最近给凯爹做的一个苦力活，统计检验这个东西说实话也挺有趣，跟算法设计一样，好的检验真的是挺难设计的，就有近似算法的那种感觉，检验很难保证size和power都很理想，所以就要做tradeoff，感觉这个假设检验的思路还是挺有趣，所以破例记录一下。

今天阳历生日（其实我每年都是过农历生日），凯爹职场情场皆得意，前脚拿到offer，后脚抱得美人归，说到底凯爹还是个挺励志的人，二战时吃那么多苦，如今终于是苦尽甘来（这不狠宰他一手哈哈哈哈哈哈哈哈

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假设检验① $H_0:A,B$是对角阵

1 生成模拟数据$X$

对于matrix normal distribution，$M{N}_{pq}(0,A,B)$，$0$代表零均值，$A,B$分别是行与列的协方差。从分布中抽取两组模拟数据，$X^{(1)}=(X_1^{(1)},...,X_{n1}^{(1)}),X^{(2)}=(X_1^{(2)},...,X_{n2}^{(2)})$，$X_1^{(1)}$维度为$p\times q$。两组数据的分布中$A$不一样，$B$一样，$n_1=n_2=n$

参数设置：p={10,20},q={10,20},n={5,8}p=\{10,20\},q=\{10,20\},n=\{5,8\}p={10,20},q={10,20},n={5,8}

矩阵$A_{p\times p}=$
$$\{\begin{array}{rl}& H_0:I\\ & H_1:\frac{I+U+\delta I}{1+\delta }\end{array}$$

矩阵$B_{q\times q}:b_{ij}=0.4^{|i-j|},1\le i,j\le q$

其中$\delta =|{\lambda }_{\min }(I+U)|+0.05$，${\lambda }_{\min }(I+U)$表示取矩阵$I+U$的最小特征根的绝对值。$U$是稀疏对称矩阵，有z={2}z=\{2\}z={2}个非零元素。有$\frac{z}{2}$个非零元素在下/上三角中（不在对角线上），服从$U(2{\left(\frac{\log p}{nq}\right)}^{1/2},4{\left(\frac{\log p}{nq}\right)}^{1/2})$均匀分布，正负随机，位置随机。

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2 虽然生成模拟数据时已知$A,B$，但假设$A,B$未知，对其进行估计。

对于每种估计方法，需要重复第3部分。

2.1 第一种估计方法：Naive

直接代入真实的$A,B$

2.2 第二种估计方法：Sample

$A_{p\times p}$的朴素估计：$\tilde{A}\propto \frac{1}{nq}{\sum }_{k=1}^n{X}_k{X}_k^{\top }$，$B_{q\times q}$的朴素估计：$\tilde{B}\propto \frac{1}{np}{\sum }_{k=1}^n{X}_k^{\top }{X}_k$，注意，此处估计值差了常数倍，不可直接调用。

当$A$已知时（用$\tilde{A}$代替），可以改进$B$的估计：
$$\hat{B}=\frac{1}{np}\sum _{k=1}^n{X}_k^{\top }{\left(\frac{\tilde{A}}{c}\right)}^{-1}{X}_k$$

$c$是一个未知常数，后续计算中会被抵消。

当$B$已知时（用$\tilde{B}$代替），可以改进$A$的估计：
$$\hat{A}=\frac{1}{nq}\sum _{k=1}^n{X}_k{\left(\frac{\tilde{A}}{c}\right)}^{-1}{X}_k^{\top }$$

$c$是一个未知常数，后续计算中会被抵消。

2.3 第三种估计方法：Banded（只可对代表时间维度的矩阵$B$使用）

只保留$\hat{B}$对角线以及两侧副对角线上的值：
$$\bar{B}=\{\begin{array}{rlrl}& {\hat{b}}_{i,j}& & \text{if }|i-j|\le 2\\ & 0& & \text{otherwise}\end{array}$$

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3 对$A,B$的估计值进行假设检验

3.1 假设检验① $H_0:B$是对角阵

记$M_n={\max }_{1\le i<j\le q}{M}_{ij},M_{ij}$为$b_{ij}$标准化后的值，$M_{ij}=\frac{{\hat{b}}_{ij}^2}{{\hat{\theta }}_{ij}/(np)}$，此处常数$c$会相互抵消，其中：
$${\hat{\theta }}_{ij}=\frac{1}{np}\sum _{k=1}^n\sum _{l=1}^p{\left[{\left(X_k^{\top }{\left(\frac{\tilde{A}}{c}\right)}^{-1/2}\right)}_{i,l}{\left(X_k^{\top }{\left(\frac{\tilde{A}}{c}\right)}^{-1/2}\right)}_{j,l}-{\hat{b}}_{i,j}\right]}^2$$

