数理统计学复习

格式:\(\newcommand{\dif}{\mathop{}\\!\mathrm{d}}\)
/ 连接的是同一个概念的两个名称
,表示列举
?表示TODO
()表示注解

chap 1

概念

  • 概率空间

A probability space is a triple \((\Omega, \mathcal{F}, P)\) where \(\Omega\) is a set of "outcomes," \(\mathcal{F}\) is a set of "events," and $P\colon \mathcal{F} \to [0,1] $ is a function that assigns probabilities to events. We assume that \(\mathcal{F}\) is a \(\sigma\)-field (or \(\sigma\)-algebra), i.e., a (nonempty) collection of subsets of \(\Omega\) that satisfy

(i) if \(A\in\mathcal{F}\) then \(A^c \in\mathcal{F}\), and
(ii) if \(A_i\in\mathcal{F}\) is a countable sequence of sets then \(\bigcup_iA_i\in\mathcal{F}\).

Here and in what follows, countable means finite or countably infinite. Since \(\bigcap_iA_i = (\bigcup_iA_i^c)^c\), it follows that a \(\sigma\)-field is closed under countable intersections.

Without \(P\), \((\Omega,\mathcal{F})\) is called a measurable space, i.e., it is a space on which we can put a measure. A measure is a nonnegative countably additive set of function; that is, a function \(\mu\colon\mathcal{F}\to\mathbb{R}\) with

(i) \(\mu(A)\ge\mu(\emptyset) = 0\) for all \(A\in\mathcal{F}\), and
(ii) if \(A_i\in\mathcal{F}\) is a countable sequence of disjoint sets, then
\[
\mu(\bigcup_iA_i) = \sum_i\mu(A_i)
\]
If \(\mu(\Omega) = 1\), we call \(\mu\) a probability measure. In this book, probability measures are usually denoted by \(P\).

  • 随机试验/随机现象
  • 样本空间,样本点
  • 随机事件 abbr. 事件
  • 事件 \(A\)\(B\) 相互独立
  • \(n\) 个事件相互独立,两两独立

  • ?随机变量 abbr. RV: \(X\)\(Y\),……

A real valued function \(X\) defined on \(\Omega\) is said to be a random variable if for every Borel set \(B\in\mathbb R\) we have \(X^{-1}(B) = \\{\omega\colon X(\omega)\in B\\}\in\mathcal{F}\). When we need to emphasize the \(\sigma\)-field, we will say that \(X\) is \(\mathcal{F}\)-measurable or write \(X\in\mathcal{F}\).(A Borel set is an element of a Borel sigma-algebra.)

这个定义我还不能完全理解,我不理解 Berel set 究竟是什么。我本科概统教材上给出的随机变量的定义是:

\(\Omega\) 为一个样本空间,若对任意 \(\omega\in\Omega\),都有一个实数 \(X(\omega)\) 与之对应,则称 \(X(\omega)\) 为一个随机变量,并简记为 \(X\)

  • 分布函数/DF \(F(x) := P(X\le x)\quad x\in\mathbb R\)
  • 概率密度函数 abbr. 密度函数/PDF(在下文中,对于连续 RV,“分布”一词一般指概率密度函数)

正态分布:$X\sim N(\mu,\sigma^2)\quad f(x) = \dfrac{1}{\sqrt{2\pi}\sigma} e{-\dfrac{(x-\mu)2}{2\sigma^2}}, x\in\mathbb R, \mu\in\mathbb R, \sigma > 0 $

高斯积分

考虑广义积分

\[A = \int_{-\infty}^{\infty} \dfrac{1}{\sqrt{2\pi}\sigma} e^{-\dfrac{(x-\mu)^2}{2\sigma^2}} \dif x \]

做变量替换,令 $ t = \frac{x - \mu} {\sqrt2\sigma}$,得

\[A = \frac1{\sqrt\pi} \boxed{\color{blue} {\int_{-\infty}^{\infty} e^{-t^2} \dif t} } \]

\(\int_{-\infty}^{\infty} e^{-t^2} \dif t\) 称作高斯积分,也称概率积分。需要证明

\[I = \int_{-\infty}^{\infty} e^{-t^2} \dif t = \sqrt{\pi} \]

