Eric Yang

二项检验笔记

Thu, 03 Oct 2024 13:15:00 GMT

定义

二项检验是一种非参数检验(model-free), 用于检验一个具有二元结果的变量是否满足要求。例如: 产品的次品率是否符合要求，一个硬币是否是公平的...

二项检验的假设如下

感兴趣的变量应该是一个二元结果，意味着它只能取两个值中的一个（例如，抛硬币（头/尾），是否有疾病（是/否），道德（死/活））。这有时也被称为二分法变量。
观察应该是独立的，也就是说一个观察对另一个观察的概率没有任何影响。
实验应该有一个固定的样本量，表示为n。
所有独立的观察结果应该具有相同的概率。这类似于独立假设，可以通过随机抽样来实现。

简单来说就是 BINS: Binary, Independent, N samples and Same probability.

与z检验的对比

z检验是一种参数检验，即我们知道总体的均值和标准差，然后用样本的均值和标准差来推断总体的均值是否满足要求。

我们用统计量 $ z = \frac{\bar{x} - \mu}{\sigma/\sqrt{n}} $ 来检验总体均值是否满足要求。

相较于 z 检验，二项检验更加灵活，因为它不需要知道总体的均值和标准差，只需要知道样本的二元结果即可。他们的对比如下表所示:

	二项检验	z检验
总体	二元结果	连续结果
参数	无	有
样本	二元结果	连续结果
假设	二元结果是否满足要求	总体均值是否满足要求

总的来说，二项检验容易实施，也是exact test，但是它的置信区间比较宽，而且只能用于二元结果的检验。

步骤

先看一个例子

!!! note 某机器生产一种产品，当次品率小于等于 5% 时可以认为该机器工作正常:当大于 5% 时对机器进行检修.某天抽出 15 件产品，发现 3 件次品，问该天机器工作是否正常?^1

我们可能会直接使用 3/15 = 0.2 > 0.05 来判断机器是否正常，但是这种方法是不准确的，因为我们只抽取了 15 件产品，引入了偶然性，无法判断到底是随机效应还是真的次品率超过了 5%.

我们可以使用二项检验来解决这个问题。设每个产品为次品的概率为 $p$，且是否为次品相互独立.困此，这个假设检验问题的原假设 $H_0$ 和备择假设 $H_1$ 可以写为 $$ H_0: p \leq 0.05 \leftrightarrow H_1: p > 0.05 $$ 我们知道，如果次品太多，就要拒绝 $H_0$. 所以取检验统计量 T 为次品的总个数. 可以得到 T 服从二项分布 B(15, 0.05). 查二项分布表可得 $$ P{T\leq 2}=0.9638 $$ 如果取显著性水平 $\alpha= 0.05$，那么拒绝域为 $T>2$. 现 T 的观测值为 3，所以拒绝 $H_0$ ，即认为该天机器工作不正常.检验的 $p$ 值为 $$ P{T\geq 3}=1-P{T\leq 2}=0.0362 $$

在这里，查阅二项分布表是一个很麻烦的事情，并且，如果我们计算实际的置信水平，可以发现，实际的置信水平并不是 0.05，而是略有偏差，这是因为二项分布表是离散的，而我们的检验统计量是连续的，所以我们需要使用连续的分布来近似二项分布。

双边检验

设检验统计量 $T$ 为 $n$ 重 Bernoulli 试验中成功的次数，则 $T$ 服从二项分布 $B(n,p)$ 对于 $n\leqslant20$ 和选定的 $p$ ,附表 2 中列出了 $T$ 的零分布的值. 对于 $n>20$, 我们可以用正态分布逼近，即 $T$ 的 $q$ 分位数 $x_q$ 可以由下式近似给出.

$$x_q=np+z_q\sqrt{np(1-p)},$$

其中 $z_q$是标准正态分布的 $q$ 分位数. 显著性水平 $\alpha$ 的拒绝域对应于 $T$ 的零分布的两边，其中左边水平为 $\alpha/2$ ,右边水平为 $1-\alpha/2$ 对于 $n\leqslant20$ 和给定的 $p_0$ ,我们可以查附表2而得到二项分布 $B(n,p_{0})$ 的 $\alpha/2$ 分位数 $b(\alpha/2,n)$ 和 $1-\alpha/2$ 分位数 $b(1-\alpha/2,n)$ ,使得

$$P{Y\leqslant b(\alpha/2,n)}\leqslant\frac{\alpha}{2},\P{Y\leqslant b(1-\alpha/2,n)}\geqslant1-\frac{\alpha}{2},$$

单边检验

类似于双边检验，只需要考察

$$ P{Y\leqslant b(\alpha,n)}\leqslant\alpha $$ 或 $$ P{Y\leqslant b(1-\alpha,n)}\geqslant1-\alpha $$ 即可

国庆节间写于6教6A214, 是日晴空万里，心情愉悦。早上看了很久《唐朝诡事录》，闲来无事，写写笔记，也是一种放松。

统计计算复习-第一章

Thu, 03 Oct 2024 13:15:00 GMT

浮点数

浮点数表示如下 $$ \text{sign} \times 2^{\text{exponent}} \times \text{mantissa} $$ 其中，sign表示符号位(1位)，exponent表示指数位(8位)，mantissa表示尾数位(23位)。

双精度浮点数三个部分的位数分别为1, 11, 52。

规格数

8位指数位的取值范围为$[0, 255]$，但是指数位不能全为0或全为1，所以实际的指数范围为$[-126, 127]$。即 $$ \text{exponent} = \text{exponent} - 127 $$

隐藏位: 尾数位的第一位默认为1，所以实际的尾数位为24位。

舍入误差

浮点数的表示是有限的，所以会有舍入误差。例如，$0.1$在二进制中是无限循环的，所以在计算机中表示为一个近似值。

绝对舍入误差 $$ \text{abs_error} = \left| \text{true_value} - \text{approx_value} \right| $$

相对舍入误差 $$ \text{rel_error} = \frac{\text{abs_error}}{\text{true_value}} $$ 对于较大的数，绝对舍入误差可能很大，以至于gap中存在新的整数。

