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Sebastian Raschka
last modified: 03/31/2014

Problem Category

  • Statistical Pattern Recognition
  • Supervised Learning
  • Parametric Learning
  • Bayes Decision Theory
  • Multivariate data (2-dimensional)
  • 2-class problem
  • different variances
  • equal prior probabilities
  • Gaussian model (2 parameters)
  • with conditional Risk (1-0 loss functions)


Sections

• Given information
• Deriving the decision boundary
• Classifying some random example data
• Calculating the empirical error rate


Given information:

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model: continuous univariate normal (Gaussian) model for the class-conditional densities

$ p(\vec{x} | \omega_j) \sim N(\vec{\mu}|\Sigma) $

$ p(\vec{x} | \omega_j) \sim \frac{1}{(2\pi)^{d/2} |\Sigma|^{1/2}} \exp{ \bigg[-\frac{1}{2} (\vec{x}-\vec{\mu})^t \Sigma^{-1}(\vec{x}-\vec{\mu}) \bigg] } $

Prior probabilities:

$ P(\omega_1) = P(\omega_2) = 0.5 $


Loss functions:

where

$ \lambda(\alpha_i|\omega_j) = \lambda_{ij}$,

the loss occured if $action_i$ is taken if the actual true class is $ \omega_j$ (assuming that $action_i$ classifies sample as $\omega_i$)

\begin{equation} \lambda = \begin{pmatrix} \lambda_{11} \quad \lambda_{12} \\ \lambda_{21} \quad \lambda_{22} \end{pmatrix} = \begin{pmatrix} 0 \quad 1 \\ 1 \quad 0 \\ \end{pmatrix}\\ \end{equation}

The samples are of 2-dimensional feature vectors:

$ \vec{x} = \bigg[ \begin{array}{c} x_1 \\ x_2 \\ \end{array} \bigg] $

Means of the sample distributions for 2-dimensional features:

$ \vec{\mu}_{\,1} = \bigg[ \begin{array}{c} 0 \\ 0 \\ \end{array} \bigg] $, $ \; \vec{\mu}_{\,2} = \bigg[ \begin{array}{c} 1 \\ 2 \\ \end{array} \bigg] $

Covariance matrices for the statistically independend and identically distributed ('i.i.d') features:

$ \Sigma_i = \bigg[ \begin{array}{cc} \sigma_{11}^2 & \sigma_{12}^2\\ \sigma_{21}^2 & \sigma_{22}^2 \\ \end{array} \bigg], \; \Sigma_1 = \bigg[ \begin{array}{cc} 1 & 0\\ 0 & 1 \\ \end{array} \bigg], \; \Sigma_1 = \bigg[ \begin{array}{cc} 2 & 0\\ 0 & 2 \\ \end{array} \bigg] $


Deriving the decision boundary

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Bayes' Rule:

$ P(\omega_j|x) = \frac{p(x|\omega_j) * P(\omega_j)}{p(x)}$

Risk Functions:

$ R(\alpha_1|\;\vec{x}) = \lambda_{11}P(\omega_1|\;\vec{x}) + \lambda_{12}P(\omega_2|\;\vec{x}) $

$ R(\alpha_2|\;\vec{x}) = \lambda_{21}P(\omega_1|\;\vec{x}) + \lambda_{22}P(\omega_2|\;\vec{x}) $


with 1-0 loss function:

$ R(\alpha_1|\; \vec{x}) = P(\omega_2|\;\vec{x}) = 1 - P(\omega_1|\;\vec{x}) $

$ R(\alpha_2|\;\vec{x}) = P(\omega_1|\;\vec{x}) = 1 - P(\omega_2|\;\vec{x}) $

Discriminant Functions:

The goal is to maximize the discriminant function, which we define as the posterior probability here to perform a minimum-error classification (Bayes classifier).