为证此式,考虑

\[I^2 = \int_{-\infty}^{\infty} e^{-t^2} \dif t \int_{-\infty}^{\infty} e^{-u^2} \dif u = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(t^2+u^2)}\dif t\dif u \]

转化成极坐标 $ t = r\cos\theta, u = r\sin\theta$,上式转化为

\[I^2 = \int_0^{2\pi}\dif\theta\int_0^\infty e^{-r^2}r\dif r = \pi, \]

明所欲证。

  • 二维随机变量 \((X,Y)\)(可以理解为两个随机变量)
  • \((X,Y)\) 的联合分布函数 \(F(X,Y) := P(X\le x, Y\le y)\)
  • 边际分布函数:\(F_X(x)\)\(F_Y(y)\)
  • 随机变量的函数的分布:\(X\) 的 PDF 为 \(f(x)\)\(Y=g(X)\),求 \(Y\) 的 PDF。

随机变量的数字特征

  • 数学期望 abbr. 期望:\(E(X)\)

和的期望等于期望的和,不论独立不独立。

  • \(k\) 阶原点矩: \(E(X^k)\)
  • \(k\) 阶中心矩:\(E\\{[X-E(X)]^k \\}\)
  • 方差:\(\mathrm{Var}(X) := E\\{[X-E(X)]^2 \\} =\boxed{\color{blue}{ E(X^2) - [E(X)]^2 }}\)(二阶中心矩)
  • 标准差:\(\sigma_X = \sqrt{\mathrm{Var}(X)}\)
  • \(X\)\(Y\) 的协方差:\(\mathrm{Cov}(X,Y) := E\\{[X-E(X)] [Y-E(Y)] \\} = \boxed{E(XY) - E(X)E(Y)}\)

\begin{align*}
\mathrm{Var}(X+Y)& = E[(X+Y)^2] - [E(X+Y)]^2 \\
&= E(X^2) - [E(X)]^2 + E(Y^2) - [E(Y)]^2 + 2[E(XY)- E(X)E(Y)] \\
&= \mathrm{Var}(X) + \mathrm{Var}(Y) + 2\mathrm{Cov}(X,Y)
\end{align*}

  • \(X\)\(Y\) 的线性相关系数:\(\rho_{XY} = \dfrac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)}\sqrt{\mathrm{Var}(Y)}}\)
  • \(X\) 的标准化随机变量:\(X^\* := \dfrac{X-E(X)} {\sqrt{\mathrm{Var}(X)}}\)
  • \(n\) 个随机变量的协方差矩阵:\(\Sigma := (\sigma_{ij})\_{n\times n}\)\(\sigma_{ij} = \mathrm{Cov}(X_i,X_j)\)

泊松分布:\(P(X = k) = \dfrac{\lambda^k}{k!}e^{-\lambda}\),记做 \(X\sim P(\lambda)\)\(\lambda > 0\)
\(E(X) = e^{-\lambda}\sum_{k\ge 0} k\dfrac{\lambda^k}{k!} = e^{-\lambda}\sum_{k\ge 1} k\dfrac{\lambda^k}{k!} = \lambda e^{-\lambda}\sum_{k\ge 1} \dfrac{\lambda^{k-1}}{(k-1)!} = \lambda e^{-\lambda}\sum_{k\ge 0} \dfrac{\lambda^{k}}{k!} = \lambda\)

$ E(X^2) = e^{-\lambda} \sum_{k>=0} k^2 \dfrac{\lambda^k}{k!} = \lambda e^{-\lambda} \sum_{k\ge 0} (k+1) \dfrac{\lambda^{k}}{k!} = \lambda(\lambda + 1)$

从而 $ \mathrm{Var}(X) = E(X^2) - [E(X)]^2 = \lambda $

chap 2

  • 总体,个体,个体的数量指标(个体的出现是随机的 \(\implies\) 个体的数量指标是随机变量,记做 \(X\)
  • 抽样
  • 样本,样本容量 \(n\)(样本是一个复数(plural)概念,抽到的个体是随机得到的,其数量指标是 \(n\) 个随机变量 \(X_1, \dots, X_n\)
  • 样本容量 \(n\),样本观测值 \(x_1, \dots x_n\)
  • 简单随机抽样,简单随机样本