Cancelation

相近的数相减可能会导致有效数字的丢失，这种现象称为Cancelation。

!!!note 定理1 With one guard digit, the relative rounding error in the result is less than $2\varepsilon_{machine}$ 加减都会导致Cancelation，乘除不会。

尽量避免大数加减小数，以及大数之间的加减
尽量避免相近的数相减

误差

算法误差分为前向误差和后向误差。

前向误差

!!!note 前向误差是指一种算法的输出与真实值之间的差异。即 $$ \Delta y = \tilde{y} - y = \tilde{f}(x) - f(x) $$

后向误差

!!!note 后向误差是指输入值的舍入误差对输出值的影响。即使 $$ f(\tilde{x})=\tilde{f}(x) $$ 的 $\Delta x = \tilde{x} - x$

稳定性

!!!note 算法是稳定的，如果对于任何输入 $$ \frac{|f(\tilde{x})-\tilde{f}(x)|}{|f(\tilde{x})|}=O(\varepsilon_{machine}) $$ 对于$\tilde{x}$ 满足 $|\tilde{x}-x| = |x|O(\varepsilon_{machine})$

一个稳定的算法对于几乎正确的问题给出几乎正确的答案。

后向稳定

即$ f(\tilde{x})=\tilde{f}(x) $，对于 $\tilde{x}$ 满足 $|\tilde{x}-x| = |x|O(\varepsilon_{machine})$

后向稳定算法对于几乎正确的输入给出完全正确的输出。

条件数

绝对条件数定义为:

$$ \hat{\kappa}=\hat{\kappa}(x)=\operatorname*{lim}{\varepsilon\rightarrow0}\operatorname*{sup}{|\delta x|\leq\varepsilon}{\frac{|\delta f|}{|\delta x|}}\ \triangleq\operatorname*{sup}_{\delta x}{\frac{|\delta f|}{|\delta x|}} $$ 如果 $f$ 是可导的，那么 $$ \hat{\kappa}=|Jacobi(f)| $$
相对条件数定义为:

$$ \kappa=\kappa(x)=\operatorname*{lim}{\varepsilon\rightarrow0}\operatorname*{sup}{||\delta x||\leq\varepsilon}\frac{||\delta f||/|f(x)||}{||\delta x||/|x||}\ \triangleq\ S_{\Delta x}\frac{||\delta f||/|f(x)||}{||\delta x||/|x||} $$

Addition Formular

Sat, 27 Jul 2024 08:00:00 GMT

The addition formular

Consider the map

$$ z\to (1,\wp(z),\wp(z)^\prime) $$

It parametrizes points on the cubic curve A defined by the equation

$$ y^2=4x^3-g_2x-g_3 $$

The map is actually defined on the torus $\mathbb{C}/\Lambda$

Now for any complex number $\alpha$, $\wp(z)-\alpha$ has at most two zeros and at least one zero, so that already under $\wp$ we cover each complex number $\alpha$

Furthermore, $\mathbb{C}/\Lambda$ has a natural group structure, and we now want to see what it looks like when transported to A. We shall see that it's algebraic. In other words, if

$$ P_1=(x_1,y_1), \quad P_2=(x_2,y_2), \quad P_3=(x_3,y_3) $$

and

$$ P_{3}=P_{1}+P_{2} $$

then we shall express $x_3 , y_3$ as rational functions of $x_1 , y_1$ and $x_2 , y_2$. We shall see that $P_3$ is obtained by taking the line through $P_1,P_2$, intersecting it with the curve, and reflecting the point of intersection through the x-axis. Select $u_1,u_2\in C$ and $\notin L$, and assume $u_1= u_2 (mod \lambda)$. Let a, b be complex numbers such that

$$ \begin{gathered} \wp^{\prime}(u_{1}) =a\wp(u_1)+b \ \wp^{\prime}(u_{2}) =a\wp(u_2)+b \end{gathered} $$

in other words $ y = ax + b $ is the line through $\wp^{\prime}(u_{1}) =a\wp(u_1)+b$,and $\wp^{\prime}(u_{2}) =a\wp(u_2)+b$. Then

$$ \wp^{\prime}(z)-a\wp(z)-b $$

has a pole of order 3 at 0, whence it has 3 zeros, counting multiplicities, and tow of these at $u_1,u_2$. If say $u_1$ had multiplicity 2, then

$$ 2u_1+u_2=0 (mod\space \lambda) $$

and

$$ u_3=-(u_1+u_2)(mod \space\lambda) $$

the equation

$$ 4x^{3}-g_{2}x-g_{3}-(a x+b)^{2}=0 $$

has 3 roots. and

$$ LHS=4(x-\wp(u_{1}))(x-\wp(u_{2}))(x-\wp(u_{3})). $$

and you can get

$$ \wp(u_1)+\wp(u_2)+\wp(u_3)=\frac{a^2}{4} $$

we also have

$$ a(\wp(u*{1})-\wp(u_{2}))=\wp^{\prime}(u_{1})-\wp^{\prime}(u_{2}) $$

$$ \wp(u_{3})=\wp(-(u_{1}+u_{2}))=\wp(u_{1}+u_{2}) $$

and we get

$$ \wp(u_1+u_2)=-\wp(u_1)-\wp(u_2)+\frac{1}{4}\bigg(\frac{\wp'(u_1)-\wp'(u_2)}{\wp(u_1)-\wp(u_2)}\bigg)^2 $$

if we take limit as $u_1\to u_2$ and get

$$ \wp(2u)=-2\wp(u)+\frac{1}{4}\bigg(\frac{\wp^{\prime\prime}(u)}{\wp^{\prime}(u)} \bigg)^2 $$

geometric meaning of additional formulae

Theorem The sum of 3 distinct points $ a, b, c $ on the elliptic curve is zero.

$$ a+b+c=0 $$

iff $ a, b, c $ lie on the projective line.