$ g_1(\vec{x}) = P(\omega_1 | \; \vec{x}), \quad g_2(\vec{x}) = P(\omega_2 | \; \vec{x})$

$ \Rightarrow g_1(\vec{x}) = P(\vec{x}|\;\omega_1) \;\cdot\; P(\omega_1) \quad | \; ln \\ \quad g_2(\vec{x}) = P(\vec{x}|\;\omega_2) \;\cdot\; P(\omega_2) \quad | \; ln $


We can drop the prior probabilities (since we have eual priors in this case):

$ \Rightarrow g_1(\vec{x}) = ln(P(\vec{x}|\;\omega_1)) \\ \quad g_2(\vec{x}) = ln(P(\vec{x}|\;\omega_2)) $

$ \Rightarrow g_1(\vec{x}) = \vec{x}^{\,t} - \frac{1}{2} \Sigma_1^{-1} \vec{x} + \bigg( \Sigma_1^{-1} \vec{\mu}_{\,1}\bigg)^t + \bigg( -\frac{1}{2} \vec{\mu}_{\,1}^{\,t} \Sigma_{1}^{-1} \vec{\mu}_{\,1} -\frac{1}{2} ln(|\Sigma_1|)\bigg) \\ \quad g_2(\vec{x}) = \vec{x}^{\,t} - \frac{1}{2} \Sigma_2^{-1} \vec{x} + \bigg( \Sigma_2^{-1} \vec{\mu}_{\,2}\bigg)^t + \bigg( -\frac{1}{2} \vec{\mu}_{\,2}^{\,t} \Sigma_{2}^{-1} \vec{\mu}_{\,2} -\frac{1}{2} ln(|\Sigma_2|)\bigg) $

Let:

$\vec{W}_{\,i} = - \frac{1}{2} \Sigma_i^{-1}\\ \vec{w}{\,i} = \bigg( \Sigma_i^{-1} \vec{\mu}_{\,i} \bigg)^t\\ \omega_{i0} = \bigg( -\frac{1}{2} \vec{\mu}_{\,i}^{\,t} \Sigma_{i}^{-1} \vec{\mu}_{\,i} -\frac{1}{2} ln(|\Sigma_i|)\bigg)$

$ \vec{W}_{\,1} = \bigg[ \begin{array}{cc} (1/4) & 0\\ 0 & (1/4) \\ \end{array} \bigg] $

$ \vec{w}_{\,1} = \bigg[ \begin{array}{cc} (1/4) & 0\\ 0 & (1/4) \\ \end{array} \bigg] \cdot \bigg[ \begin{array}{c} 0 \\ 0 \\ \end{array} \bigg] = \bigg[ \begin{array}{c} 0 \\ 0 \\ \end{array} \bigg]$

$ \omega_{10} = -\frac{1}{2} [0 \quad 0 ] \bigg[ \begin{array}{cc} (1/4) & 0\\ 0 & (1/4) \\ \end{array} \bigg] \cdot \bigg[ \begin{array}{c} 0 \\ 0 \\ \end{array} \bigg]

  • ln(2) = -ln(2)$

$ \vec{W}_{\,2} = \bigg[ \begin{array}{cc} (-1/2) & 0\\ 0 & (-1/2) \\ \end{array} \bigg] $

$ \vec{w}_{\,2} = \bigg[ \begin{array}{c} 1 \\ 2 \\ \end{array} \bigg]$

$ \omega_{20} = -\frac{1}{2} [1 \quad 2 ] \bigg[ \begin{array}{cc} (1/4) & 0\\ 0 & (1/4) \\ \end{array} \bigg] \cdot \bigg[ \begin{array}{c} 1 \\ 2 \\ \end{array} \bigg]

  • \frac{1}{2}ln(1) = -2.5$

$ \Rightarrow g_1(\vec{x}) = \vec{x}^{\,t} \bigg[ \begin{array}{cc} (1/4) & 0\\ 0 & (1/4) \\ \end{array} \bigg] \vec{x} + \bigg[ \begin{array}{c} 0 \\ 0 \\ \end{array} \bigg]^t - ln(2) = \vec{x}^{\,t} - \frac{1}{4} \vec{x} - ln(2)$

$ \Rightarrow g_2(\vec{x}) = \vec{x}^{\,t} \bigg[ \begin{array}{cc} (-1/2) & 0\\ 0 & (-1/2) \\ \end{array} \bigg] \vec{x} + \bigg[ \begin{array}{c} 1 \\ 2 \\ \end{array} \bigg]^t \vec{x} -2.5 = \vec{x}^{\,t} - \frac{1}{2} \vec{x} + [1 \quad 2] \; \vec{x} - 2.5$

Decision Boundary

$g_1(\vec{x}) = g_2(\vec{x}) $

$ \Rightarrow \vec{x}^{\,t} - \frac{1}{4} \vec{x} - ln(2) = \vec{x}^{\,t} - \frac{1}{2} \vec{x} + [1 \quad 2] \; \vec{x} - 2.5 \quad \bigg| \; \cdot \; 4$

