luixiao1220的博客

原创 勒贝格外侧度为0的集合勒贝格可测.

这里，我们使用Carathéodory’s criterion来证明这个结论。勒贝格可测， 我们需要证明下面的等式对所有的。接下来我们就需要证明。

原创 如果{Eₙ}都是可测集，且他们处处收敛于E. 那么E也是可测集.

本证明是一个简要的证明。这两个结论这里没有直接给出。因为如果给出这两个结论的证明就需要更多篇幅。其中1和3需要读者自行查找资料证明。NOTE: 所谓集合的处处收敛，是说得他们的特征函数处处收敛.

原创 两个不相交的闭集并不能保证两个集合可分

现在我们可以来讨论，如果两个集合不相交且两个集合都是闭集的时候，并不能说明两个集合不可分。也就是说即便两个集合都是闭集且不相交，他们也可能不可分。或者我们用一种不太严谨的说法，两个集合的并集没有缝隙，那么就说明两个集合不可分。而这个缝隙的数学含义就是。本文就构造出两个例子，一个是在有理数集合里面构造出两个集合， 一个是在。如果一个集合的聚点都属于这个集合本身吗，那么这个集合是一个闭集。也就是说这两个例子都说明，即便两个都是闭集且交集为空，这两个集合依然可能是不可分。同理，这两个集合也是闭集，且交集为空集。

原创 A+B的闭包不一定也是A的闭包+B的闭包

闭包

原创 A的闭包+B的闭包包含于A+B的闭包

Let X be a topological vector space. If A⊂XA\subset XA⊂X and B⊂XB\subset XB⊂X, then A‾+B‾⊂A+B‾\overline{A}+\overline{B}\subset \overline{A+B}A+B⊂A+B​.Take a∈A‾a\in \overline{A}a∈A, b∈B‾b\in \overline{B}b∈B; Let WWW be a neighborhood of a+ba+ba+b.There ar

原创 如果一个集合的Lebesgue测度为0, 那么它的子集也是Lebesgue可测的并且其测度也为0.

a subset of a zero(0) Lebesgue measure set is Lebesgue measurable

原创 Egoroff‘s Theorem

Assume E has finite Lebesgue measure. Let {fn}\{f_n\}{fn​} be a sequence of Lebesgue measurable functions on EEE that converges pointwise on EEE to a Real-Valued function fff. Then for each ϵ>0\epsilon>0ϵ>0, there is a closed set FFF contained in EEE for

原创 If B is a local base for X, then every member of B contains the closure of some member of B

Theorem: If B is a local base for a topological vector space X, then every member of B contains the closure of some member of B::: {.CJK}UTF8gbsnIf B\mathcal{B}B is a local base for a topological vector space XXX, then every member of B\mathcal{B}B contai

原创 解线性方程组的各种情况

解线性方程组的各种情况

原创 三门问题（一个有趣的概率题）

::: CJKUTF8gbsn三门问题是一个十分有趣的问题。 它讲述的是这样一个问题。 假如有一档节目。节目里面有设置有三道门， 其中一道门里面有奖品。主持人会让你选一道门。当你做出选择之后，主持人会将你没有选择的两扇门中选择一个空门打开（主持人是知道哪扇门没有礼物的）。接下来主持人会问你，需要改变自己的选择么？那么问题也就来了， 假如你想使得自己的获奖概率最大化，你是选择更换自己的选择还是保持自己的选择？ 而中奖概率是多少呢？穷举法礼物在1号门第一次选择1号门主持人选择开

原创 组合数学中将物体放入盒子中的四种情况

在实现生活中， 如何将物体分配到盒子里面是一个非常普通且常见的一个问题。要解决这个问题需要考虑几种清空。首先我们把这个问题分成四个类别的的问题。将不同的物体分配到不同的盒子中将相同的物体分配到不同的盒子中将不同的物体分配到相同的盒子中将相同的物体分配到相同的盒子中将不同的物体分配到不同的盒子中举例：如果将52张扑克开（一套扑克牌）分配给4个玩家， 每人5张牌。有多少种分配方法？解答：这个问题就是典型的将不同的物体分配到不同的盒子中的问题。要解决这个问题其实很

原创 有限值函数 定义（Finite Valued Function）

In the Principles of Mathematical Analysis, Rudin introduced a concept named finite valued function. Many students get confused with this terminology. Does finite valued mean bounded? The answer is no. Before we introduce this concept, we have to get to kn

原创 Simpson‘s rule Error analysis (辛普森积分方法的误差分析)

IntroductionSimpson’s rule is an integral approximate method. Instead of using theoriginal function f(x)f(x)f(x) to compute the integration, it uses apolynomial function. If we have three points a,m,ba,m,ba,m,b and their valuesf(a),f(m),f(b)f(a), f(m),

原创 数学分析第七课(复数)

The Complex Field. We firstly introduce the definition of complexnumbers.Definiton: A complex numbers is an ordered pair (a,b)(a,b)(a,b) of realnumbers. (a,b)(a,b)(a,b) and (b,a)(b,a)(b,a) are regarded as distinct if a≠ba\ne ba​=b. Letx=(a,b),y=(c,d)x