  • 统计量:函数 \(T = T(X_1, \dots, X_n)\),其中不含未知参数。
  • 样本均值:\(\overline{X} = \frac1n \sum_{1\le i\le n} X_i\)
  • 样本方差:\(S^2 = \frac{1}{n-1}\sum_{1\le i\le n} \left(X_i - \overline{X}\right)^2\),样本标准差 \(S\)
  • 样本 \(k\) 阶原点矩:\(A_k = \frac{1}{n}\sum_{1\le i\le n} X_i^k\)
  • 极大次序统计量:\(X_{(n)} = \max\\{X_1, \dots, X_n\\}\)
  • 极小次序统计量:\(X_{(1)} = \min\\{X_1, \dots, X_n\\}\)

  • 抽样分布:统计量的分布(统计量也是一个随机变量)
  • \(\chi^2\) 分布:\(\chi^2 = X_1^2 + \dots + X_n^2, \quad X_i\sim N(0,1)\)

\(\chi^2\) 分布的推导

设 RV \(X\sim \chi^2(n)\) 。考虑 \(X\) 的 DF

\[P(X \le x) = \frac1{\left(\sqrt{2\pi}\right)^n} \int_{V} e^{-\frac12(\sum_i x_i^2)} \dif x_1 \dif x_2 \dots \dif x_n \]

其中积分区域 \(V\)\(n\) 维球 \(\sum_i x_i^2 \le x\)

由于积分区域和被积函数具有球对称性,上述积分在 \(n\) 维球坐标系下表示为

\[P(X \le x) = c_n \int_{0}^{\sqrt{x}} e^{-\frac{r^2}{2}} r^{n-1}\dif r \]

其中 \(c_n\) 是与 \(n\) 有关的常数。

考虑上述积分在 \(x\to \infty\) 时的极限,有

\[1 = c_n \boxed{\color{blue}{\int_{0}^{\infty} e^{-\frac{r^2}{2}} r^{n-1}\dif r}} \]

形如 $ \int_{0}^{\infty} e{-\frac{r2}{2}} r^{n-1}\dif r $ 的广义积分没有解析形式,引入一种特殊函数来表示这一类积分。

\(\Gamma\) 函数

\[\Gamma(x) = \int_0^\infty t^{x-1} e^{-t} \dif t , \quad x > 0 \]

注意:并非对任意 \(x\in\mathbb{R}\) 上述广义积分都收敛,\(\Gamma(0)\) 就不收敛。

做变量替换,令 \(u = r^2/2\) ,则 \(r = (2u)^{\frac12}, \dif r = (2u)^{-\frac12}\dif u\) 。于是

\[\int_{0}^{\infty} e^{-\frac{r^2}{2}} r^{n-1}\dif r = \int_{0}^{\infty} e^{-u} (2u)^{n/2-1} \dif u = 2^{n/2-1} \Gamma(\frac n2) \]

于是

\[
c_n = \frac1{2^{n/2-1} \Gamma(\frac n2)}
\]

从而 \(\chi^2(n)\) 的 PDF 为

\[ f(x) = \frac{\dif P(X\le x)}{\dif x} = c_n \frac { e^{-\frac{x}{2}} x^{\frac{n-1}{2}} \dif\sqrt{x} } {\dif x} = \frac1{2^{n/2} \Gamma(\frac n2)} e^{-\frac{x}{2}} x^{\frac{n}{2}-1} \]


设 RV \(X\) 的 PDF 为 \(f(x)\)\(Y\) 的 PDF 为 \(g(y)\),求 \(Z=X+Y\) 的 PDF \(h(z)\),考虑下面几种解法是否正确。
(1) $ h(z) = \int_{-\infty}^\infty f(x)g(z-x)\dif x$
(2) $ h(z) = \int_{-\infty}^\infty f(x)g_{Y\mid X}(z-x\mid x)\dif x$
其中 \(g_{Y\mid X}(z-x\mid x)\) 为给定 \(X=x\) 的条件下 \(Y\) 的条件密度函数,\(g_{Y\mid X}(y\mid x) = \dfrac{f(x,y)}{f_X(x)}\)
注意:这里的 \(f(x,y)\) 不能由 \(f(x)\)\(g(y)\) 算出来,需要另外给出。

posted @ 2018-04-12 22:04  Pat  阅读(726)  评论(0编辑  收藏  举报