multidimensional confidence intervals


Question

I have numerous tuples (par1,par2), i.e. points in a 2 dimensional parameter space obtained from repeating an experiment multiple times.

I'm looking for a possibility to calculate and visualize confidence ellipses (not sure if thats the correct term for this). Here an example plot that I found in the web to show what I mean:

enter image description here

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

So in principle one has to fit a multivariate normal distribution to a 2D histogram of data points I guess. Can somebody help me with this?

1
19
9/6/2012 1:20:39 PM

Accepted Answer

It sounds like you just want the 2-sigma ellipse of the scatter of points?

If so, consider something like this (From some code for a paper here: 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()

enter image description here

36
9/7/2012 3:39:26 PM

Refer the post How to draw a covariance error ellipse.

Here's the python realization:

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

The meaning of standard deviation is wrong in the answer of Joe Kington. Usually we use 1, 2 sigma for 68%, 95% confidence levels, but the 2 sigma ellipse in his answer does not contain 95% probability of the total distribution. The correct way is using a chi square distribution to esimate the ellipse size as shown in the post.


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