Proposition: The following formular holds:

$$ \begin{vmatrix} 1 &\wp(u+v) & \wp^{\prime }(u+v)\ 1 &\wp(u) & \wp^{\prime }(u)\ 1 &\wp(v) &\wp^{\prime }(v) \end{vmatrix}=0 $$

and

$$ \begin{array}{l}(x_1,y_1)=(\wp(u),\wp'(u)),\ (x_2,y_2)=(\wp(v),\wp'(v)),\ (x_3,y_3)=(\wp(u+v),-\wp'(u+v)).\end{array} $$

Schwarz Christoffel Theorem

Fri, 26 May 2023 21:42:41 GMT

初等函数

sinz

将区域

$$ A={|Re(z)|<\frac{\pi}{2}} $$

映到上半平面

arcsinz

将区域

$$ A={|Im(z)|<\frac{\pi}{2},Re(z)>0} $$

映到上半平面

分式线性变换 $$ F(z)=\frac{i-z}{z+i}:H\to D\ G(z)=i\frac{z-1}{z+1}: D\to H $$
logz $$ H\to {z|0<Imz<\pi} $$
$ z^{\alpha} $ 上半平面到扇形

Schwarz Lemma

if a holomorphic function $ f:D\to D $ satisfies |f(0)|=0, then

|f(z)|<|z|
for some $ z_0 $, $ f(z_0)=z_0 $, then f is rotation
|f^{'}(0)| ≤ 1, and if equality holds, then f is a rotation

Riemann Mapping Theorem

Suppose $ \Omega $ is proper and simply connected. If $ z_0\in \Omega $, then there exists a unique comformal map $ F:\Omega\to D $ such that

$$ F(z_0)=0\quad and \quad F^{\prime}(z_0)>0 $$

Proof

step 1. 任意单连通区域可以共形等价到包含原点的单位圆的子集

$$ f=log(z-\alpha)\Longrightarrow e^{f}=z-\alpha $$

选定$ \omega\in \Omega $ 且

$$ f(z)\neq f(\omega)+2\pi i\quad \forall z\in\Omega $$

则有$ f(z)=f(\omega) $ 定理设 $ a_1<a_2 <a_3 $ 为实数，$ \omega_1，\omega_2，\omega_3 $ 为多角形域 $P$ 上按顺序排列的某三个顶点。则存在唯一的函数 $ f $ ，把 $ \mathbb{H} $ 双全纯地映射到 $ P $ ，把实轴一一连续地映射为 $ \partial P $ ，且

$$ f(a_k)=w_k(k=1,2,3) $$

要将单位圆映到多边形，需要 Schwarz Christoffel Integral,形式如下

$$ F(\zeta)=C\int_0^\zeta\prod\limits_{k=1}^n\left(\omega-\omega_k\right)^{-\beta_k}d\omega+C' $$

$\beta_k=1-\alpha_k$,对应内角为 $\alpha_k\pi$

定理：存在从上半平面到多边形的映射，将实轴上的点$A_1,A_2\cdots,A_k$ 映到多边形的顶点$\alpha_1,\alpha_2,\cdots,\alpha_k$ 映射为

$$ F(\zeta)=C\int_0^\zeta\prod\limits_{k=1}^n\frac{d\zeta}{\left(\zeta-A_k\right)^{\beta_k}}+C' $$

由此，映射并不是唯一的。

如果考虑将上半平面映成长方形，即考虑 Jacobi First Kind Elliptic Function, 可以证明其结果最终是椭圆曲线。

从拓扑的意义上，$f(z)$ 在环面 $\mathbb{X}=\mathbb{C}/ \mathbb{L}$ 上是双周期的有理函数，即 $\mathbb{X}$ 是 $\mathbb{P}^1$ 的分歧覆盖(ramified cover)

进一步考虑这个问题。我们换一种记号，考虑环面上的积分，令$\mathbf{y}=\sqrt{(1-\mathbf{x})(1-k^2\mathbf{x}^2)}$ 视为环面上的单值函数，固定环面上的原点 $\mathfrak{o}=\left(\mathbf{x}=\mathfrak{o},\mathbf{y}=+1\right)\in\mathbb{X}$ ，令 $\mathfrak{p}=(\mathbf{x},\mathbf{y})$ 为环面上的一点，积分 $z=\int_o^p\mathbf{y}^{-1}d\mathbf{x}$ 依赖于从 $\mathfrak{o}$ 到$\mathfrak{p}$ 的路径，对于同一形变类（deformation class）里的路径，积分值相同。这里的术语形变类可以简单的认为是经过小扰动保持不变的两个物体。沿不同的形变类中的路径有不同的值，随如下的周期变化

$$ 4Kn_1+2\sqrt{-1}K'n_2 $$

$4K$, 和 $2\sqrt{-1}K^{\prime}$ 都是 $\mathbf{y}^{-1}d\mathbf{x}$ 绕环面上的洞的积分。可以认为 $z$ 是 $\mathbb{X}$ 的万有覆盖 $\mathbb{C}$ 上的单值函数。$z$ 扫过这整个万有覆盖，我们有两个投影映射

$$ \pi:\mathfrak{p}\longrightarrow \mathbf{x} $$

和

$$ \pi_2:\mathbf{x}\longrightarrow z $$

简单来说，如下图

Montel Theorem

Thu, 25 May 2023 19:54:16 GMT

A function family $ \mathcal{F} $ is called normal if every sequence of function $ {f_n}\in \mathcal{F} $ is uniformly convergent in every compact subset of $ \Omega $. The limit of $ {f_n} $ does not need to be a member of $ \mathcal{F} $. If the limit is $ \infty $, it does not matter.

In order to verify a function family is normal, we actually have methods below:

Montel Theorem

Montel Theorem tells us that a locally bounded function family is normal.

If the family $ \mathcal{F} $ is uniformly bounded in every compact subset of a region $ \Omega $, then it is a normal family.

The proof is based on Arzela-Ascoli Theorem.

If $ f $ is bounded, then by Cauchy-Integral, we get $ \mathcal{F} $ is equicontinuous on the subset. Then by AA Theorem, it's easy to show.