由于$M_n-4\log p+\log \log p$服从Gumble分布，设统计量${\Phi }_{\alpha }=I(M_n{\ge }_{\alpha }+4\log q-\log \log q)$，其中$q_{\alpha }=-\log (8\pi )-2\log \log (1-\alpha {)}^{-1}$

当${\Phi }_{\alpha }=1$时拒绝$B$是对角阵的原假设。

3.2 类似地，假设检验① $H_0:A$是对角阵

相当于转置$X^{(1)}$，再重复3.1操作，即，对应$p$换成$q$，$X_k^{\top }$换成$X_k$，$\tilde{A}$换成$\tilde{B}$

3.3 检验协方差$A$哪些位置不为0

$$\begin{gathered} \Psi (A)=\{(i,j):a_{i,j}\ne 0,1\le i<j\le p\} \\ \Psi (\tau =4)=\{(i,j):M_{i,j}\ge \tau p,1\le i<j\le p\} \end{gathered}$$

3.4 检验协方差$B$哪些位置不为0

$$\begin{gathered} \Psi (B)=\{(i,j):b_{i,j}\ne 0,1\le i<j\le q\} \\ \Psi (\tau =4)=\{(i,j):M_{i,j}\ge \tau q,1\le i<j\le q\} \end{gathered}$$

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4 上述1-3步骤，每组参数设置{p,q,n,z}\{p,q,n,z\}{p,q,n,z}（共8种组合）重复1000次

参数设置：p={10,20},q={10,20},n1=n2=n={5,8},z={2},α=5%p=\{10,20\},q=\{10,20\},n_1=n_2=n=\{5,8\},z=\{2\},\alpha=5\%p={10,20},q={10,20},n1​=n2​=n={5,8},z={2},α=5%，需要满足$np\ge q,nq\ge p$

论文参数设置：p={50,200},q={50,200},n1=n2=n={10,50},z={8},α=5%p=\{50,200\},q=\{50,200\},n_1=n_2=n=\{10,50\},z=\{8\},\alpha=5\%p={50,200},q={50,200},n1​=n2​=n={10,50},z={8},α=5%

