Prediction ellipse and prediction ellipsoid¶

Marcos Duarte

To describe the dispersion of data or to make a prediction for a sample, we can calculate a prediction interval or tolerance interval. For the specific case of a univariate random variable, see Confidence and prediction intervals. For a multivariate random variable, we need to calculate a prediction hyperellipsoid (Chew, 1966):

The 95% prediction hyperellipsoid is a prediction interval for a sample of a multivariate random variable such that there is 95% of probability that a new observation will lie inside the hyperellipsoid.

The 95% prediction interval (a p-dimensional hyperellipsoid) for a sample of a multivariate random variable with $p$ dimensions and size $n$ from a population with normal distribution is given by the following probabilistic bounds which holds 95% of the time (Chew, 1966):

$$(x_{n+1}-\bar{x})'\: S^{-1} (x_{n+1}-\bar{x}) \leq \frac{F_{(0.95,\: p,\: n-p)}(n-1)\:p\:(n+1)}{n\:(n-p)}$$

Where $x_{n+i}$ (with dimension $p$) is the new observation for which we want to calculate the prediction interval; $F$ is the F-distribution; $\bar{x}$ is the sample mean (with dimension $p$), and $S$ is the sample covariance matrix defined as:

$$S_{j,k} = \frac{1}{n-1}\sum_{i=1}^{n}(x_{ij}-\bar{x}_j)(x_{ik}-\bar{x}_k)$$

This is the formula for the prediction of the next single observation where the population mean and variance are unknown (we only have estimates of these parameters from the sample). This formula is also equivalent to the formula for calculation of a tolerance region (called Type 2 tolerance region by Chew (1966)):

The 95% tolerance hyperellipsoid is a tolerance region for a sample of a multivariate random variable such that the average or expected value of the proportion of the population contained in the hyperellipsoid is 95%.

In this context, we can refer to the calculated hyperellipsoid as a prediction or tolerance hyperellipsoid (but prediction and tolerance intervals are in general different concepts in statistics).

The function hyperellipsoid.py (code at the end of this text) calculates the prediction or tolerance hyperellipsoid, some related parameters, and plots the results for a given multivariate random variable. The function signature is:

hypervol, saxes, angles, p0, R = hyperellipsoid(P, y=None, z=None, pvalue=.95,
units=None, show=True, ax=None):


Before showing how to use this code, let's look a bit more at the statistical principles behind the estimation of a prediction or tolerance interval.

Univariate variable¶

For $p$ = 1 (univariate case) the 95% prediction interval formula above reduces to:

$$\left|\frac{x_{n+1}-\bar{x}}{s}\right| \leq \sqrt{F_{(0.95,\: 1,\: n-1)}(1+1/n)}$$

Where $s$ is the square root of the sample covariance, the sample standard deviation.

And considering that $F_{(0.95,\: 1,\: n-1)}$ is equal to $T^2_{(0.975,\: n-1)}$, the square of the Student's t-distribution, finally we have:

$$\left|x_{n+1}-\bar{x}\right| \leq s\:T_{(0.975,\: n-1)}\sqrt{1+1/n}$$

Indeed, this results in the 95% prediction interval used for a sample of a univariate variable, see Confidence and prediction intervals.

For large $n$, according to the central limit theorem, we can use the fact that $T(\infty)=N$, where $N$ is the normal distribution, and the above statement can be simplified to:

$$\left|x_{n+1}-\bar{x}\right| \leq s\:N_{0.975}$$

$N_{0.975}$ is equal to 1.96, a commonly used value for calculating 95%-probability intervals.
Let's calculate the error of this approximation for different sample sizes.
First let's import the necessary Python libraries:

In [1]:
# import the necessary libraries
import numpy as np
from scipy import stats

In [2]:
F = lambda p, n: np.sqrt(stats.f.ppf(0.95, p, n-p)*(n-1)*p*(n+1)/n/(n-p))  # F distribution
N = stats.norm.ppf(0.975)                                                  # Normal distribution
for i in [1000, 100, 30, 10]:
print('\nApproximation error for n = %d' %i)
print('Using Normal distribution: %.1f%%' % (100*(N-F(1, i))/F(1, i)))

Approximation error for n = 1000
Using Normal distribution: -0.2%

Approximation error for n = 100
Using Normal distribution: -1.7%

Approximation error for n = 30
Using Normal distribution: -5.7%

Approximation error for n = 10
Using Normal distribution: -17.4%


For n=1000, the approximation is probably good enough, for n=10 it is bad, and it always underestimates.