Marty Theorem

A family of analytic or meromorphic functions f is normal in the classical sense if and only if the expressions

$$ \rho(f)=\frac{2|f^{'}(z)|}{1+|f(z)|^2} $$

are locally bounded

Some problems in Ahlfors

Problem P227 T1

Prove that in any region n the family of analytic functions with positive real part is normal. Under what added condition is it locally bounded ? Hint: Consider the functions $ e^{-f} $

Solution

Actually, we may use Marty Theorem by letting

$$ g(z)=\frac{f(z)-1}{f(z)+1} $$

which is bounded. Then $ g^{'} $ is also locally bounded by using Cauchy Formula. Now you do the calculation.

Problem P227 T2

Show that the functions $ z^n $ , $ n $ a nonnegative integer, form a normal family in $ |z| < 1 $ , also in $ |z| > 1 $ , but not in any region that contains a point on the unit circle.

Solution

It's easy to show the first part. For the second, consider $ |a|=1 $. As soon as you show it's a normal family in region $ |z|<1 $ and $ |z|>1 $, then there exists a subsequence $ z^{n_k} $ which is convergent in the region contains a point on the unit circle. And this subsequence converges to $ \infty $ when $ |z|>1 $ and converges to $ 0 $ when $ |z|<1 $, which is contradict.

Problem P227 T3

If $ f(z) $ is analytic in the whole plane, show that the family formed by all functions $f(kz) $ with constant $ k $ is normal in the annulus $ r_l < |z| < r_2 $ if and only if $ f(z) $ is a polynomial.

Solution

We can still use Marty Theorem. Since $ f(kz) $ is normal in the annulus. Then, by Marty Theorem, the expression

$$ \rho(f)=\frac{2k|f^{'}(z)|}{1+|f(z)|^{2}} < M\quad for \space some \space M $$

It means $ |f^{'}| $ is bounded. Then it's trivial.

Problem P227 T4

If the family $ \mathcal{F} $ of analytic (or meromorphic) functions is not normal in n, show that there exists a point $ Z_0 $ such that $ \mathcal{F} $ is not normal in any neighborhood of $Z_0$ Hint: A compactness argument.

Solution

we prove by contradiction. That is near every point $ z $, $ \mathcal{F} $ is normal. Use exhaustion: $ \Omega=\bigcup_{n=1}^{\infty}\Omega_n $, where $ \Omega_{n}\in \Omega_{n+1}$.We have a subsequence converging uniformly on $ \Omega_1 $ from it choose a subsequence converging uniformly on $ \Omega_2 $ , the process is infinite but the diagonal selection produces the desired subsequence.

Singularity, Residue and some theorems

Riemann Theorem

If $f$ is a holomorphic function in an open set $\Omega$ except a point $z_0$. If f is bounded in $\Omega/{z_0}$ , then $z_0$ is the removable singularity of $f$

Corollory

Suppose $f$ has an isolated singularity $z_0$, and $|f|\to \infty$ . Then, $z_0$ is a pole of $f$

Casorati-Weierstrass

If $f$ is holomorphic in the punctured region $\Omega/ {z_0}$, where $z_0$ is a essential singularity. Then the image of $\Omega/ {z_0}$ under $f$ is dense.

The meromorphic functions in the extended complex planeare the rational functions

Argument Principle

$$ \frac{1}{2\pi i}\int_{\gamma}\frac{f(z)^{'}}{f(z)}dz=n(zeros)-n(poles) $$

Rouche Theorem

if function $f$ is holomorphic in region $\Omega$ containing a circle $C$ and its interior. Then if

$$ |f|>|g|,\quad \forall z\in \partial C $$

then $f$ and $f+g$ has the same number of zeros in $C$

Open mapping theorem

If $f$ is holomorphic and nonconstant in a region $\Omega$, then $f $ is open.

Maximum modulus principle

If $f $ is a non-constantholomorphic function in a region $\Omega$, then $f$ cannot attain a maximum in $\Omega$.

Theta Function

Sat, 13 May 2023 21:05:19 GMT

Background

As the speed of convergence of $ \wp(z) $ is too slow, we need a new function.

We can represent $ \wp(z) $ as a ration of two functions.

Consider the case of lattice being generated by $ 2\pi $ and $ i $. We fix a positive integer k and consider the conditions on a holomorphic function on $ \mathbb{Z} $ $$ \varphi(z+2\pi)=\varphi(z)\\ \varphi(z+i)=\varphi(z)e^{k(-iz+1/2)} $$

Actually, it's constructed through Fourier Transformation. By taking Fourier Transformation, you get $$ \varphi(w)=\sum\limits_{n=-\infty}^\infty c_n e^{2n i w}. $$ Then, you also have $$ \varphi(z+i)=\varphi(z)e^{k(-iz+1/2)} $$ and you can solve $ c_n $ and get $$ \theta(z)=\sum e^{-p^2/2}e^{ipz} $$ For general lattice$ \tau \mathbb{Z}\sum 2\pi\mathbb{Z} $ , we have$$ \theta(z,\tau)=\sum_{p}\sqrt{\frac{i\tau}{2\pi}}e^{i\tau p^{2}/2}e^{i p z} $$

Theta function plays an important role in many areas, which I don't know. Here are some fun facts

Matrix Description of Theta Function

Consider a orthonormal basis of $ \Lambda $, actually to obtain a symmetric bilinear form which is positive non-degenerate, then setting $ a_{ij}=e_i\cdot e_j $, then the matrix $ A=(a_{ij}) $must be positive and non-degenerate and symmetric. the corresponding theta function is
$$ \varTheta_{\Lambda}(t)=\sum_{x\in \mathbb{Z}^{n}}e^{-\pi t\sum a_{ij}x_ix_j} $$ The volume of lattice $ \Lambda $ is $det\sqrt{A} $

Higher dimension Lattice

We consider lattice $ L=\mathbb{Z}^n $ in space $ \mathbb{Q}^n $ equiped with usual dot product bilinear form, where $ \mathbb{Q} $ denotes rational numbers. However, it does not satisfy the property$ x\cdot x=0 \space mod 2 $.