4.1 对于每组参数设置，计算1000次试验后假设检验①的size

Size=P(原假设为真，拒绝原假设)=P(犯第一类错误)，即假设检验①种矩阵$A_0$被拒绝的概率，好的方法需要将size控制在0.05以内。

4.2 对于每组参数设置，计算1000次试验后假设检验①的power

Power=P(原假设为假，拒绝原假设)，即假设检验①种$A_1,B$被拒绝的概率。

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Appendix 代码实现

这个仿真实现并不难，当然最好是用matlab写，这里给出numpy的示例代码。

• 矩阵正态分布随机变量的生成可以用scipy.stats里封装好的方法，也可以用cholesky分解来做。

• 测试结果表明小规模数据上的size还行，但是power明显不太好，但是原论文的效果就很漂亮：

• 但是这个量级的参数跑起来会特别慢。

# -*- coding: UTF-8 -*-
# @author: caoyang
# @email: caoyang@163.sufe.edu.cnimport math
import time
import numpy as np
from scipy.linalg import sqrtm
from scipy import stats# Randomly generate matrix normal distributed matrix.
# M is a p-by-q matrix, U is a p-by-p matrix, and V is a q-by-q matrix.
def randomize_matrix_normal_variable(M, U, V):# X_rand = np.random.randn(*M.shape)# P = np.linalg.cholesky(U)# Q = np.linalg.cholesky(V)# return M + P @ X_rand @ Q.Treturn stats.matrix_normal(M, U, V).rvs()# Randomly generate matrix U
def randomize_matrix_u(p, q, n, z=2):temp = np.sqrt((np.log(p) / n / q))low = 2 * temphigh = 4 * tempU = np.zeros((p, p))	# initializetotal_index = [(i, j) for i in range(p) for j in range(p)]np.random.shuffle(total_index)upper_index = []lower_index = []i = 0while len(upper_index) < int(z / 2) and len(lower_index) < int(z / 2):if total_index[i][0] < total_index[i][1] and len(upper_index) < int(z / 2):upper_index.append((total_index[i][0], total_index[i][1]))lower_index.append((total_index[i][1], total_index[i][0]))elif total_index[i][0] > total_index[i][1] and len(lower_index) < int(z / 2):lower_index.append((total_index[i][0], total_index[i][1]))upper_index.append((total_index[i][1], total_index[i][0]))i += 1for upper_indice, lower_indice in zip(upper_index, lower_index):sign = 2 * np.random.randint(0, 2) - 1	# random 1 and -1 for signvalue = sign * np.random.uniform(low, high)U[upper_indice] = valueU[lower_indice] = valuereturn U# Randomly generate matrix A
def randomize_matrix_a(hypothesis, p, q, n, z=2):if hypothesis == 0:return np.eye(p)elif hypothesis == 1:U = randomize_matrix_u(p, q, n, z)delta = abs(np.min(np.linalg.eigvals(np.eye(p) + U))) + .05return (np.eye(p) + U + delta * np.eye(p)) / (1 + delta)assert False, f'Hypothesis should be 0 or 1 but got {hypothesis} !'# Randomly generate matrix B
def randomize_matrix_b(q):# return np.eye(q)return np.array([[0.4 ** (abs(i - j)) for j in range(q)] for i in range(q)])# Calculate tilde A and tilde B
def calc_tilde_A_and_B(p, q, n, sample):tilde_A = np.zeros((p, p))for X in sample:tilde_A += X @ X.Ttilde_A /= (n * q)tilde_B = np.zeros((q, q))for X in sample:tilde_B += X.T @ Xtilde_B /= (n * p)return tilde_A, tilde_B# Method 1
def estimate_method_1(A, B):hat_A = Ahat_B = Breturn hat_A, hat_B# Method 2	
def estimate_method_2(p, q, n, sample):tilde_A, tilde_B = calc_tilde_A_and_B(p, q, n, sample)hat_A = np.zeros((p, p))for X in sample:hat_A += X @ np.linalg.inv(tilde_B) @ X.That_A /= (n * q)hat_B = np.zeros((q, q))for X in sample:hat_B += X.T @ np.linalg.inv(tilde_A) @ Xhat_B /= (n * q)return hat_A, hat_B# Method 3	
def estimate_method_3(p, q, n, sample):hat_A, hat_B = estimate_method_2(p, q, n, sample)for i in range(p):for j in range(p):if abs(i - j) > 2:hat_B[i, j] = 0return hat_A, hat_B# Hypothesis 1: B is diagonal matrix