$ \Rightarrow \vec{x}^{\,t} - \vec{x} - 4ln(2) = \vec{x}^{\,t} - 2 \vec{x}+ 4 \;( [1 \quad 2] \; \vec{x}) - 10 $

$ \Rightarrow [x_1 \quad x_2] \cdot \bigg(-\bigg[ \begin{array}{c} x_1 \\ x_2 \\ \end{array}\bigg] \bigg) -4ln(2) = [x_1 \quad x_2] \cdot \bigg(-\bigg[ \begin{array}{c} 2x_1 \\ 2x_2 \\ \end{array}\bigg] \bigg) + [4 \quad 8] \cdot \bigg[ \begin{array}{c} x_1 \\ x_2 \\ \end{array}\bigg] -10$

$ \Rightarrow -x_1^2 - x_2^2 - 4ln(2) = -2x_1^2 - 2x_2^2 + 4x_1 + 8x_2 -10 $

$ \Rightarrow x_1^2 + x_2^2 -4x_1 - 8x_2 - 4ln(2) + 10 = 0 $


Classifying some random example data

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In [1]:



    


%pylab inline

import numpy as np
from matplotlib import pyplot as plt

def decision_boundary(x_1):
    """ Calculates the x_2 value for plotting the decision boundary."""
    return 4 - np.sqrt(-x_1**2 + 4*x_1 + 6 + np.log(16))

def decision_rule(x_vec)
x_1^2 + x_2^2 - -4x_1 - 8x_2 - 4ln(2) + 10 = 0


# Generate 100 random patterns for class1
mu_vec1 = np.array([0,0])
cov_mat1 = np.array([[2,0],[0,2]])
x1_samples = np.random.multivariate_normal(mu_vec1, cov_mat1, 100)
mu_vec1 = mu_vec1.reshape(1,2).T # to 1-col vector

# Generate 100 random patterns for class2
mu_vec2 = np.array([1,2])
cov_mat2 = np.array([[1,0],[0,1]])
x2_samples = np.random.multivariate_normal(mu_vec2, cov_mat2, 100)
mu_vec2 = mu_vec2.reshape(1,2).T # to 1-col vector

# Scatter plot
f, ax = plt.subplots(figsize=(7, 7))
ax.scatter(x1_samples[:,0], x1_samples[:,1], marker='o', color='green', s=40, alpha=0.5)
ax.scatter(x2_samples[:,0], x2_samples[:,1], marker='^', color='blue', s=40, alpha=0.5)
plt.legend(['Class1 (w1)', 'Class2 (w2)'], loc='upper right') 
plt.title('Densities of 2 classes with 100 bivariate random patterns each')
plt.ylabel('x2')
plt.xlabel('x1')
ftext = 'p(x|w1) ~ N(mu1=(0,0)^t, cov1=2*I)\np(x|w2) ~ N(mu2=(1,2)^t, cov2=I)'
plt.figtext(.15,.8, ftext, fontsize=11, ha='left')

# Plot decision boundary
x_1 = np.arange(-5, 5, 0.1)
bound = decision_boundary(x_1)
plt.plot(x_1, bound, 'r--', lw=4)

x_vec = np.linspace(*ax.get_xlim())
x_1 = np.arange(0, 100, 0.05)

plt.show()



    


















    






Populating the interactive namespace from numpy and matplotlib










    






-c:8: RuntimeWarning: invalid value encountered in sqrt


Calculating the empirical error rate

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In [17]:



    


def decision_rule(x_vec):
    """ Returns value for the decision rule of 2-d row vectors """
    x_1 = x_vec[0]
    x_2 = x_vec[1]
    return x_1**2 + x_2**2 -4*x_1 - 8*x_2 - 4*np.log(2) + 10

w1_as_w2, w2_as_w1 = 0, 0

for x in x1_samples:
    if decision_rule(x) < 0:
        w1_as_w2 += 1
for x in x2_samples:
    if decision_rule(x) > 0:
        w2_as_w1 += 1

emp_err = (w1_as_w2 + w2_as_w1) / float(len(x1_samples) + len(x2_samples))
    
print('Empirical Error: {}%'.format(emp_err * 100))



    


















    






Empirical Error: 19.5%

Content source: voxlol/pattern_classification

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