多维置信区间

Question

我有许多元组(par1,par2),即通过多次重复实验获得的二维参数空间中的点.

我正在寻找计算和可视化置信椭圆的可能性(不确定这是否是正确的术语).这是我在网上找到的一个示例图,用于显示我的意思:

来源:blogspot.ch/2011/07/classification-and-discrimination-with.html

所以原则上我必须将多元正态分布拟合到数据点的二维直方图.有人可以帮我这个吗？

Answer 1

听起来你只是想要分散点的2-sigma椭圆？

如果是这样,请考虑这样的事情(从这里的一些代码:https://github.com/joferkington/oost_paper_code/blob/master/error_ellipse.py):

import numpy as np

import matplotlib.pyplot as plt
from matplotlib.patches import Ellipse

def plot_point_cov(points, nstd=2, ax=None, **kwargs):
    """
    Plots an `nstd` sigma ellipse based on the mean and covariance of a point
    "cloud" (points, an Nx2 array).

    Parameters
    ----------
        points : An Nx2 array of the data points.
        nstd : The radius of the ellipse in numbers of standard deviations.
            Defaults to 2 standard deviations.
        ax : The axis that the ellipse will be plotted on. Defaults to the 
            current axis.
        Additional keyword arguments are pass on to the ellipse patch.

    Returns
    -------
        A matplotlib ellipse artist
    """
    pos = points.mean(axis=0)
    cov = np.cov(points, rowvar=False)
    return plot_cov_ellipse(cov, pos, nstd, ax, **kwargs)

def plot_cov_ellipse(cov, pos, nstd=2, ax=None, **kwargs):
    """
    Plots an `nstd` sigma error ellipse based on the specified covariance
    matrix (`cov`). Additional keyword arguments are passed on to the 
    ellipse patch artist.

    Parameters
    ----------
        cov : The 2x2 covariance matrix to base the ellipse on
        pos : The location of the center of the ellipse. Expects a 2-element
            sequence of [x0, y0].
        nstd : The radius of the ellipse in numbers of standard deviations.
            Defaults to 2 standard deviations.
        ax : The axis that the ellipse will be plotted on. Defaults to the 
            current axis.
        Additional keyword arguments are pass on to the ellipse patch.

    Returns
    -------
        A matplotlib ellipse artist
    """
    def eigsorted(cov):
        vals, vecs = np.linalg.eigh(cov)
        order = vals.argsort()[::-1]
        return vals[order], vecs[:,order]

    if ax is None:
        ax = plt.gca()

    vals, vecs = eigsorted(cov)
    theta = np.degrees(np.arctan2(*vecs[:,0][::-1]))

    # Width and height are "full" widths, not radius
    width, height = 2 * nstd * np.sqrt(vals)
    ellip = Ellipse(xy=pos, width=width, height=height, angle=theta, **kwargs)

    ax.add_artist(ellip)
    return ellip

if __name__ == '__main__':
    #-- Example usage -----------------------
    # Generate some random, correlated data
    points = np.random.multivariate_normal(
            mean=(1,1), cov=[[0.4, 9],[9, 10]], size=1000
            )
    # Plot the raw points...
    x, y = points.T
    plt.plot(x, y, 'ro')

    # Plot a transparent 3 standard deviation covariance ellipse
    plot_point_cov(points, nstd=3, alpha=0.5, color='green')

    plt.show()

Answer 2

请参阅如何绘制协方差误差椭圆.

这是python的实现:

import numpy as np
from scipy.stats import norm, chi2

def cov_ellipse(cov, q=None, nsig=None, **kwargs):
    """
    Parameters
    ----------
    cov : (2, 2) array
        Covariance matrix.
    q : float, optional
        Confidence level, should be in (0, 1)
    nsig : int, optional
        Confidence level in unit of standard deviations. 
        E.g. 1 stands for 68.3% and 2 stands for 95.4%.

    Returns
    -------
    width, height, rotation :
         The lengths of two axises and the rotation angle in degree
    for the ellipse.
    """

    if q is not None:
        q = np.asarray(q)
    elif nsig is not None:
        q = 2 * norm.cdf(nsig) - 1
    else:
        raise ValueError('One of `q` and `nsig` should be specified.')
    r2 = chi2.ppf(q, 2)

    val, vec = np.linalg.eigh(cov)
    width, height = 2 * sqrt(val[:, None] * r2)
    rotation = np.degrees(arctan2(*vec[::-1, 0]))

    return width, height, rotation

在Joe Kington的回答中,标准偏差的含义是错误的.通常我们使用1,2西格玛为68%,95%置信水平,但他的答案中的2西格玛椭圆不包含95%的总分布概率.正确的方法是使用卡方分布来估算椭圆大小,如帖子所示.