Question

要生成具有多变量t分布的样本，我使用此函数：

def multivariatet(mu,Sigma,N,M):
    '''
    Output:
    Produce M samples of d-dimensional multivariate t distribution
    Input:
    mu = mean (d dimensional numpy array or scalar)
    Sigma = scale matrix (dxd numpy array)
    N = degrees of freedom
    M = # of samples to produce
    '''
    d = len(Sigma)
    g = np.tile(np.random.gamma(N/2.,2./N,M),(d,1)).T
    Z = np.random.multivariate_normal(np.zeros(d),Sigma,M)
    return mu + Z/np.sqrt(g)

但我现在正在寻找的是multivariate student t-distribution它自己，所以我可以计算dimension > 1的元素密度。

这将类似于包scipy的stats.t.pdf(x, df, loc, scale)，但在多维空间中。

Answer 1

我自己编码密度：

import numpy as np
from math import *

def multivariate_t_distribution(x,mu,Sigma,df,d):
    '''
    Multivariate t-student density:
    output:
        the density of the given element
    input:
        x = parameter (d dimensional numpy array or scalar)
        mu = mean (d dimensional numpy array or scalar)
        Sigma = scale matrix (dxd numpy array)
        df = degrees of freedom
        d: dimension
    '''
    Num = gamma(1. * (d+df)/2)
    Denom = ( gamma(1.*df/2) * pow(df*pi,1.*d/2) * pow(np.linalg.det(Sigma),1./2) * pow(1 + (1./df)*np.dot(np.dot((x - mu),np.linalg.inv(Sigma)), (x - mu)),1.* (d+df)/2))
    d = 1. * Num / Denom 
    return d

Answer 2

这将通过d数据矩阵X评估多元学生-T分布的log pdf：

from scipy.special import gamma
from numpy.linalg import slogdet

def multivariate_student_t(X, mu, Sigma, df):    
    #multivariate student T distribution

    [n,d] = X.shape
    Xm = X-mu
    V = df * Sigma
    V_inv = np.linalg.inv(V)
    (sign, logdet) = slogdet(np.pi * V)

    logz = -gamma(df/2.0 + d/2.0) + gamma(df/2.0) + 0.5*logdet
    logp = -0.5*(df+d)*np.log(1+ np.sum(np.dot(Xm,V_inv)*Xm,axis=1))

    logp = logp - logz            

    return logp

Answer 3

我概括了@ farhawa的代码以允许x中的多个条目（我发现我想一次查询多个点）。

import numpy as np
from math import gamma

def multivariate_t_distribution(x, mu, Sigma, df):
    '''
    Multivariate t-student density. Returns the density
    of the function at points specified by x.

    input:
        x = parameter (n-d numpy array; will be forced to 2d)
        mu = mean (d dimensional numpy array)
        Sigma = scale matrix (dxd numpy array)
        df = degrees of freedom

    Edited from: http://stackoverflow.com/a/29804411/3521179
    '''

    x = np.atleast_2d(x) # requires x as 2d
    nD = Sigma.shape[0] # dimensionality

    numerator = gamma(1.0 * (nD + df) / 2.0)

    denominator = (
            gamma(1.0 * df / 2.0) * 
            np.power(df * np.pi, 1.0 * nD / 2.0) *  
            np.power(np.linalg.det(Sigma), 1.0 / 2.0) * 
            np.power(
                1.0 + (1.0 / df) *
                np.diagonal(
                    np.dot( np.dot(x - mu, np.linalg.inv(Sigma)), (x - mu).T)
                ), 
                1.0 * (nD + df) / 2.0
                )
            )

    return 1.0 * numerator / denominator

Answer 4

我尝试了上述答案并得到了不同的结果，但我不确定为什么/可能出错。以下是我基于高斯混合的scikit-learn code，我认为有效（对于任意大小的输入numpy数组X，以及 c t-distribution与list中的参数意味着和covars）：

import numpy as np
from scipy import linalg
try:  # SciPy >= 0.19
    from scipy.special import gammaln as sp_gammaln
except ImportError:
    from scipy.misc import gammaln as sp_gammaln

def log_multivariate_t_density(X, means, covars, nu = 1):
    n_samples, n_dim = X.shape
    nmix = len(means)
    log_prob = np.empty((n_samples, nmix))
    for c, (mu, cv) in enumerate(zip(means, covars)):
        try:
            cv_chol = linalg.cholesky(cv, lower=True)
        except linalg.LinAlgError:

            try:
                cv_chol = linalg.cholesky(cv + min_covar * np.eye(n_dim),
                                  lower=True)
            except linalg.LinAlgError:
                raise ValueError("'covars' must be symmetric, "
                         "positive-definite")

        cv_log_det = 2 * np.sum(np.log(np.diagonal(cv_chol)))
        cv_sol = linalg.solve_triangular(cv_chol, (X - mu).T, lower=True).T

        norm = (sp_gammaln((nu + n_dim) / 2.) - sp_gammaln(nu / 2.)
                - 0.5 * n_dim * np.log(nu * np.pi))
        inner = - (nu + n_dim) / 2. * np.log1p(np.sum(cv_sol ** 2, axis=1) / nu)
        log_prob[:, c] = norm + inner - cv_log_det

    return log_prob

使用python进行多变量学生t分布

4 个答案: