-
Notifications
You must be signed in to change notification settings - Fork 4
/
Copy path05_naive_bayes.tex
executable file
·275 lines (209 loc) · 9.01 KB
/
05_naive_bayes.tex
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
\section{Na{\"i}ve Bayes}
\subsection{Independency Assumptions}
\begin{frame}
\frametitle{Na{\"i}ve Bayes and Statistical Independency}
\structure{Na{\"i}ve Bayes is}
\begin{itemize}
\item still widely (and successfully) used \\[.5cm]
\item often outperforming much more advanced classifiers \\[.5cm]
\item appropriate in the presence of high dimensional features\\ (curse of dimensionality) \\[.5cm]
\item also called \structure{``Idiot's Bayes''}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes and Statistical Independency \cont}
For the class dependent pdf we can do the following factorization:
%
\begin{eqnarray*}
p(\vec x| y) &=& p(x_1, x_2, \dots, x_d |y) \\[.3cm] \pause
&=& p(x_1|y)p(x_2, x_3,\dots, x_d | y, x_1) \\[.3cm] \pause
&=& p(x_1|y)p(x_2|y,x_1)p(x_3,x_4,\dots, x_d | y, x_1, x_2) \\[.3cm] \pause
&=& p(x_1|y)\prod_{i=2}^d p(x_i|y, x_1,\dots,x_{i-1})
\end{eqnarray*}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes and Statistical Independency \cont}
\begin{itemize}
\item The Na{\"i}ve Bayes classifier makes a very strong -- so to call \structure{na{\"i}ve} -- \structure{independency assumption}. \\[.5cm]
\item All $d$ components of the feature vector $\vec x$ are assumed to be mutually independent. \\[.5cm] \pause
\item This independency assumption implies:
\begin{eqnarray*}
p(\vec x| y) &=& \prod_{i=1}^d p(x_i|y)
\end{eqnarray*}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes and Statistical Independency \cont}
The decision rule of na{\"i}ve Bayes reads as follows:
\begin{eqnarray*}
y^* &=& \argmax\limits_{y} p(y|\vec x) \\ \pause
&=& \argmax\limits_{y} p(y) p(\vec x| y) \\ \pause
&=& \argmax\limits_{y} p(y) \prod_{i=1}^d p(x_i | y)
\end{eqnarray*}
\end{frame}
\subsection{An Example: Gaussians}
\begin{frame}
\frametitle{An Example: Na{\"i}ve Bayes and Gaussians}
\begin{ovalblock}{Example}
Assume the $100$--dimensional feature vector $\vec x\in \real^{100}$ belonging to class $y$ is normally distributed and all components are {\em mutually dependent}:
\begin{eqnarray*}
\vec\mu_y &\in& \real^{100} \\
\mat\Sigma &=& \mat\Sigma^T \in \real^{100\times 100}
\end{eqnarray*}
The total number of parameters to be estimated for each class is \pause $$100+100\cdot (100+1)/2= 5150.$$
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{An Example: Na{\"i}ve Bayes and Gaussians \cont}
\begin{ovalblock}{Example cont.}
Assume the $100$--dimensional feature vector $\vec x\in \real^{100}$ belonging to class $y$ is normally distributed and all components are \emph{mutually independent}.
\begin{displaymath}
p(\vec x| y) = \prod_{i=1}^{100} p(x_i | y) \quad = \quad \prod_{i=1}^{100} {\cal N}(x_i;\mu_i, \sigma^2_i).
\end{displaymath}
For each component $i=\{1,2,3,\dots, 100\}$ we have to estimate mean $\mu_i\in \real$ and variance $\sigma_i^2\in \real$.
The total number of parameters to be estimated for each class is \pause $$100+100= 200.$$
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{An Example: Na{\"i}ve Bayes and Gaussians \cont}
\begin{ovalblock}{Example cont.}
\begin{figure}
\resizebox{.75\linewidth}{!}{
\input{\texfigdir/naive_bayes_parameters.pstex_t}
}
\end{figure}
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{An Example: Na{\"i}ve Bayes and Gaussians \cont}
\begin{ovalblock}{Example cont.}
\begin{figure}
\resizebox{.5\linewidth}{!}{
\input{\texfigdir/gaussian1.pstex_t}
}
\caption{Quadratic decision boundary that considers statistical dependency}
\end{figure}
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{An Example: Na{\"i}ve Bayes and Gaussians \cont}
\begin{ovalblock} {Example cont.}
\begin{figure}
\resizebox{.5\linewidth}{!}{
\input{\texfigdir/gaussian4.pstex_t}
}
\caption{Quadratic decision boundary assuming independency of $x_1$ and $x_2$}
\end{figure}
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes}
Let us consider the \structure{logit transform}
\footnotesize
\begin{eqnarray*}
\log \frac{p(y=0|\vec x)}{p(y=1|\vec x)}
&=& \pause \log\frac{p(y=0) p(\vec x | y=0)}{p(y=1) p(\vec x | y=1)}\\[.3cm] \pause
&=& \log\frac{p(y=0)}{p(y=1)} + \log\frac{p(\vec x | y=0)}{p(\vec x | y=1)}\\[.3cm] \pause
&=& \log\frac{p(y=0)}{p(y=1)} + \log\frac{\prod_{i=1}^d p(x_i | y=0)}{\prod_{i=1}^d p(x_i | y=1)} \\[.3cm] \pause
&=& \underbrace{\alpha_0 + \sum_{i=1}^d \alpha_{0,i}( x_i)}_{\mbox {\structure{generalized additive model}}}
\end{eqnarray*}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes \cont}
\begin{center}
Is there anything between Bayes and Na{\"i}ve Bayes?