def test_B(p, q, n, sample, tilde_A, hat_B, alpha=.05, tau=4):hat_theta = np.zeros((q, q))tilde_A_inv = sqrtm(np.linalg.inv(tilde_A))for i in range(q):for j in range(q):res = 0for k in range(n):X_k_tilde_A_inv = sample[k].T @ tilde_A_invfor l in range(p):res += (X_k_tilde_A_inv[i, l] * X_k_tilde_A_inv[j, l] - hat_B[i, j]) ** 2hat_theta[i, j] = reshat_theta /= (n * p)M = n * p * hat_B * hat_B / (hat_theta)for i in range(q):for j in range(i + 1):M[i, i] = -999M_n = np.max(M)q_alpha = -np.log(8 * math.pi) - 2 * np.log(-np.log(1 - alpha))Phi_alpha = 1 * (M_n > q_alpha + 4 * np.log(q) - np.log(np.log(q)))return Phi_alpha, M# Hypothesis 2: A is diagonal matrix
def test_A(p, q, n, sample, tilde_B, hat_A, alpha=.05, tau=4):sample = list(map(lambda X: X.T, sample))return test_B(q, p, n, sample, tilde_B, hat_A, alpha)def run():p_choices = [10, 20]q_choices = [10, 20]n_choices = [5, 8]# p_choices = [50, 200]# q_choices = [50, 200]# n_choices = [10, 50]N = 1000alpha = .05z = 2tau = 4time_string = time.strftime('%Y%m%d%H%M%S')filename = f'res1-{time_string}.txt'with open(filename, 'w') as f:passfor p in p_choices:for q in q_choices:for n in n_choices:print(f'p = {p}, q = {q}, n = {n}')count_Phi_alpha_B_0 = 0count_Phi_alpha_A_0 = 0count_Phi_alpha_B_1 = 0count_Phi_alpha_A_1 = 0for _ in range(N):A_0 = randomize_matrix_a(0, p, q, n, z)A_1 = randomize_matrix_a(1, p, q, n, z)B = randomize_matrix_b(q)sample_0 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_0, B) for _ in range(n)]sample_1 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_1, B) for _ in range(n)]tilde_A_0, tilde_B_0 = calc_tilde_A_and_B(p, q, n, sample_0)tilde_A_1, tilde_B_1 = calc_tilde_A_and_B(p, q, n, sample_0)hat_A_0, hat_B_0 = estimate_method_2(p, q, n, sample_0)hat_A_1, hat_B_1 = estimate_method_2(p, q, n, sample_1)# hat_A_0, hat_B_0 = estimate_method_1(A_0, B)# hat_A_1, hat_B_1 = estimate_method_1(A_1, B)Phi_alpha_B_0, M_B0 = test_B(p, q, n, sample_0, tilde_A_0, hat_B_0, alpha, tau)Phi_alpha_A_0, M_A0 = test_A(p, q, n, sample_0, tilde_B_0, hat_A_0, alpha, tau)Phi_alpha_B_1, M_B1 = test_B(p, q, n, sample_1, tilde_A_1, hat_B_1, alpha, tau)Phi_alpha_A_1, M_A1 = test_A(p, q, n, sample_1, tilde_B_1, hat_A_1, alpha, tau)count_Phi_alpha_B_0 += Phi_alpha_B_0count_Phi_alpha_A_0 += Phi_alpha_A_0count_Phi_alpha_B_1 += Phi_alpha_B_1count_Phi_alpha_A_1 += Phi_alpha_A_1###################################### 3.3 & 3.4Psi_B0 = (B != 0) * 1												# 得到零一矩阵（可用于画热力图）Psi_tau_B0 = (M_B0 >= tau * p) * 1									# 得到零一矩阵（可用于画热力图）where_B0 = np.where(Psi_B0 == 1)print([(x, y) for (x, y) in zip(where_B0[0], where_B0[1])])			# 得到零一矩阵中元素1的坐标where_tau_B0 = np.where(Psi_tau_B0 == 1)print([(x, y) for (x, y) in zip(where_tau_B0[0], where_tau_B0[1])])	# 得到零一矩阵中元素1的坐标# ----------------------------------Psi_A0 = (A_0 != 0) * 1												# 得到零一矩阵（可用于画热力图）Psi_tau_A0 = (M_A0 >= tau * q) * 1									# 得到零一矩阵（可用于画热力图）where_A0 = np.where(Psi_A0 == 1)print([(x, y) for (x, y) in zip(where_A0[0], where_A0[1])])			# 得到零一矩阵中元素1的坐标where_tau_A0 = np.where(Psi_tau_A0 == 1)print([(x, y) for (x, y) in zip(where_tau_A0[0], where_tau_A0[1])])	# 得到零一矩阵中元素1的坐标# ----------------------------------# 以下类似Psi_B1 = (B != 0) * 1Psi_tau_B1 = (M_B1 >= tau * p) * 1# ----------------------------------Psi_A1 = (A_1 != 0) * 1Psi_tau_A1 = (M_A1 >= tau * q) * 1#####################################print('Phi_alpha_B_0: ', count_Phi_alpha_B_0)print('Phi_alpha_A_0: ', count_Phi_alpha_A_0)print('Phi_alpha_B_1: ', count_Phi_alpha_B_1)print('Phi_alpha_A_1: ', count_Phi_alpha_A_1)with open(filename, 'a') as f:f.write(f'Phi_alpha_B_0: {count_Phi_alpha_B_0}\n')f.write(f'Phi_alpha_A_0: {count_Phi_alpha_A_0}\n')		f.write(f'Phi_alpha_B_1: {count_Phi_alpha_B_1}\n')		f.write(f'Phi_alpha_A_1: {count_Phi_alpha_A_1}\n')		
run()