Bivariate variable¶

For $p$ = 2 (bivariate case) the 95% prediction interval formula above reduces to (again, these probabilistic bounds hold 95% of the time (Chew, 1966):

$$(x_{n+1}-\bar{x})'\: S^{-1} (x_{n+1}-\bar{x}) \leq \frac{2(n-1)(n+1)}{n\:(n-2)}F_{(0.95,\: 2,\: n-2)}$$

This formula is also equivalent to the formula for calculation of a tolerance region such that the average or expected value of the proportion of the population contained in this region is exactly 95% (called Type 2 tolerance region by Chew (1966)).

For large $n$, according to the central limit theorem, we can use the fact that $2F(2,\infty)=\chi^2(2)$, where $\chi^2$ is the chi-square distribution, and the formula above for the 95% prediction interval for the bivariate case can be simplified to:

$$(x_{n+1}-\bar{x})'\: S^{-1} (x_{n+1}-\bar{x}) \leq \chi^2_{0.95,\: 2}$$

Or, equivalently, we can use the fact that $\sqrt{2F(2,\infty)}=R$, where $R$ is the Rayleigh distribution, and:

$$(x_{n+1}-\bar{x})'\: S^{-1} (x_{n+1}-\bar{x}) \leq (R_{0.95})^2$$

Let's calulate the error of this approximation for different sample sizes:

In [3]:
C = lambda p: np.sqrt(stats.chi2.ppf(0.95, p))  # Chi-square distribution
R = stats.rayleigh.ppf(0.95)                    # Rayleigh distribution
for i in [1000, 100, 30, 10]:
print('\nApproximation error for n = %d' %i)
print('Using Chi-square distribution: %.1f%%' % (100*(C(2)-F(2, i))/F(2, i)))
print('Using Rayleigh distribution: %.1f%%' % (100*(R-F(2, i))/F(2, i)))

Approximation error for n = 1000
Using Chi-square distribution: -0.2%
Using Rayleigh distribution: -0.2%

Approximation error for n = 100
Using Chi-square distribution: -2.5%
Using Rayleigh distribution: -2.5%

Approximation error for n = 30
Using Chi-square distribution: -8.5%
Using Rayleigh distribution: -8.5%

Approximation error for n = 10
Using Chi-square distribution: -26.3%
Using Rayleigh distribution: -26.3%


For n=1000, the approximation is probably good enough, for n=10 it is bad, and it always underestimates.
To read more about multivariate prediction and confidence intervals, see Chew (1966), and about the use of prediction ellipse in the field of posturography, see Schubert and Kirchner (2014).

A Python code to compute the hyperellipsoid¶

We can write a code to calculate the ellipse, ellipsoid, and in a more general case, the hyperellipsoid using the formula for the prediction hyperellipsoid (this code is published in Duarte (2015)). The directions and lengths of the hyperellipsoid semi-axes are found, respectively, as the eigenvectors and eigenvalues of the covariance matrix of the data using the concept of principal components analysis (PCA) or singular value decomposition (SVD) and the length of the semi-axes are adjusted to account for the necessary prediction probability. The volume of the hyperellipsoid is calculated with the same equation for the volume of a n-dimensional ball with the radius replaced by the semi-axes of the hyperellipsoid.
In Python, such code to compute the hypervolume of the 95% prediction hyperellipsoid is:

import numpy as np                            # import Numpy package
from scipy.stats import f as F                # import F distribution
from scipy.special import gamma               # import Gamma function
n, p = np.asarray(data).shape                 # 2-D array dimensions
cov = np.cov(data, rowvar=0)                  # covariance matrix of data
U, s, Vt = np.linalg.svd(cov)                 # singular value decomposition
f95 = F.ppf(.95,p,n-p)*(n-1)*p*(n+1)/n/(n-p)  # F 95 percent point function
saxes = np.sqrt(s*f95)                        # semi-axes lengths
hypervolume = np.pi**(p/2)/gamma(p/2+1)*np.prod(saxes)
hypervolume


For instance, if data has two columns, hyperellipsoid will contain the area of the 95% prediction ellipse.