原创 数学分析第六课(域以及域的性质)

In this lecture, we shall introduce an important concept which is calledfield.Definition: A field is a set FFF with two operations, called addtionand multiplication, which satisfy the following so-called fieldaxionms.Axioms for addition∀x,y(x,y∈F

原创 数学分析第五课(关于实数的4个定理)

We will introduce four theorems in R\mathbb{R}R, and they are suchobvious that you even don’t realise that they need proofs.Theorem 1: If x∈R,y∈Rx\in \mathbb{R}, y\in \mathbb{R}x∈R,y∈R and x>0x>0x>0, thenthere is a positive integer nnn such tha

原创 数学分析第四课(从有理数开始构建实数)

Here goes the most important theory, and it is a theorem about realnumbers. This theorem explains the superiority of R\mathbb{R}R.Theorem There exists an ordered field R\mathbb{R}R which has theleast-upper-bound property. Moreover, R\mathbb{R}R contains

原创 数学分析第三课(上下确界的概念)

Definition Suppose SSS is an ordered set, E⊂SE\subset SE⊂S, and EEE isbounded above. Supose there exists an α∈S\alpha\in Sα∈S with the followingproperties:α\alphaα is an upper bound of EEEIf γ<α\gamma<\alphaγ<α then γ\gammaγ is not an uppe

原创 数学分析第二课(有理数的缺点)

As all we know that Expanding natural numbers N\mathbb{N}N a little bityields integers Z\mathbb{Z}Z. Similarily, Expanding integersZ\mathbb{Z}Z a little bit generates Q\mathbb{Q}Q. But, what about realnumbers R\mathbb{R}R ? How to get real numbers R\mat

原创 数学分析第一课(数学分析和微积分的桥梁)

What is the foundation of differential calculus? We will discover itfrom one of the most important theorem in differential calculus whichcalled Mean-Value theorem. In order to prove this theorem, we only needseveral steps.THEOREM 1—The Extreme Value

原创 Hermitian Matrix可以对角化证明。 艾尔米特矩阵可以对角化

Schur triangularization: Let AAA be a n×nn\times nn×n matrix in thecomplex field. We use λ1,λ2,⋯ ,λn\lambda_1, \lambda_2, \cdots, \lambda_nλ1​,λ2​,⋯,λn​ todenote the eigenvalues of AAA. Then there is an unitary UUU such thatU∗AU=TU^{*}AU=TU∗AU=T, where

原创 克莱蒙法则简要证明

UTF8gbsn克莱蒙法则：对于形如 (a11a12⋯a1na21a21⋯a2n⋮⋮⋱⋮an1an2⋯ann)(x1x2⋮xn)=(b1b2⋮bn)\left( \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{21} & \cdots & a_{2n} \\ \vdots & \vdots &

原创 多元正态分布积分为1

原创 n个不同对象聚类为k个类别有多少种可能性? 李航博士,统计学习方法2nd.公式14.21的修正.

UTF8gbsnn个不同的对象分到k个相同的盒子里面, 要求每个盒子至少有一个对象.有多少种分法. 这是在k均值聚类里面的一个组合数学问题.在k均值聚类里面有n个对象各不相同,要把这个n个对象分到k个类别里面并要求每个类别必须至少含有一个对象.总共的分法有多少中? 这道题的答案是第二类的stirling number.我们来看看如何来求解.我们把原问题定义为 P(n,k)P(n,k)P(n,k)初探如果将n个不同的对象放到k个不同的盒子里面总共有多少种方法?这个问题不再限制每个盒子必须含有

原创 HMM(隐马尔科夫模型的预测算法)

UTF8gbsnThe predication of HMM has two different methods. the approximation wayand the Viterbi algorithm. The predication problem of HMM is to find thestate sequence based on λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) and observationO=(o1,o2,⋯ ,oT)O=(o_1,o_2,

原创 HMM(隐马尔科夫模型的无监督学习方法)

UTF8gbsnInductionThe main topic of this article is about Baum-Welch algorithm. We willestimate λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) with {O1,O2,⋯ ,OS}\{O_1,O_2,\cdots,O_S\}{O1​,O2​,⋯,OS​} and thelenght of OiO_iOi​ is TTT. The object of our probability m

原创 HMM(监督学习)

UTF8gbsnThe subject of today is the supervised learning of HMMλ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π). The task of the supervised learning of HMM is toestimate λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) from these known observation sequences OOOand their correspo

原创 HMM(几种期望)

UTF8gbsnAfter we studied the HMM λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) and the backward andforward methods of computing P(O∣λ)P(O|\lambda)P(O∣λ), we should be able tocompute the following probabilities and expectations.For a given λ\lambdaλ and OOO, th

原创 HMM(backward法求观察序列的概率)

UTF8gbsnbackward probability: HMM λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) is known. We definethe backward probability at time ttt and its state qiq_iqi​ as theprobability of sequence (ot+1,ot+2,⋯ ,oT)\left( o_{t+1}, o_{t+2}, \cdots, o_T \right)(ot+1​,ot+2​,

原创 HHM(forward法求观察序列的概率)