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假设检验② $H_0:R_A^1=R_A^2$

5 生成模拟数据$X$

数据从matrix normal distribution，$M{N}_{pq}(0,A^{(g)},B^{(g)})$中生成。$B^{(g)}$为$AR(1)$过程的自相关系数矩阵，系数为0.8和0.9.此处简化使用协方差矩阵$B_{q\times q}^g:b_{ij}=0.4^{|i-j|},1\le i,j\le q$的相关系数矩阵，同**1 生成模拟数据$X$**中一样。

$H_0$下，$A^{(1)}=A^{(2)}={\Sigma }^{(1)}=D^{1/2}{{\Sigma }^{\ast }}^{(1)}{D}^{1/2}$，其中：

$${{\Sigma }^{\ast }}^{(1)}=({\sigma }_{i,j}^{\ast (1)})=\{\begin{array}{rlrl}& 1& & \text{if }i=j\\ & 0.5& & \text{if }5(k-1)+1\le i\ne j\le 5k\text{ with }k=1,...,p/5\\ & 0& & \text{otherwise}\end{array}$$

即非零值集中在对角线附近，$D$为对焦矩阵，$d_{i,i}=Unif(0.5,2.5)$

$H_1$下，$(A^{(1)}{)}^{-1}=({\Sigma }^{(1)}+\delta I)/(1+\delta )$，$(A^{(2)}{)}^{-1}=({\Sigma }^{(1)}+U+\delta I)/(1+\delta )$，两者相差一个稀疏矩阵$U$，其中δ=∣min⁡{λmin⁡(Σ(1)),λmin⁡(Σ(1)+U)}∣+0.05\delta=|\min\{\lambda_{\min}(\Sigma^{(1)}),\lambda_{\min}(\Sigma^{(1)}+U)\}|+0.05δ=∣min{λmin​(Σ(1)),λmin​(Σ(1)+U)}∣+0.05。$U$是稀疏对称矩阵，有$z=2$个非零元素，有$z/2$个非零元素在下/上三角中（不在对角线上），服从$U(3{\left(\frac{\log p}{nq}\right)}^{1/2},5{\left(\frac{\log p}{nq}\right)}^{1/2})$的均匀分布，正负随机，位置随机。

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6 估计${\tilde{B}}^{(g)}$的3种方法

对于每种估计方法，需要重复第7部分

6.1 第一种估计方法：Naive

直接代入真实的$B^{(g)}$

6.2 第二种估计方法：Sample

$${\tilde{B}}^{(g)}=\frac{1}{n_{g}p}\sum _{k=1}^{n_g}(X_k^{(g)}{)}^{\top }{X}_k^{(g)}$$

6.3 第三种估计方法：Banded

只保留${\tilde{B}}^{(g)}$对角线以及两侧副对角线上的值：
$${\bar{B}}^{(g)}=\{\begin{array}{rlrl}& {\tilde{b}}_{i,j}^{(g)}& & \text{if }|i-j|\le 2\\ & 0& & \text{otherwise}\end{array}$$

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7 对两个协相关矩阵$A$的相关系数矩阵$R_A^{(g)}$进行假设检验

7.1 假设检验② $H_0:R_A^1=R_A^2$，即第一组和第二组的相关系数矩阵是否相同

公式太长，直接截图了：

7.2 检验相关系数矩阵$R_A^1$和$R_A^2$哪些位置不同

$$\begin{gathered} {\Psi }^{\ast }(R_A^1,R_A^2)=\{(i,j):{\hat{r}}_{i,j}^{(1)}\ne {\hat{r}}_{i,j}^{(2)},1\le i<j\le p\} \\ {\Psi }^{\ast }(\tau =4)=\{(i,j):M_{i,j}^{\ast }\ge \tau \log (p),1\le i<j\le p\} \end{gathered}$$

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8 上述5-7步骤，每组参数设置{p,q,n,z}\{p,q,n,z\}{p,q,n,z}（共8种组合），重复1000次