Equivalent code for a Matlab-like environment¶

The equivalent code for a Matlab-like environment to compute the hypervolume of the 95% prediction hyperellipsoid is:

[n, p] = size(data);                          % 2-D array dimensions
covar = cov(data);                            % covariance matrix of data
[U, S, V] = svd(covar);                       % singular value decomposition
f95 = finv(.95,p,n-p)*(n-1)*p*(n+1)/n/(n-p);  % F 95 percent point function
saxes = sqrt(diag(S)*f95);                    % semi-axes lengths
hypervolume = pi^(p/2)/gamma(p/2+1)*prod(saxes);
hypervolume


Python function hyperellipsoid.py¶

Let's see now a more complete function, hyperellipsoid.py (code at the end of this text), to compute the prediction hyperellipsoid and related variables:

In [4]:
import matplotlib.pyplot as plt
%matplotlib inline
import sys
sys.path.insert(1, r'./../functions')  # add to pythonpath
from hyperellipsoid import hyperellipsoid


Prediction ellipse¶

For a bivariate random data with normal distribution with mean 0 and variance 1, the prediction ellipse should be a circle:

In [5]:
P = np.random.randn(10000, 2)
area, axes, angles, center, R = hyperellipsoid(P, units='cm', show=True)
print('Area =', area)
print('Semi-axes =', axes)
print('Angles =', angles)
print('Center =', center)
print('Rotation matrix =\n', R)

Area = 18.7053747592
Semi-axes = [ 2.45774169  2.42259214]
Angles = [ 50.19869787 -39.80130213]
Center = [-0.00042923 -0.0083536 ]
Rotation matrix =
[[ 0.64012716  0.76826898]
[ 0.76826898 -0.64012716]]


For other data:

In [7]:
y = np.cumsum(np.random.randn(3000)) / 50
x = np.cumsum(np.random.randn(3000)) / 100
area, axes, angles, center, R = hyperellipsoid(x, y, units='cm', show=True)
print('Area =', area)
print('Semi-axes =', axes)
print('Angles =', angles)
print('Center =', center)
print('Rotation matrix =\n', R)

Area = 1.32689696924
Semi-axes = [ 0.86222193  0.48985581]
Angles = [ 115.33248985   25.33248985]
Center = [-0.21831605 -0.07911865]
Rotation matrix =
[[-0.42787046  0.90384007]
[ 0.90384007  0.42787046]]


Prediction ellipsoid¶

For a trivariate random data with normal distribution with mean 0 and variance 1, the prediction ellipsoid should be a sphere:

In [7]:
P = np.random.randn(10000, 3)
volume, axes, angles, center, R = hyperellipsoid(P, units='cm', show=True)
print('Volume =', volume)
print('Semi-axes =', axes)
print('Angles =', angles)
print('Center =', center)
print('Rotation matrix =\n', R)

Volume = 90.9127278047
Semi-axes = [ 2.81469786  2.78797462  2.7657663 ]
Angles = [-27.57570493 -41.60225587 -40.05866733]
Center = [ 0.0008755  -0.00746681 -0.02818246]
Rotation matrix =
[[ 0.57233408  0.80570973 -0.15253041]
[-0.48124483  0.48063078  0.73307398]
[ 0.66395565 -0.34615875  0.66282502]]


For other data:

In [8]:
P = np.random.randn(1000, 3)
P[:, 2] = P[:, 2] + P[:, 1]*.5
P[:, 1] = P[:, 1] + P[:, 0]*.5
volume, axes, angles, center, R = hyperellipsoid(P, units='cm', show=True)
print('Volume =', volume)
print('Semi-axes =', axes)
print('Angles =', angles)
print('Center =', center)
print('Rotation matrix =\n', R)