We want to construct a submodule $ L_2 $ of $ L $ satisfies this property. To fix this, we add the vector $ e = (1/2, · · · , 1/2)\in V $ and consider $ \Gamma_n $ = submodule of V generated by $ L_2 $ and e. Then $$ x · e =\frac{1}{2}\sum x_i \in \mathbb{Z} \quad \forall x \in L_2,\quad e \cdot e = 2k $$ where $ n=8k $. so $ \Gamma_n $ satisfies the mod 2 property. Moreover, $ L_2 $ has index 2 in $ \Gamma_n $ since $ 2e \in L_2 $ , so $ \Gamma_n $ has finite volume equal to 1 as desired, and defines a lattice in $ \mathbb{R}^n $

Solution to Alhfors

Thu, 11 May 2023 22:51:21 GMT

Alhfors Complex Analysis 的习题答案，主要是椭圆函数这一节。

P274 Exercise 1

The Weierstrass functions satisfy numerous identities which are best dealt with in an exercise section. They can be proved either by comparing two elliptic functions with the same zeros and poles (when $ \sigma $ -functions are involved), or by comparing elliptic functions with the same singular parts (when only $ \wp $ and $ \zeta $ -functions are involved). The following sequence of formulas is so arranged that we need to resort to this method only once.

Solution 1

$$ \wp(z)-\wp(u)=-\frac{\sigma(z-u)\sigma(z+u)}{\sigma(z)^2\sigma(u)^2} $$

依据提示，我们考虑利用3.2节的(14)式，即

$$ \begin{gathered} \sigma(z+\omega_1) =-\sigma(z)e^{\eta_1(z+\omega_1/2)} \\ \sigma(z+\omega_2) =-\sigma(z)e^{\eta_2(z+\omega_2/2)}. \end{gathered} $$

证明右侧式的周期为$ \omega_1,\omega_2 $ 事实上

$$ \begin{align*} \phi(z+\omega_1)&=-\frac{\sigma(z-u+\omega_1)\sigma(z+u+\omega_1)}{\sigma(z+\omega_1)^2\sigma(u)^2} \\ &=-\frac{\sigma(z-u)\sigma(z+u)e^{2\eta_1z+\eta_1\omega_1}}{\sigma(z)^2\sigma(u)^2e^{2\eta_1(z+\omega_1/2)}}\\ &=-\frac{\sigma(z-u)\sigma(z+u)}{\sigma(z)^2\sigma(u)^2}\\ &=\phi(z) \end{align*} $$

同理可证$ \omega_2 $ 也为$ \phi(z) $ 的周期。我们再考虑$ \phi(z) $ 的零极点显然左式的零点为$ \pm u $, 仅有的二阶极点为原点。对于右式，有\

$$ \phi(u)=\phi(-u)=0 $$

且$ \phi(z) $的极点为 $ \sigma(z)$ 的零点。由定义可知为$ z=0 $,且是二阶极点。综上可知 $$ \wp(z)-\wp(u)=-\frac{\sigma(z-u)\sigma(z+u)}{\sigma(z)^2\sigma(u)^2} $$

P274 Exercise 2

Prove $$ \frac{\wp'(z)}{\wp(z)-\wp(u)}=\zeta(z-u)+\zeta(z+u)-2\zeta(z) $$

Follows from (16) by taking logarithmic derivatives.

Solution 2

依据提示，我们考虑利用上面的结论，对两边求对数导数，即 $$ \begin{align*} &\frac{\wp'(z)}{\wp(z)-\wp(u)}\\ &=\frac{(\sigma(z-u)^{\prime}\sigma(z+u)+\sigma(z-u)\sigma(z+u)^{\prime})}{\sigma(z+u)\sigma(z-u)}\\ &-2\frac{\sigma(z+u)\sigma(z-u)\sigma(z)^{\prime}}{\sigma(z)\sigma(z+u)\sigma(z-u)}\\ &=\zeta(z-u)+\zeta(z+u)-2\zeta(z) \end{align*} $$ 证毕

P274 Exercise 3

Prove $$ \zeta(z+u)=\zeta(z)+\zeta(u)+\frac{1}{2}\frac{\wp'(z)-\wp'(u)}{\wp(z)-\wp(u)} $$

This is a symmetrized version of (17).

Solution 3

事实上，对换 Exercise 2中$ z $ 和$ u $ 的位置可以得到 $$ \frac{\wp'(u)}{\wp(u)-\wp(z)}=\zeta(u-z)+\zeta(u+z)-2\zeta(u) $$ 则有 $$ \begin{align*} \frac{1}{2}\frac{\wp'(z)-\wp'(u)}{\wp(z)-\wp(u)}&=\zeta(u+z)+\frac{1}{2}(\zeta(u-z)+\zeta(z-u))\\ &-(\zeta(u)+\zeta(z))\\ &=\zeta(u+z)-(\zeta(u)+\zeta(z)) \end{align*} $$ 移项即得。

P277 Exercise 5

Prove $$ \wp(2z)=\frac{1}{4}\left(\frac{\wp''(z)}{\wp'(z)}\right)^2-2\wp(z) $$

Solution 5

考虑使用加法公式 $$ \wp(z+u)=-\wp(z)-\wp(u)+\frac{1}{4}\bigg(\frac{\wp'(z)-\wp'(u)}{\wp(z)-\wp(u)}\bigg)^2 $$

令 $u=z+\Delta z$,则 $$ \begin{align*} \wp(2z+\Delta z)&=-\wp(z)-\wp(z+\Delta z)\\ &+\frac{1}{4}\bigg(\frac{\wp'(z)-\wp'(z+\Delta z)}{\wp(z)-\wp(z+\Delta z)}\bigg)^2\\ \end{align*} $$

而 $$ \lim_{\Delta z\to 0}-\wp(z)-\wp(z+\Delta z)=-2\wp(z) $$

$$ \begin{align*} \frac{1}{4}\bigg(\frac{\wp'(z)-\wp'(z+\Delta z)}{\wp(z)-\wp(z+\Delta z)}\bigg)^2&=\frac{1}{4}\bigg(\frac{(\wp'(z)-\wp'(z+\Delta z))/\Delta z}{(\wp(z)-\wp(z+\Delta z))/\Delta z}\bigg)^2\\ &=\frac{1}{4}\left(\frac{\wp''(z)}{\wp'(z)}\right)^2\quad \Delta z\to 0 \end{align*} $$