UTF8gbsnThe direct method of computing P(O∣λ)P(O|\lambda)P(O∣λ) is ultra expensive.Fortunately, there are some other ways to get an appropriateperformance. Let’s look at the first improvement called forward method.Definition: The forward probability o

原创 HHM(直接计算法求观察序列的概率)

UTF8gbsnAfter we known the HMM λ=(A,B,π)\lambda=(A,B,\pi)λ=(A,B,π) andO=(o1,o2,⋯ ,oT)O=\left( o_1, o_2, \cdots, o_T \right)O=(o1​,o2​,⋯,oT​), the probability P(O∣λ)P(O|\lambda)P(O∣λ)can be computed directly. We illustrate possible state sequences ofI=(

原创 隐马尔科夫模型(HMM)的基本问题.

UTF8gbsnIn the previous blog, we illustrated the definition of HMM and in thisblog we will list two important aspects of HMM.The generation of observation sequenceBased on the definition of HMM, the generation of the observationsequence O={o1,o2,⋯ ,oT

原创 隐马尔科夫模型(HMM)的基本问题.

UTF8gbsnIn the previous blog, we illustrated the definition of HMM and in thisblog we will list two important aspects of HMM.The generation of observation sequenceBased on the definition of HMM, the generation of the observationsequence O={o1,o2,⋯ ,oT

原创 HMM定义(Hidden Markov Model)

UTF8gbsnHidden Markov model(HMM) is a statistical learning model which used inlabeled problems. We will introduce the definition of HMM first here andthere will be an example followed by the HMM definition.definitionHMM is a time sequence statistics m

原创 线性最小二乘法

UTF8gbsn本文介绍线性最小二乘法.线性最小二乘法比较简单并且在很多场景里面都有应用.下面我们就来举几个例子.例子已知散点集合X={x1,x2,⋯ ,xn},Y={y1,y2,⋯ ,yn}X=\{x_1,x_2,\cdots, x_n\}, Y=\{y_1,y_2,\cdots, y_n\}X={x1​,x2​,⋯,xn​},Y={y1​,y2​,⋯,yn​},求参数方程y=ax+by=ax+by=ax+b.已知是X,YX,YX,Y未知是a,ba,ba,b. 这个问题是一个标准的可以使用

原创 多个等式束的拉格朗日乘子问题(详细证明)

UTF8gbsn多约束的拉格朗日乘子问题.f(x)h1(x)=0⋮hm(x)=0\left. \begin{aligned} \quad & f(x)\\ \quad& h_1(x)=0\\ & \quad \quad \vdots\\ & h_m(x)=0 \end{aligned} \right.​f(x)h1​(x)=0⋮hm​(x)=0​假设这个问题的解是x∗x^{*}x∗.

原创 多个约束的lagrange multiplier证明.

UTF8gbsn前面我们介绍了lagrange multiplier. 在单constraint的时候该怎么办.f(x),x∈Rng(x)=0,x∈Rn\left. \begin{aligned} &f(x), x\in R^n\\ &g(x)=0, x\in R^n \end{aligned} \right.​f(x),x∈Rng(x)=0,x∈Rn​L(x,λ)=f(x)−λg(x),(x,λ)∈Rn+1L(x,\la

原创 拉格朗日乘数法证明

UTF8gbsn介绍现在我们来介绍下拉格朗日乘数法.首先提出问题. 假如我们有一个目标函数f(x)f(\mathbf{x})f(x) 约束条件为 g(x)=0g(\mathbf{x})=0g(x)=0拉格朗日乘数法的流程是写出目标函数L(x,λ)=f(x)−λg(x)L(\mathbf{x}, \lambda)=f(\mathbf{x})-\lambda g(\mathbf{x})L(x,λ)=f(x)−λg(x)并求出稳定点{∇f(x)=λ∇g(x)g(x)=0\left\{ \

原创 MLE(最大似然), LMS(均方误差),KL(散度), H(交叉熵).之间的等效性证明

UTF8gbsn我们在机器学习中常用的损失函数, 比如MLE(极大似然估计), KL(散度),H(交叉熵), 均方误差等之间是什么关系? 本文就简单的讲解一下它们之间的关系.均方误差 vs MLE如果你的数据分布是来自指数型数据分布, 那么他们是等价的.具体证明这里就不详细去证明了. 详情参看文章.Charnes, A., Frome, E. L., Yu, P. L. (1976). The equivalence of generalized least squares and maxi

原创 单位正方形上两个随机点之间距离的期望(均值)

UTF8gbsnQ:趣味问题, 假如我有一个边长为1的正方形.随机的从正方形里面取得两个点p1,p2p_1, p_2p1​,p2​.计算dp1,p2=∥p1−p2∥22d_{p_1,p_2}=\|p_1-p_2\|^2_2dp1​,p2​​=∥p1​−p2​∥22​. 如果重复这个操作NNN次,我们就会得到一个距离的集合D={d1,d2,⋯ ,dN}D=\{d_1,d_2,\cdots, d_N\}D={d1​,d2​,⋯,dN​}. 请问E(D)=?E(D)=?E(D)=?.就是两两随机点之间