any open interval $(x, y)$ is a Borel set. It is also the case that closed intervals $[x, y]$, half-open intervals $(x, y]$ and $[x, y)$, and finite unions of interval are all Borel sets in $\mathbb{R}$. In fact, sets that are not Borel sets are hard to construct; any subset of $\mathbb{R}$ that you come across in a practical problem is likely to be a Borel set. Clearly, since $\mathcal{B}$ is a $\sigma$-algebra, $(\mathbb{R}, \mathcal{B})$ is a measurable space. The term measurable can also be applied to functions.
Definition 3.8.2 (Measurable function)
Consider measurable spaces $(\Omega, \mathcal{F})$ and $(\mathbb{R}, \mathcal{B})$. We say that a function $h: \Omega \rightarrow \mathbb{R}$ is $\mathcal{F}$-measurable if $h^{-1}(B) \in \mathcal{F}$ for all $B \in \mathcal{B}$.
We can now give an alternative definition of a random variable.

## 统计代写|统计推断代写Statistical inference代考|Further exercises

1. Let $Y$ be a random variable that has a binomial distribution with $n$ trials and probability $p$ of success for each trial, that is, $Y \sim \operatorname{Bin}(n, p)$. Without using generating functions:
(a) show that $\mathbb{E}(Y)=n p$,
(b) work out $\operatorname{Var}(Y)$,
(c) explain why $\mathbb{B}\left(Y^{r}\right)=p$ for $r=1,2, \ldots$,
(d) find the third central moment of $Y$.
2. Let $Y$ be a random variable that has a Poisson distribution with parameter $\lambda$, that is, $Y \sim \operatorname{Pois}(\lambda)$. Without using generating functions:
(a) show that $\mathbb{E}(Y)=\lambda$,
(b) find $\mathbb{E}\left(Y^{3}\right)$.
3. Let $Y$ be a random variable that has an exponential distribution with parameter $\theta$. Without using generating functions show that

(a) $\mathbb{E}(Y)=\frac{1}{\theta}$,
(b) $\operatorname{Var}(Y)=\frac{1}{\theta^{2}}$.

1. Find the cumulative distribution functions corresponding to the following density functions:
(a) Cauchy: $f_{X}(x)=1 /\left[\pi\left(1+x^{2}\right)\right]$ for $x \in \mathbb{R}$
(b) Logistic: $f_{X}(x)=e^{-x} /\left(1+e^{-x}\right)^{2}$ for $x \in \mathbb{R}$
(c) Pareto: $f_{X}(x)=(a-1) /(1+x)^{a}$ for $x>0$, where $a>1$
(d) Weibull: $f_{X}(x)=c \tau x^{\tau-1} e^{-c x^{\tau}}$ for $x>0$, where $c, \tau>0$
2. If $X$ is a positive continuous random variable with density function $f_{X}(x)$ and mean $\mu$, show that
$$g(y)= \begin{cases}y f_{X}(y) / \mu & y \geq 0 \ 0 & y<0\end{cases}$$
is a valid density function, and hence show that
$$\mathbb{E}\left(X^{3}\right) \mathbb{E}(X) \geq\left{\mathbb{E}\left(X^{2}\right)\right}^{2}$$

## 统计代写|统计推断代写Statistical inference代考|Joint and marginal distributions

The cumulative distribution function for a collection of random variables is referred to as the joint cumulative distribution function. This is a function of several variables.
Definition 4.1.1 (General joint cumulative distribution function)
If $X_{1}, \ldots, X_{n}$ are random variables, the joint cumulative distribution function is a function $F_{X_{1}, \ldots, X_{r}}: \mathbb{R}^{n} \rightarrow[0,1]$ given by
$$F_{X_{1}, \ldots, X_{n}}\left(x_{1}, x_{2}, \ldots, x_{n}\right)=\mathrm{P}\left(X_{1} \leq x_{1}, X_{2} \leq x_{2}, \ldots, X_{n} \leq x_{n}\right)$$
The notation associated with the general case of $n$ variables rapidly becomes rather

cumbersome. Most of the ideas associated with multivariate distributions are entirely explained by looking at the two-dimensional case, that is, the bivariate distribution. The generalisations to $n$ dimensions are usually obvious algebraically, although $n$ dimensional distributions are considerably more difficult to visualise. The definition of a bivariate cumulative distribution function is an immediate consequence of Definition 4.1.1.
Definition 4.1.2 (Bivariate joint cumulative distribution function)
For two random variables $X$ and $Y$, the joint cumulative distribution function is a function $F_{X, Y}: \mathbb{R}^{2} \rightarrow[0,1]$ given by
$$F_{X, Y}(x, y)=\mathrm{P}(X \leq x, Y \leq y) .$$
Notice that there is an implicit $\cap$ in the statement $\mathrm{P}(X \leq x, Y \leq y)$, so $F_{X, Y}(x, y)$ should be interpreted as the probability that $X \leq x$ and $Y \leq y$. The elementary properties of bivariate distributions are given by Claim $4.1 .3$ below. Part of Exercise $4.1$ is to generalise these to the $n$-dimensional case.

## 统计代写|统计推断代写Statistical inference代考|Further exercises

1. 让是是一个具有二项分布的随机变量n试验和概率p每次试验的成功率，即是∼垃圾桶⁡(n,p). 不使用生成函数：
（a）表明和(是)=np,
(b) 制定曾是⁡(是),
(c) 解释原因乙(是r)=p为了r=1,2,…,
(d) 求第三个中心矩是.
2. 让是是具有参数的泊松分布的随机变量λ， 那是，是∼然后⁡(λ). 不使用生成函数：
（a）表明和(是)=λ,
(b) 找到和(是3).
3. 让是是具有参数的指数分布的随机变量θ. 不使用生成函数表明

（一个）和(是)=1θ,
(b)曾是⁡(是)=1θ2.

1. 求与下列密度函数对应的累积分布函数：
(a) Cauchy：FX(X)=1/[圆周率(1+X2)]为了X∈R
(b) 后勤：FX(X)=和−X/(1+和−X)2为了X∈R
(c) 帕累托：FX(X)=(一个−1)/(1+X)一个为了X>0， 在哪里一个>1
(d) 威布尔：FX(X)=CτXτ−1和−CXτ为了X>0， 在哪里C,τ>0
2. 如果X是具有密度函数的正连续随机变量FX(X)和意思μ， 显示
G(是)={是FX(是)/μ是≥0 0是<0
是一个有效的密度函数，因此表明
\mathbb{E}\left(X^{3}\right) \mathbb{E}(​​X) \geq\left{\mathbb{E}\left(X^{2}\right)\right}^{ 2}\mathbb{E}\left(X^{3}\right) \mathbb{E}(​​X) \geq\left{\mathbb{E}\left(X^{2}\right)\right}^{ 2}

## 统计代写|统计推断代写Statistical inference代考|Joint and marginal distributions

FX1,…,Xn(X1,X2,…,Xn)=磷(X1≤X1,X2≤X2,…,Xn≤Xn)

FX,是(X,是)=磷(X≤X,是≤是).

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## MATLAB代写

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