python - 为什么即使协方差是半正定的，bivariate_normal 也会返回 NaN？

Question

我有以下正态分布点:

import numpy as np
from matplotlib import pylab as plt
from matplotlib import mlab

mean_test = np.array([0,0])
cov_test = array([[ 0.6744121 , -0.16938146],
                  [-0.16938146,  0.21243464]])

协方差矩阵是半正定的，可以作为协方差

# Semi-positive definite if all eigenvalues are 0 or 
# if there exists a Cholesky decomposition 
print np.linalg.eigvals(cov_test)
print np.linalg.cholesky(cov_test)

[0.72985988 0.15698686]

[[ 0.82122597 0.] [-0.20625439 0.41218172]]

如果我产生一些积分，我会得到:

 data_test = np.random.multivariate_normal(mean_test, cov_test, 1000)
 plt.scatter(data_test[:,0],data_test[:,1])

问题:

当我尝试绘制协方差等高线时，为什么 bivariate_normal 方法会失败(返回 NaN)？

x = np.arange(-3.0, 3.0, 0.1)
y = np.arange(-3.0, 3.0, 0.1)
X, Y = np.meshgrid(x, y)
Z = mlab.bivariate_normal(X, Y, 
                      cov_test[0,0], cov_test[1,1],
                      0, 0, cov_test[0,1])
print Z
plt.contour(X, Y, Z)

输出:

 [[ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 ..., 
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]
 [ nan  nan  nan ...,  nan  nan  nan]]

 ValueError: zero-size array to reduction operation minimum which has no identity

Answer 1

协方差矩阵的对角线是方差，但是mlab.bivariate_normal 的参数sigmax 和sigmay 是平方方差的根。改变这个:

Z = mlab.bivariate_normal(X, Y, 
                      cov_test[0,0], cov_test[1,1],
                      0, 0, cov_test[0,1])

为此:

Z = mlab.bivariate_normal(X, Y, 
                      np.sqrt(cov_test[0,0]), np.sqrt(cov_test[1,1]),
                      0, 0, cov_test[0,1])