均值向量的推断

作者: readilen | 来源:发表于2019-06-30 16:47 被阅读1次

    import numpy as np
    from operator import *
    from cytoolz import *
    from scipy import stats
    def Hotelling(X, u0_, alpha):
        n,p = X.shape
        x_ = np.array(list(map(lambda x:reduce(add, X[:,x])/n, range(X.shape[1]))))
        S = np.eye(p)
        for i in range(p):
            for j in range(p):
                if i == j:
                    S[i, j] = reduce(add, map(lambda x: pow(x-x_[i], 2), X[:,i]))/(n-1)
                else:
                    S[i, j] = reduce(add, map(lambda x: (x[0]-x_[i])*(x[1]-x_[j]), zip(X[:,i], X[:,j])))/(n-1)
        S_r = np.linalg.inv(S)
        T = n*(x_ - u0).dot(S_r).dot(x_.T - u0.T)
        F = stats.f.isf(alpha, p, n-p)
        print("u:")
        print(x_)
        print("S:")
        print(S)
        if T > F:
            print(f"拒绝H0假设  T:{T:0.3f}  >  F:{F:0.3f}")
            print('这个检验假设没通过啊')
        else:
            print(f"接受均值假设T:{T:0.3f}  <  F:{F:0.3f}")
    
    X = np.array([
        [3.7, 5.7, 3.8, 3.2, 3.1, 4.6, 2.4, 7.2, 6.7, 5.4, 3.9, 4.5, 3.5, 4.5, 1.5, 8.5, 4.5, 6.5, 4.1, 5.5], 
        [48.5, 65.1, 47.2, 53.2, 55.5, 36.1, 24.8, 33.1, 47.4, 54.1, 36.9, 58.8, 27.8, 40.2, 13.5, 56.4, 71.6, 52.8, 44.1, 40.9], 
        [9.3, 8.0, 10.9, 12.0, 9.7, 7.9, 14.0, 7.6, 8.5, 11.3, 12.7, 12.3, 9.8, 8.4, 10.1, 7.1, 8.2, 10.9, 11.2, 9.4]]).T
    
    u0 = np.array([4, 50, 10])
    Hotelling(X, u0, 0.1)
    
    u:
    [ 4.64  45.4    9.965]
    S:
    [[  2.87936842  10.01        -1.80905263]
     [ 10.01       199.78842105  -5.64      ]
     [ -1.80905263  -5.64         3.62765789]]
    拒绝H0假设  T:9.739  >  F:2.437
    这个检验假设没通过啊
    


    def confidence_ellipse(X, alpha):
        n,p = X.shape
        x_ = np.array(list(map(lambda x:reduce(add, X[:,x])/n, range(X.shape[1])))) #椭球中心
        S = np.eye(p)
        for i in range(p):
            for j in range(p):
                if i == j:
                    S[i, j] = reduce(add, map(lambda x: pow(x-x_[i], 2), X[:,i]))/(n-1)
                else:
                    S[i, j] = reduce(add, map(lambda x: (x[0]-x_[i])*(x[1]-x_[j]), zip(X[:,i], X[:,j])))/(n-1)
        S_r = np.linalg.inv(S)    
        l, e = np.linalg.eig(S)
        T = n*(x_ - u0).dot(S_r).dot(x_.T - u0.T)
        F = stats.f.isf(alpha, p, n-p)    
    
        upper = list(map(lambda x:np.sqrt(x)*np.sqrt(p*(n-1)*F/(n*(n-p))), l))# 长轴长度
        return x_, e, upper
    
    confidence_ellipse(X, 0.1)
    (array([ 4.64 , 45.4  ,  9.965]),
     array([[-0.05084144, -0.81748351, -0.57370364],
            [-0.99828352,  0.02487655,  0.05302042],
            [ 0.02907156, -0.57541452,  0.81734508]]),
     [9.05067409589578, 0.7292367091847594, 1.3607856615453258])
    
    def Joint_confidence_interval(X, alpha):
        n,p = X.shape
        x_ = np.array(list(map(lambda x:reduce(add, X[:,x])/n, range(X.shape[1])))) #椭球中心
        S = np.eye(p)
        for i in range(p):
            for j in range(p):
                if i == j:
                    S[i, j] = reduce(add, map(lambda x: pow(x-x_[i], 2), X[:,i]))/(n-1)
                else:
                    S[i, j] = reduce(add, map(lambda x: (x[0]-x_[i])*(x[1]-x_[j]), zip(X[:,i], X[:,j])))/(n-1)
        F = stats.f.isf(alpha, p, n-p)    
    
        off = list(map(lambda x:np.sqrt(p*(n-1)*F/(n-p))*np.sqrt(S[x,x]/n), range(p)))# 长轴长度
        ran = list(map(lambda x:[x_[x]-off[x],x_[x]+off[x]], range(p)))
        return ran
    
    Joint_confidence_interval(X, 0.1)
    
    def Single_confidence_interval(X, alpha):
        n,p = X.shape
        x_ = np.array(list(map(lambda x:reduce(add, X[:,x])/n, range(X.shape[1])))) #椭球中心
        S = np.eye(p)
        for i in range(p):
            for j in range(p):
                if i == j:
                    S[i, j] = reduce(add, map(lambda x: pow(x-x_[i], 2), X[:,i]))/(n-1)
                else:
                    S[i, j] = reduce(add, map(lambda x: (x[0]-x_[i])*(x[1]-x_[j]), zip(X[:,i], X[:,j])))/(n-1)
        t = stats.t.isf(alpha/2, n-1)    
    
        off = list(map(lambda x:t*np.sqrt(S[x,x]/n), range(p)))# 长轴长度
        ran = list(map(lambda x:[x_[x]-off[x],x_[x]+off[x]], range(p)))
        return ran
    
    Single_confidence_interval(X ,0.1)
    [[3.9839121839272464, 5.2960878160727525],
     [39.93489498718458, 50.86510501281542],
     [9.228578403535657, 10.701421596464343]]
    
    def Bonferroni(X, alpha):
        n,p = X.shape
        x_ = np.array(list(map(lambda x:reduce(add, X[:,x])/n, range(X.shape[1])))) #椭球中心
        S = np.eye(p)
        for i in range(p):
            for j in range(p):
                if i == j:
                    S[i, j] = reduce(add, map(lambda x: pow(x-x_[i], 2), X[:,i]))/(n-1)
                else:
                    S[i, j] = reduce(add, map(lambda x: (x[0]-x_[i])*(x[1]-x_[j]), zip(X[:,i], X[:,j])))/(n-1)
        t = stats.t.isf(alpha/(2*p), n-1)    
    
        off = list(map(lambda x:t*np.sqrt(S[x,x]/n), range(p)))# 长轴长度
        ran = list(map(lambda x:[x_[x]-off[x],x_[x]+off[x]], range(p)))
        return ran
    
    Bonferroni(X, 0.1)
    [[3.7694351029105735, 5.510564897089425],
     [38.14833571587469, 52.65166428412531],
     [8.98783996801227, 10.94216003198773]]
    

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