则 $$ \wp(2z)=\frac{1}{4}\left(\frac{\wp''(z)}{\wp'(z)}\right)^2-2\wp(z) $$

P277 Exercise 6

Prove $$ \wp'(z)=-\sigma(2z)/\sigma(z)^4 $$

Solution 6

注意到右侧极点为 $z=0$,且为3阶极点，而零点则为 $$ \frac{\omega_1}{2},\space\frac{\omega_2}{2},\space\frac{\omega_1+\omega_2}{2} $$ 事实上有 $$ -\sigma(\omega_1)/\sigma(\omega_1/2)^4=-\frac{\sigma(0)e^{\eta_1\omega_1/2}}{\sigma(\omega_1/2)^4}=0 $$

同理可证其余情况。而周期性是显然的，由式(14)立得 $$ \begin{align*} -\sigma(2z+2\omega_1)/\sigma(z+\omega_1)^4&=-\frac{\sigma(2z)e^{4\eta_1z+2\eta_1\omega_1}}{\sigma(z)^4e^{4\eta_1(z+\omega_1/2)}}\\ &=-\sigma(2z)/\sigma(z)^4 \end{align*} $$

同理可证 $\omega_2$ 也是周期，由此可得 $$ \wp'(z)=-\sigma(2z)/\sigma(z)^4 $$

What Can I Hold You With

Wed, 03 May 2023 14:30:38 GMT

What can I hold you with?

还能用什么留住你呢？

I offer you lean streets, desperate sunsets, the moon of the jagged suburbs.

我予你枯寂的街道，失落的黄昏，和残缺的月。

I offer you the bitterness of a man who has lookedlong and long at the lonely moon.

我予你我那脉脉渴望孤月的苦涩。

I offer you my ancestors, my dead men, the ghosts that living men have honoured in bronze:

我予你我逝去的先祖，他的灵魂被后人以石碑铭记。

my father's father killed in the frontier of Buenos Aires, two bullets through his lungs, bearded and dead, wrapped by his soldiers in the hide of a cow;

我祖父战死在布宜诺斯艾利斯的边境，两颗子弹穿过他的肺。他蓄着胡子死了，他的尸体被士兵用牛皮包裹；

my mother's grandfather --just twentyfour-- heading a charge of three hundred men in Peru, now ghosts on vanished horses.

我的外祖父，那时只有24岁，在秘鲁率领三百将士冲锋陷阵，如今灵魂消失于萧萧马鸣间。

I offer you whatever insight my books may hold, whatever manliness or humour my life.

我予你我的书中所有的深刻洞见，我生命中所有的刚毅坚卓和幽默风趣。

I offer you the loyalty of a man who has never been loyal.

我予你我的全然的忠诚纵使我从未获得。

I offer you that kernel of myself that I have saved, somehow --the central heart that deals not in words, traffics not with dreams, and is untouched by time, by joy, by adversities.

我予你我所尽力存留的最柔软的内心。不知为何，这内心无法以言语表达，从未被时间，被欢乐，或被苦难染指。

I offer you the memory of a yellow rose seen at sunset, years before you were born.

我予你落日下玫瑰上余晖的记忆，那时是你还未出生的遥远年岁。

I offer you explanations of yourself, theories about yourself, authentic and surprising news of yourself.

我予你你的表达，你的理论，你的真实而令人惊奇的消息。

I can give you my loneliness, my darkness, the hunger of my heart; I am trying to bribe you with uncertainty, with danger, with defeat.

我可以给你我的孤独，我的黑暗，我内心的饥渴；我试图用困惑，用危险和挫败来打动你。

Jorge Luis Borges

Elliptic Function

Tue, 02 May 2023 19:59:45 GMT

Approach from Jacobi

Jacobi considered the exact solution of Motion of Simple Pendulum. Let $ m $ be the mass of the bob at the end of the pendulum, $ a $ be the length of the pendulum, $ \theta $ be the angle of inclination which the pendulum makes with a vertical line,$ \alpha $ be the initial angle of inclination when the pendulum is released from rest position at time zero, $ t $ be the time variable, and $ g $ be the constant of the gravity of the earth. The equation of the conservation of energy is $$ \frac{1}{2}ma^2\left(\frac{d\theta}{dt}\right)-mgacos\theta=-mgacos\alpha $$

Replace $ cos\theta $ by $ 1-\dfrac{\theta^2}{2} $ to give $$ \left(\frac{d\theta}{dt}\right)^2+A\theta^2=B $$
solution is $$ t=\rho sin(\sigma \theta+\tau) $$

But what is the exact solution?

Exchange function by using double angle formula $$ \left(\frac{d\theta}{dt}\right)^2=2\frac{g}{a}(cos\theta-cos\alpha)=4\frac{g}{a}(sin^2\dfrac{\alpha}{2}-sin^2\dfrac{\theta}{2}) $$ $$ =4\frac{gsin^2\frac{\alpha}{2}}{a}(1-\left(\frac{sin\frac{\theta}{2}}{sin\frac{\alpha}{2}}\right)^2) $$

What does it mean?

$$ cos\theta\approx 1-\frac{\theta^2}{2}\space and\space cos\theta=1-2sin^2(\theta) $$ gives out the motivation. Exchange $ sin\frac{\alpha}{2} $ as $ sin\varphi $, then$$ \left(\dfrac{d\varphi}{dt}\right)^2=\dfrac{g}{a}\left(1-\sin^2\dfrac{\alpha}{2}\sin^2\varphi\right). $$ and the solution is$$ t=\sqrt{\dfrac{a}{g}}\int_{\psi=0}^{\varphi}\dfrac{d\psi}{\sqrt{1-\sin^2\frac{\alpha}{2}\sin^2\psi}}. $$ We also know $$ \int_{0}^{\frac{\pi}{2}}\sin^{2n}\varphi d\varphi=\dfrac{\pi}{2}\cdot\dfrac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots(2n)} $$ by residue theorem, we can take Taylor expansion of the solution which is $$ \begin{gathered} =2\pi\sqrt{\frac{a}{g}}\left(1+\left(\frac{1}{2}\right)^{2}k^{2}+\left(\frac{1\cdot3}{2\cdot4}\right)^{2}k^{4}+\cdots\right) \end{gathered} $$

That is where elliptic function araises

We call integrals like $$ I=\int_{z_1}^{z}\frac{dz}{\sqrt{(1-z^2)(1-k^2z^2)}} $$ as Jacobi Elliptic Function.