Volume = 90.6457486623
Semi-axes = [ 3.79021862  2.93119572  1.94782418]
Angles = [ -57.58550257  -40.19619377  117.23828851]
Center = [ 0.02000714 -0.03737589 -0.04519863]
Rotation matrix =
[[-0.3496031  -0.72597176 -0.59223532]
[ 0.67913735  0.23908832 -0.69398071]
[ 0.64540695 -0.64482694  0.40944852]]


To be able to rotate and zoom the figure, for now you will have to plot it in a separate window:

In [9]:
# plot figure in a separate window
%matplotlib qt

In [11]:
hyperellipsoid(P, units='cm', show=True);


Let's generate an animation where the ellipsoid is rotated to better visualize it in this notebook:

In [12]:
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.animation as animation
from hyperellipsoid import _plot

fig = plt.figure(figsize=(7, 7))
ax = fig.add_axes([0, 0, 1, 1], projection='3d')
pts = ax.plot([], [], [], 'o')

volume, axes, angles, center, R = hyperellipsoid(P, units='cm', show=False, ax=None)
_plot(P, volume, axes, center, R, .95, 'cm', ax)

# animation function
def animate(i):
# rotate the point of view
ax.view_init(20, 3*i)
fig.canvas.draw()
return pts

# Animation object
anim = animation.FuncAnimation(fig, func=animate, frames=120, interval=30, blit=True)

plt.show()


This is the animation you see in a separate window when you run the code above:

To save the animation generated above as a video file (you need to have FFmpeg or ImageMagick installed):

In [13]:
anim.save('ellipsoid.gif', writer='imagemagick', fps=12, dpi=36)
# or
# anim.save('ellipsoid.mp4', writer='ffmpeg', fps=50)

In [14]:
# get back the inline plot
%matplotlib inline


Function ellipseoid.py¶

In [ ]:
# %load './../functions/hyperellipsoid.py'
"""Prediction hyperellipsoid for multivariate data."""

from __future__ import division, print_function
import numpy as np

__author__ = 'Marcos Duarte, https://github.com/demotu/BMC'
__version__ = "1.0.3"

def hyperellipsoid(P, y=None, z=None, pvalue=.95, units=None, show=True, ax=None):
"""
Prediction hyperellipsoid for multivariate data.

The hyperellipsoid is a prediction interval for a sample of a multivariate
random variable and is such that there is pvalue*100% of probability that a
new observation will be contained inside the hyperellipsoid [1]_.
The hyperellipsoid is also a tolerance region such that the average or
expected value of the proportion of the population contained in this region
is exactly pvalue*100% (called Type 2 tolerance region by Chew (1966) [1]_).

The directions and lengths of the semi-axes are found, respectively, as the
eigenvectors and eigenvalues of the covariance matrix of the data using
the concept of principal components analysis (PCA) [2]_ or singular value
decomposition (SVD) [3]_ and the length of the semi-axes are adjusted to
account for the necessary prediction probability.

The volume of the hyperellipsoid is calculated with the same equation for
the volume of a n-dimensional ball [4]_ with the radius replaced by the
semi-axes of the hyperellipsoid.

This function calculates the prediction hyperellipsoid for the data,
which is considered a (finite) sample of a multivariate random variable
with normal distribution (i.e., the F distribution is used and not
the approximation by the chi-square distribution).

Parameters
----------
P : 1-D or 2-D array_like
For a 1-D array, P is the abscissa values of the [x,y] or [x,y,z] data.
For a 2-D array, P is the joined values of the multivariate data.
The shape of the 2-D array should be (n, p) where n is the number of
observations (rows) and p the number of dimensions (columns).
y : 1-D array_like, optional (default = None)
Ordinate values of the [x, y] or [x, y, z] data.
z : 1-D array_like, optional (default = None)
Ordinate values of the [x, y] or [x, y, z] data.
pvalue : float, optional (default = .95)
Desired prediction probability of the hyperellipsoid.
units : str, optional (default = None)
Units of the input data.
show : bool, optional (default = True)
True (1) plots data in a matplotlib figure, False (0) to not plot.
Only the results for p=2 (ellipse) or p=3 (ellipsoid) will be plotted.
ax : a matplotlib.axes.Axes instance (default = None)