\end{center}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes \cont}
There are multiple techniques to beat the curse of dimensionality, \\
for example:
\begin{itemize}
\item Reduction of the parameter space\\
\begin{itemize}
\item Introduction of independency assumptions \\
(from complete dependency to mutual independency)
\item Parameter tying
\end{itemize}
\item Reduction of the dimension of the feature vectors
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes \cont}
First order dependency
\begin{eqnarray*}
p(\vec x| y) &=& p(x_1, x_2, \dots, x_d |y) \\[.3cm] \pause
&=& p(x_1|y)p(x_2, x_3,\dots, x_d | y, x_1) \\[.3cm] \pause
&=& p(x_1|y)p(x_2|y,x_1)p(x_3,x_4,\dots, x_d | y, x_1, x_2) \\[.3cm] \pause
&=& p(x_1|y)\prod_{i=2}^d p(x_i|y, x_{i-1})
\end{eqnarray*}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes \cont}
\begin{ovalblock}{Example}
\structure{First order dependency} in a Gaussian random vector can be identified through the covariance matrix $\mat\Sigma$. It has the following structure:
\begin{eqnarray*}
\mat\Sigma &=& \left(
\begin{array}{ccccccc}
\sigma_{1,1} & \sigma_{2,1} & 0 & 0 & \cdots & 0 & 0 \\
\sigma_{2,1} & \sigma_{2,2} & \sigma_{3,2} & 0 & \cdots & 0 & 0 \\
0 & \sigma_{3,2} & \sigma_{3,3} & \sigma_{4,3} & \cdots & 0 & 0 \\
0 & 0 & \sigma_{4,3} & \sigma_{4,4} & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \sigma_{d,d-1} \\
0 & 0 & 0 & 0 & \cdots & \sigma_{d,d-1} & \sigma_{d,d} \\
\end{array}
\right)
\end{eqnarray*}
\end{ovalblock}
\end{frame}
\begin{frame}
\frametitle{Na{\"i}ve Bayes \cont}
\begin{ovalblock}{Example}
First order dependency in Gaussian random vector with \\
\structure{tied diagonal elements}, i.\,e.\ $\sigma_{i,i}=\sigma$:
\begin{eqnarray*}
\mat\Sigma &=& \left(
\begin{array}{ccccccc}
\sigma & \sigma_{2,1} & 0 & 0 & \cdots & 0 & 0 \\
\sigma_{2,1} & \sigma & \sigma_{3,2} & 0 & \cdots & 0 & 0 \\
0 & \sigma_{3,2} & \sigma & \sigma_{4,3} & \cdots & 0 & 0 \\
0 & 0 & \sigma_{4,3} & \sigma & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \sigma_{d,d-1} \\
0 & 0 & 0 & 0 & \cdots & \sigma_{d,d-1} & \sigma \\
\end{array}
\right)
\end{eqnarray*}
\end{ovalblock}
\end{frame}
\subsection{Lessons Learned}
\begin{frame}
\frametitle{Lessons Learned}
\begin{itemize}
\item Na{\"i}ve Bayes is rather successful. \\[.5cm]
\item Na{\"i}ve Bayes does not require a huge set of training data. \\[.5cm]
\item Statistical dependency vs. dimension of the search space. \\[.5cm]
\item Na{\"i}ve Bayes: give it a try!
\end{itemize}
\end{frame}
\input{nextTime.tex}
\subsection{Further Readings}
\begin{frame}
\frametitle{Further Readings}
\begin{itemize}
\item Brian D. Ripley: \\
\structure{Pattern Recognition and Neural Networks}, \\
Cambridge University Press, Cambridge, 1996.\\[.3cm]
\item Christopher M.\ Bishop: \\
\structure{Pattern Recognition and Machine Learning}, \\
Springer, New York, 2006
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Comprehensive Questions}
\begin{itemize}
\item What is the assumption of Na{\"i}ve Bayes? \\[1cm]
\item How does the assumption affect the class dependent pdf? \\[1cm]
\item What is the structure of the covariance matrix of normal-distributed classes in Na{\"i}ve Bayes? \\[1cm]
\item How can Na{\"i}ve Bayes be extended to first-order statistical dependencies?
\end{itemize}
\end{frame}