Abel's approach

Consider the integration of the 1-form$$ \omega:=\frac{dz}{\sqrt{(1-z^2)(1-k^2z^2)}},\quad z\in \mathbb{C} $$ function$$ \sqrt{(1-z^2)(1-k^2z^2)} $$ has four roots and it's double-valued. We want to turn it into single-valued. The idea is construct a new structure replace a single point by two points(except $ z=\pm 1,\pm 1/k $ )

on the domain obtained by removing the two branch-cuts [−1/k,−1] and [1,1/k] from the Riemann sphere $ \mathbb{P}_1 =\mathbb{C} \bigcup {\infty} $ . The two branches are the negative of each other. Any of the two branches take values of opposite sign on both edges of each of the two branch-cuts. To get the Riemann surface X we can take two copies of $ \mathbb{P}_1-\left(\left[\frac{-1}{k},-1\right]\cup\left[1,\frac{1}{k}\right]\right) $ and join them by identifying the upper edge of $ \left[\frac{-1}{k},-1\right] $ in one copy identified with the lower edge of $ \left[\frac{-1}{k},-1\right] $ in the other copy and at the same time identifying the upper edge of $ \left[1,\frac{1}{k}\right] $ in onecopy identified with the lower edge of $ \left[1,\frac{1}{k}\right] $ in the other copy.

Weierstrass $ \wp(z) $ function

We all know the three fundamental properties of $\wp(z)$

The sum of the residues of the function inside a fundamental parallelogram is zero.
The number of zeroes of the function equals the number of poles inside a fundamental parallelogram.
Inside a fundamental parallelogram the sum of the coordinates of the zeroes equals the sum of the coordinates of the poles modulo a period.

Which can be proved through Argument Principle and Residue Theorem.

Now we want to know some other properties of $ \wp $

1. Convergence

Consider $$ \wp(z)=\frac{1}{z^2}+\sum_{(n_1,n_2)\in\mathbb{Z}^2-(0,0)}\left(\frac{1}{(z-(n_1\omega_1+n_2\omega_2))^2}-\frac{1}{(n_1\omega_1+n_2\omega_2)^2}\right) $$ Its convergence can be argued by considering the infinite sum as the limit of the sequence of partial sums $$ \sum_{\ell\in L,|\ell|\leq p}\left(\frac{1}{(z-\ell)^2}-\frac{1}{\ell^2}\right) $$ as $ p \to \infty $ and using the Cauchy criterion for the convergence of a sequence as follows. The difference between the q-th and the p-th partial sums $$ \sum_{\begin{array}{c}\ell\in L,\ p\leq|\ell|\leq q\end{array}}\left(\frac{1}{(z-\ell)^2}-\frac{1}{\ell^2}\right) $$ is comparable to $$ \sum\limits_{k=p}^q\left(\sum\limits_{k-1\leq|\ell|\leq k+1}\frac{1}{|\ell|^3}\right)\sim\sum\limits_{k=p}^q k\frac{1}{k^3}=\sum\limits_{k=p}^q\frac{1}{k^2} $$ which goes to 0.

2. Taylor Expansion

$$ \begin{gathered} F(z) =\frac{1}{z^2}+3s_4z^2+5s_6z^4+\cdots, \ F'(z) =-\frac2{z^3}+6s_4z+20s_6z^3+\cdots, \ F'(z)^2 =\frac4{z^6}-\frac{24s_4}{z^2}-80s_6+\cdots, \ F(z)^3 =\frac{1}{z^6}+\frac{9s_4}{z^2}+15s_6+\cdots \end{gathered} $$

Solution to Stein Complex Analysis Chapter 9💩

Exercise 4

By rearranging the series $$ \frac{1}{z^2}+\sum\limits_{\omega\in\Lambda^*}\left[\frac{1}{(z+\omega)^2}-\frac{1}{\omega^2}\right] $$ show directly, without differentiation, that$ \wp(z + \omega) = \wp(z) $ whenever $ \omega\in \Lambda $ .

Solution

First we show $$ \begin{align*} \wp(z)&=\wp^R(z)+O(\frac{1}{R})\ &=z^{-2}+\sum_{0<|\omega|< R}((z+\omega) ^{-2}-\omega^{-2})+O(\frac{1}{R}) \end{align*} $$ Actually, $$ \begin{align*} \wp(z)=z^{-2}&+\sum_{0<|\omega|< R}((z+\omega) ^{-2}-\omega^{-2})\ &+\sum_{|\omega|> R}((z+\omega) ^{-2}-\omega^{-2}) \end{align*} $$ where $ |z|<\sqrt{R} $. Then$$ \left|((z+\omega) ^{-2}-\omega^{-2})\right|=|\frac{2z\omega+z^2}{\omega^2(z-\omega)^2}|<C\frac{1}{|\omega|^3} $$ When $ R\to \infty $, the summention can be estimated by integration$$ \sum_{|\omega|> R}((z+\omega) ^{-2}-\omega^{-2})<\sum_{n=R}^{+\infty}\sum_{n-1<|\omega|<n+1}\frac{C}{|\omega|^3}=O( \frac{1}{R}) $$ A fact is used that is $ {n − 1 ≤ |y| ≤ n + 1} $ is almost $ kn $ where $ k $ is a constant. So $ \wp(z)=\wp^R(z)+O(\frac{1}{R}) $

Then, to show $ \wp^R(z+1)-\wp^R(z) $ is $ O(\sum_{R-c<|\omega|<R+c}|\omega|^{-2})=O(\dfrac{1}{R}) $ still use the fact: $$ |\wp^R(z+1)-\wp^R(z)|<C+\sum_{R-1\leq|\omega|\leq R+1}\frac{1}{|(z+\omega)^{2}|}\sim O(\frac{1}{R}) $$ where $ |z|<\sqrt{R} $.similarly we can prove it for $ \wp^R(z+\tau)-\wp^R(z) $

Finally, let$ R\to \infty $, we get the conclusion

Exercise 20

Other examples of elliptic integrals providing conformal maps from the upper half-plane to rectangles are given below (a)The function $$ \int_0^z\frac{d\zeta}{\sqrt{\zeta(\zeta-1)(\zeta-\lambda)}},\quad\text{with}\lambda\in\mathbb{R}\text{and}\lambda\neq1 $$ maps the upper half-plane conformally to a rectangle, one of whose vertices is the image of the point at infinity.