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代码实现

跟第一个检验的代码有很大的共通处，其实我看到第二个才知道这个检验在做什么事情，大概是自相关时间序列上的协方差和相关系数检验：

# -*- coding: UTF-8 -*-
# @author: caoyang
# @email: caoyang@163.sufe.edu.cnimport math
import time
import numpy as np
from scipy.linalg import sqrtm
from scipy import stats# Randomly generate matrix normal distributed matrix.
# M is a p-by-q matrix, U is a p-by-p matrix, and V is a q-by-q matrix.
def randomize_matrix_normal_variable(M, U, V):return stats.matrix_normal(M, U, V).rvs()# Randomly generate matrix U
def randomize_matrix_u(p, q, n, z=2):temp = np.sqrt((np.log(p) / n / q))low = 3 * temphigh = 5 * tempU = np.zeros((p, p))	# initializetotal_index = [(i, j) for i in range(p) for j in range(p)]np.random.shuffle(total_index)upper_index = []lower_index = []i = 0while len(upper_index) < int(z / 2) and len(lower_index) < int(z / 2):if total_index[i][0] < total_index[i][1] and len(upper_index) < int(z / 2):upper_index.append((total_index[i][0], total_index[i][1]))lower_index.append((total_index[i][1], total_index[i][0]))elif total_index[i][0] > total_index[i][1] and len(lower_index) < int(z / 2):lower_index.append((total_index[i][0], total_index[i][1]))upper_index.append((total_index[i][1], total_index[i][0]))i += 1for upper_indice, lower_indice in zip(upper_index, lower_index):sign = 2 * np.random.randint(0, 2) - 1	# random 1 and -1 for signvalue = sign * np.random.uniform(low, high)U[upper_indice] = valueU[lower_indice] = valuereturn U# Randomly generate matrix A
def randomize_matrix_a(hypothesis, p, q, n, z=2):Sigma_star = np.zeros((p, p))def _check(_i, _j):if _i == _j:return Falsefor _k in range(p // 5):if 5 * (_k - 1) <= _i < 5 * _k and 5 * (_k - 1) <= _j < 5 * _k:return Truereturn Falsefor i in range(p):for j in range(p):if i == j:Sigma_star[i, j] = 1elif _check(i, j):Sigma_star[i, j] = .5else:Sigma_star[i, j] = 0D = np.diag(np.random.uniform(.5, 2.5, p))D_sqrt = np.sqrt(D)Sigma = D_sqrt @ Sigma_star @ D_sqrt	if hypothesis == 0:A_1 = Sigma[:, :]A_2 = Sigma[:, :]return A_1, A_2elif hypothesis == 1:U = randomize_matrix_u(p, q, n, z)delta = abs(min(np.min(np.linalg.eigvals(Sigma + U)),np.min(np.linalg.eigvals(Sigma)),)) + .05A_1 = np.linalg.inv((Sigma + delta * np.eye(p)) / (1 + delta))A_2 = np.linalg.inv((Sigma + U + delta * np.eye(p)) / (1 + delta))# A_1 = (Sigma + delta * np.eye(p)) / (1 + delta)# A_2 = (Sigma + U + delta * np.eye(p)) / (1 + delta)return A_1, A_2assert False, f'Hypothesis should be 0 or 1 but got {hypothesis} !'# Randomly generate matrix B
def randomize_matrix_b(q):# return np.eye(q)return np.array([[0.4 ** (abs(i - j)) for j in range(q)] for i in range(q)])# Calculate tilde B
def calc_tilde_B(p, q, n, sample):tilde_B = np.zeros((q, q))for X in sample:tilde_B += X.T @ Xtilde_B /= (n * p)return tilde_B# Calculate hat A
def calc_hat_A(p, q, n, sample, tilde_B):hat_A = np.zeros((p, p))tilde_B_inv = np.linalg.inv(tilde_B)for X in sample:hat_A += X @ tilde_B_inv @ X.That_A /= (n * q)return hat_A# Calculate hat R
def calc_hat_R(p, q, n, hat_A):hat_R = np.zeros((p, p))for i in range(p):for j in range(p):hat_R[i, j] = hat_A[i, j] / np.sqrt(hat_A[i, i] * hat_A[j, j])return hat_R# Method 1
def estimate_method_1(B):hat_B = Breturn hat_B# Method 2	
def estimate_method_2(p, q, n, sample):tilde_B = calc_tilde_B(p, q, n, sample)return tilde_B# Method 3	
def estimate_method_3(p, q, n, sample):bar_B = calc_tilde_B(p, q, n, sample)for i in range(p):for j in range(p):if abs(i - j) > 2:bar_B[i, j] = 0return bar_B# Hypothesis: R_A^1 = R_A^2