Returns
-------
hypervolume : float
Hypervolume (e.g., area of the ellipse or volume of the ellipsoid).
axes : 1-D array
Lengths of the semi-axes hyperellipsoid (largest first).
angles : 1-D array
Angles of the semi-axes hyperellipsoid (only for 2D or 3D data).
For the ellipsoid (3D data), the angles are the Euler angles
calculated in the XYZ sequence.
center : 1-D array
Centroid of the hyperellipsoid.
rotation : 2-D array
Rotation matrix for hyperellipsoid semi-axes (only for 2D or 3D data).

References
----------
.. [1] http://www.jstor.org/stable/2282774
.. [2] http://en.wikipedia.org/wiki/Principal_component_analysis
.. [3] http://en.wikipedia.org/wiki/Singular_value_decomposition
.. [4] http://en.wikipedia.org/wiki/Volume_of_an_n-ball

Examples
--------
>>> from hyperellipsoid import hyperellipsoid
>>> y = np.cumsum(np.random.randn(3000)) / 50
>>> x = np.cumsum(np.random.randn(3000)) / 100
>>> area, axes, angles, center, R = hyperellipsoid(x, y, units='cm')
>>> print('Area =', area)
>>> print('Semi-axes =', axes)
>>> print('Angles =', angles)
>>> print('Center =', center)
>>> print('Rotation matrix =\n', R)

>>> P = np.random.randn(1000, 3)
>>> P[:, 2] = P[:, 2] + P[:, 1]*.5
>>> P[:, 1] = P[:, 1] + P[:, 0]*.5
>>> volume, axes, angles, center, R = hyperellipsoid(P, units='cm')
"""

from scipy.stats import f as F
from scipy.special import gamma

P = np.array(P, ndmin=2, dtype=float)
if P.shape[0] == 1:
P = P.T
if y is not None:
y = np.array(y, copy=False, ndmin=2, dtype=float)
if y.shape[0] == 1:
y = y.T
P = np.concatenate((P, y), axis=1)
if z is not None:
z = np.array(z, copy=False, ndmin=2, dtype=float)
if z.shape[0] == 1:
z = z.T
P = np.concatenate((P, z), axis=1)
# covariance matrix
cov = np.cov(P, rowvar=0)
# singular value decomposition
U, s, Vt = np.linalg.svd(cov)
p, n = s.size, P.shape[0]
# F percent point function
fppf = F.ppf(pvalue, p, n-p)*(n-1)*p*(n+1)/n/(n-p)
# semi-axes (largest first)
saxes = np.sqrt(s*fppf)
hypervolume = np.pi**(p/2)/gamma(p/2+1)*np.prod(saxes)
# rotation matrix
if p == 2 or p == 3:
R = Vt
if s.size == 2:
angles = np.array([np.rad2deg(np.arctan2(R[1, 0], R[0, 0])),
else:
angles = rotXYZ(R, unit='deg')
# centroid of the hyperellipsoid
center = np.mean(P, axis=0)
else:
R, angles = None, None

if show and (p == 2 or p == 3):
_plot(P, hypervolume, saxes, center, R, pvalue, units, ax)

return hypervolume, saxes, angles, center, R

def _plot(P, hypervolume, saxes, center, R, pvalue, units, ax):
"""Plot results of the hyperellipsoid function, see its help."""