(b) In the case $ \lambda = −1 $ , the image of $$ F(z)=\int_0^z \frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}} $$ is a square whose side lengths are $ \frac{\Gamma^2(1/4)}{2\sqrt{2\pi}} $

Solution

(a) by applying Prop. 4.1 from the chapter, which tells you about the vertices and angles of the polygon that's mapped to by the real axis.

(b)The image of $ F(z) $ is a rectangle and has four vertices $ F(1),F(\infty),F(0),F(-1) $

If $$ |F(1)-F(0)|=|F(\infty)-F(1)| $$ then it's a square.

Note $$ \begin{aligned} F(1)-F(0) &=\int_0^1\frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}=\frac1i\int_0^1\frac{d\zeta}{\sqrt{\zeta(1-\zeta^2)}}\ F(\infty)-F(1) &=\int_1^\infty\frac{d\zeta}{\sqrt{\zeta(\zeta^2-1)}}=\int_0^1\frac{dt}{\sqrt{t(1-t^2)}}\quad(t=1/\zeta)\ \end{aligned} $$ So each side of the rectangle is the same length, which means it's a square. And then $$ \begin{aligned}\int_0^1\frac{dt}{\sqrt{t(1-t^2)}}=\int_0^{\frac{\pi}{2}}\frac{d\theta}{\sqrt{\sin\theta}}=\frac{1}{2}\text{B}\left(\frac{1}{2},\frac{1}{4}\right),\end{aligned} $$ where $$ B(\alpha,\beta)=\int_0^1t^{\alpha-1}(1-t)^{\beta-1}dt=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} $$ We will use $ \Gamma(z)\Gamma\left(z+\frac{1}{2}\right)=2^{1-2z}\sqrt{\pi}\Gamma(2z) $ to derive the result.

数理统计2

Thu, 22 Dec 2022 20:19:53 GMT

今天总结几个估计，矩估计(MOM)，最大似然估计(MLE),无偏估计和UMVUE

矩估计

矩估计是利用样本的k阶矩来估计参数，由于样本均值是样本k阶矩的无偏估计，一般就直接这么写了。样本矩是总体矩的相合估计量，这是本质。

简单来说，先求出样本总体的k阶矩，解方程组就能得到参数的矩估计。

最大似然估计

要点是求出似然函数，这里分两类：离散和连续

连续版本连续很好求，你先获得 $ p(x, \theta) $, 接着把他们乘在一起即 $$ L(X, \theta)=\Pi_{i=1}^{n}p(x_i, \theta) $$ 最后再令 $$\frac{\partial log L(X,\theta )}{\partial \theta}=0$$即可
离散版本离散稍微麻烦一些，仍然是先求出 $ p(x, \theta) $。需要注意的是满足条件的 $ p(x, \theta) $ 并不唯一。之后的步骤和前面一样。

值得注意的是，有时候似然函数是严格单调函数，这时候需要考虑支撑集，利用支撑集的范围求出 MLE , 一般是顺序统计量。

UMVUE

UMVUE 的求法主要是利用 L-S 定理和 C-R 不等式，这两者都需要用到完备统计量的概念，而完备统计量又和分布族相关，

我们称一个统计量是完备的，如果它服从的分布是一个完备分布。常见的完备分布从属于指数分布族，一般有幂分布、指数分布、正态分布、二项分布(两点分布)、泊松分布、Gamma分布

拥有完备统计量的概念，我们只需要先找出完备统计量，此时也一般是充分统计量，也即

先利用 MOM 和 MLE 找到一个充分统计量，要求是无偏的(非无偏的可以修正)
再判断这个充分统计量服从的分布是不是完备分布
1. 求统计量的期望，纠偏
2. L-S，求条件期望，构造构造示性函数作为无偏估计量

概率论2——正态分布的性质总结

Sat, 03 Dec 2022 16:08:03 GMT

一维正态分布

表示为 $N( \mu $, $ \sigma^2 )$, $ p(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^2}{2\sigma^2}} $
$ P(X < x)=\Phi(x) $, $P(|X| < x)=2( \Phi(x)- \frac{1}{2} )=2 \Phi(x)-1$;
若$ X, Y $ 独立, 那么 $ X\sim N(\mu_1, \sigma_1^2)\quad Y\sim N(\mu_2, \sigma_2^2) $, 则 $$ X+Y\sim N(\mu_1+\mu_2, \sigma_1^2+\sigma_2^2)$$$$X-Y\sim N(\mu_1-\mu_2, \sigma_1^2-\sigma_2^2)$$
若 $ X\sim N(\mu, \sigma^2) $, 那么 $ X^2 $ 服从 $ \chi^2 $分布。

二维正态分布

表示为 $ N(\mu_1,\mu_2,\sigma_1^2,\sigma_2^2, \rho)$
$ p(x)=\frac{1}{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}e^{-\frac{1}{2(1-\rho^2)}(\frac{(x-\mu_1)^2}{\sigma_1^2}-2\rho\frac{(x-\mu_1)(y-\mu_2)}{\sigma_1^2\sigma_2^2}+\frac{(y-\mu_2)^2}{\sigma_2^2})}$
$$\begin{align}

E(X_2|x_1)=\mu_2+\rho\sigma_2 \frac{x_1-\mu_1}{\sigma_1}\\ Var(X_2|x_1)=(1-\rho^2)\sigma_2 \end{align}$$ 4. 设若两随机变量服从正态分布，且相关系数为 $ \rho $, 那么$$ aX+bY+c \sim N(a\mu_1+b\mu_2+c, a^2\sigma_1^2+b^2\sigma_2^2+ 2ab\rho\sigma_1\sigma_2) $$