def test(p, q, n, A_1, A_2, B, sample_1, sample_2, alpha=0.05, tau=4):tilde_B_1 = calc_tilde_B(p, q, n, sample_1)tilde_B_2 = calc_tilde_B(p, q, n, sample_2)# 上面是用的estimate_method_2, 可以用另外两种# tilde_B_1 = estimate_method_1(B)# tilde_B_2 = estimate_method_1(B)# tilde_B_1 = estimate_method_2(p, q, n, sample_1)# tilde_B_2 = estimate_method_2(p, q, n, sample_2)# tilde_B_1 = estimate_method_3(p, q, n, sample_1)# tilde_B_2 = estimate_method_3(p, q, n, sample_2)hat_A_1 = calc_hat_A(p, q, n, sample_1, tilde_B_1)hat_A_2 = calc_hat_A(p, q, n, sample_2, tilde_B_2)hat_R_1 = calc_hat_R(p, q, n, hat_A_1)hat_R_2 = calc_hat_R(p, q, n, hat_A_2)tilde_B_1_inv = sqrtm(np.linalg.inv(tilde_B_1))tilde_B_2_inv = sqrtm(np.linalg.inv(tilde_B_2))M = np.zeros((p, p))for i in range(p):for j in range(p):theta_1 = 0theta_2 = 0for k in range(n):X_k_tilde_B_1_inv = sample_1[k] @ tilde_B_1_invX_k_tilde_B_2_inv = sample_2[k] @ tilde_B_2_invfor l in range(q):theta_1 += (X_k_tilde_B_1_inv[i, l] * X_k_tilde_B_1_inv[j, l] - hat_A_1[i, j]) ** 2theta_2 += (X_k_tilde_B_2_inv[i, l] * X_k_tilde_B_2_inv[j, l] - hat_A_2[i, j]) ** 2theta_1 /= (n * q)theta_2 /= (n * q)pretty_theta_1 = theta_1 / hat_A_1[i, i] / hat_A_1[j, j]pretty_theta_2 = theta_2 / hat_A_2[i, i] / hat_A_2[j, j]M[i, j] = ((hat_R_1[i, j] - hat_R_2[i, j]) ** 2) / (pretty_theta_1 / n / q + pretty_theta_2 / n / q)for i in range(p):for j in range(i + 1):M[i, i] = -999M_n = np.max(M)q_alpha = -np.log(8 * math.pi) - 2 * np.log(-np.log(1 - alpha))Phi_alpha = 1 * (M_n > q_alpha + 4 * np.log(q) - np.log(np.log(q)))########################### 7.2# ------------------------Psi_star = 1 * (hat_R_1 != hat_R_2)												# 得到零一矩阵（可用于画热力图）hat_Psi_star = 1 * (M > tau * np.log(p))										# 得到零一矩阵（可用于画热力图）where_Psi_star = np.where(Psi_star == 1)									# print([(x, y) for (x, y) in zip(where_Psi_star[0], where_Psi_star[1])])			# 得到零一矩阵中元素1的坐标where_hat_Psi_star = np.where(hat_Psi_star == 1)# print([(x, y) for (x, y) in zip(where_hat_Psi_star[0], where_hat_Psi_star[1])])	# 得到零一矩阵中元素1的坐标##########################return Phi_alpha, Psi_star, hat_Psi_stardef run():p_choices = [10, 20]q_choices = [10, 20]n_choices = [5, 8]z = 2alpha = .05N = 1000tau = 4time_string = time.strftime('%Y%m%d%H%M%S')filename = f'res2-{time_string}.txt'with open(filename, 'w') as f:passfor p in p_choices:for q in q_choices:for n in n_choices:print(f'p = {p}, q = {q}, n = {n}')count_Phi_alpha_0 = 0count_Phi_alpha_1 = 0for _ in range(N):A_01, A_02 = randomize_matrix_a(0, p, q, n, z)A_11, A_12 = randomize_matrix_a(1, p, q, n, z)B = randomize_matrix_b(q)sample_01 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_01, B) for _ in range(n)]sample_02 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_02, B) for _ in range(n)]sample_11 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_11, B) for _ in range(n)]sample_12 = [randomize_matrix_normal_variable(np.zeros((p, q)), A_12, B) for _ in range(n)]# 7.2的Psi在这里返回Phi_alpha_0, Psi_star_0, hat_Psi_star_0 = test(p, q, n, A_01, A_02, B, sample_01, sample_02, alpha, tau)Phi_alpha_1, Psi_star_1, hat_Psi_star_1 = test(p, q, n, A_11, A_12, B, sample_11, sample_12, alpha, tau)count_Phi_alpha_0 += Phi_alpha_0count_Phi_alpha_1 += Phi_alpha_1print('Phi_alpha_0: ', count_Phi_alpha_0)print('Phi_alpha_1: ', count_Phi_alpha_1)with open(filename, 'a') as f:f.write(f'Phi_alpha_0: {count_Phi_alpha_0}\n')f.write(f'Phi_alpha_1: {count_Phi_alpha_1}\n')if __name__ == '__main__':run()