try:
import matplotlib.pyplot as plt
except ImportError:
print('matplotlib is not available.')
else:
# code based on https://github.com/minillinim/ellipsoid:
# parametric equations
u = np.linspace(0, 2*np.pi, 100)
if saxes.size == 2:
x = saxes[0]*np.cos(u)
y = saxes[1]*np.sin(u)
# rotate data
for i in range(len(x)):
[x[i], y[i]] = np.dot([x[i], y[i]], R) + center
else:
v = np.linspace(0, np.pi, 100)
x = saxes[0]*np.outer(np.cos(u), np.sin(v))
y = saxes[1]*np.outer(np.sin(u), np.sin(v))
z = saxes[2]*np.outer(np.ones_like(u), np.cos(v))
# rotate data
for i in range(len(x)):
for j in range(len(x)):
[x[i,j],y[i,j],z[i,j]] = np.dot([x[i,j],y[i,j],z[i,j]], R) + center

if saxes.size == 2:
if ax is None:
fig, ax = plt.subplots(1, 1, figsize=(5, 5))
# plot raw data
ax.plot(P[:, 0], P[:, 1], '.-', color=[0, 0, 1, .5])
# plot ellipse
ax.plot(x, y, color=[0, 1, 0, 1], linewidth=2)
# plot axes
for i in range(saxes.size):
# rotate axes
a = np.dot(np.diag(saxes)[i], R).reshape(2, 1)
# points for the axes extremities
a = np.dot(a, np.array([-1, 1], ndmin=2))+center.reshape(2, 1)
ax.plot(a[0], a[1], color=[1, 0, 0, .6], linewidth=2)
ax.text(a[0, 1], a[1, 1], '%d' % (i + 1),
fontsize=20, color='r')
plt.axis('equal')
plt.grid()
title = r'Prediction ellipse (p=%4.2f): Area=' % pvalue
if units is not None:
units2 = ' [%s]' % units
units = units + r'$^2$'
title = title + r'%.2f %s' % (hypervolume, units)
else:
units2 = ''
title = title + r'%.2f' % hypervolume
else:
from mpl_toolkits.mplot3d import Axes3D
if ax is None:
fig = plt.figure(figsize=(7, 7))
ax = fig.add_axes([0, 0, 1, 1], projection='3d')
ax.view_init(20, 30)
# plot raw data
ax.plot(P[:, 0], P[:, 1], P[:, 2], '.-', color=[0, 0, 1, .4])
# plot ellipsoid
ax.plot_surface(x, y, z, rstride=5, cstride=5, color=[0, 1, 0, .1],
linewidth=1, edgecolor=[.1, .9, .1, .4])
# ax.plot_wireframe(x, y, z, color=[0, 1, 0, .5], linewidth=1)
#                  rstride=3, cstride=3, edgecolor=[0, 1, 0, .5])
# plot axes
for i in range(saxes.size):
# rotate axes
a = np.dot(np.diag(saxes)[i], R).reshape(3, 1)
# points for the axes extremities
a = np.dot(a, np.array([-1, 1], ndmin=2))+center.reshape(3, 1)
ax.plot(a[0], a[1], a[2], color=[1, 0, 0, .6], linewidth=2)
ax.text(a[0, 1], a[1, 1], a[2, 1], '%d' % (i+1),
fontsize=20, color='r')
lims = [np.min([P.min(), x.min(), y.min(), z.min()]),
np.max([P.max(), x.max(), y.max(), z.max()])]
ax.set_xlim(lims)
ax.set_ylim(lims)
ax.set_zlim(lims)
title = r'Prediction ellipsoid (p=%4.2f): Volume=' % pvalue
if units is not None:
units2 = ' [%s]' % units
units = units + r'$^3$'
title = title + r'%.2f %s' % (hypervolume, units)
else:
units2 = ''
title = title + r'%.2f' % hypervolume
ax.set_zlabel('Z' + units2, fontsize=18)

ax.set_xlabel('X' + units2, fontsize=18)
ax.set_ylabel('Y' + units2, fontsize=18)
plt.title(title)
plt.show()

return ax

def rotXYZ(R, unit='deg'):
""" Compute Euler angles from matrix R using XYZ sequence."""

angles = np.zeros(3)
angles[0] = np.arctan2(R[2, 1], R[2, 2])
angles[1] = np.arctan2(-R[2, 0], np.sqrt(R[0, 0]**2 + R[1, 0]**2))
angles[2] = np.arctan2(R[1, 0], R[0, 0])

if unit[:3].lower() == 'deg':  # convert from rad to degree