naive Bayes

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Why does it work?
We have not addressed the question of why does
this classifier performs well, given that the
assumptions are unlikely to be satisfied.
The linear form of the classifiers provides some hints.
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Naïve Bayes: Two Classes
•In the case of two classes we have that:
We have:
log
  w i x i  b A = 1-B; Log(B/A) = -C.
i
P(v

0
|
x
)
Then:
j
•but since
Exp(-C) = B/A =
P(v j  1 | x )  1 - P(v j  0 | x ) = (1-A)/A = 1/A – 1
= + Exp(-C) = 1/A
•We get (plug in (2) in (1); some algebra):
A = 1/(1+Exp(-C))
P(v
P(v
j
j
 1|x )
 1|x ) 
1
1  exp(-

i
w i x i  b)
•Which is simply the logistic (sigmoid) function
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Note this is a bit different than
Graphical model. It encodes
the previous linearization.
the NB independence
Rather than a single function,
assumption in the edge
here we have argmax over
structure (siblings are
several different functions.
independent given parents)
Another look at Naive Bayes
Pr(l)
l
Pr(  2 | l)
Pr(  1 | l)
Pr(  n | l)
1  2  3
x1xx21 xx2 x2 3 xx33 x 4
n
x n-1 x n
D
ˆ
c[ 0 ,l]  logPr[ 0 ,l]  log | { (x, j) | j  l } | / | S |
c[  ,l]
| { (x, j) |  (x)  1  j  l } |
D
D
ˆ
ˆ
 logPr[  ,l] / logPr[  0 ,l]  log
| { (x, j) | j  l } |
n
prediction (x)  argmax {l0.1}  c[xi ,l]  i (x)
Linear Statistical Queries
Model (Roth’99)
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Another view of Markov Models
Input:
x  ( (w 1 : t 1 ),...(w i-1 : t i-1 ), (w i : ?), (w i 1 : t i 1 )...)
States:
Observations:
t i -1
t
w i -1 w i
t i1
w i1
T
W
Assumptions: Pr(w i | t i )  Pr(w i | x)
Pr(t i 1 | t i )  Pr(t i 1 | t i , t i-1 ,..., t 1 )
Prediction: predict tT that maximizes
Pr(t | t i-1 )  Pr(w i | t)  Pr(t i 1 | t)
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Another View of Markov Models
Input:
x  ( (w 1 : t 1 ),...(w i-1 : t i-1 ), (w i : ?), (w i 1 : t i 1 )...)
States:
Observations:
t i -1
t
w i -1 w i
t i1
w i1
T
W
As for NB: features are pairs and singletons of t‘s, w’s
| { (t, t' ) | t  t 1  t'  t 2 } |
D
D
ˆ
ˆ
c[tHMM
logPmodel
r[tc1 ,[tt 21]t/2 log
Otherwise, c[  ,t]  0
is 
a linear
,t 2 ] Pr[1,t 1 ]c
[ wlog
t, t]
1 t 2 ,t 1 ]
| { (t, t' ) | t  t 1 } |
(over pairs of states and states/obs)
prediction (x)  argmax {tT}  c[xi ,t]  (x) Only 3 active features
This can be extended to an argmax that maximizes the prediction of
the whole state sequence and computed, as before, via Viterbi.
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Learning with Probabilistic Classifiers
Learning Theory
Err S (h) | { x  S | h(x)  l } | / | S |
We showed that probabilistic predictions can be viewed as predictions via
Linear Statistical Queries Models (Roth’99).
The low expressivity explains Generalization+Robustness
Is that all?
It does not explain why is it possible to (approximately) fit the data with
these models. Namely, is there a reason to believe that these hypotheses
minimize the empirical error on the sample?
In General, No. (Unless it corresponds to some probabilistic assumptions
that hold).
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Example: probabilistic classifiers
If hypothesis does not fit the
training data - augment set of
features (forget assumptions)
Pr(l)
l
Pr(  2 | l)
Pr(  1 | l)
Pr(  n | l)
1  2  3
x1xx12 x 2 x 3 x 3x 3 x 4
States:
t i -1
Observations:
n
xxnn-1 x n
t
w i -1 w i
t i1
w i1
T
W
Features are pairs and singletons of t‘s, w’s
Additional features are included
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Learning Protocol: Practice
LSQ hypotheses are computed directly:


Choose features
Compute coefficients (NB way, HMM way, etc.)
If hypothesis does not fit the training data


Augment set of features
(Assumptions will not be satisfied)
But now, you actually follow the Learning Theory Protocol:


Try to learn a hypothesis that is consistent with the data
Generalization will be a function of the low expressivity
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Robustness of Probabilistic Predictors
Remaining Question: While low expressivity explains
generalization, why is it relatively easy to fit the data?
Consider all distributions with the same marginals
Pr(  i | l )
(That is, a naïve Bayes classifier will predict the same
regardless of which distribution really generated the data.)
Garg&Roth (ECML’01):

Product distributions are “dense” in the space of all distributions.
Consequently, for most generating distributions the resulting
predictor’s error is close to optimal classifier (that is, given the correct
distribution)
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Summary: Probabilistic Modeling
Classifiers derived from probability density
estimation models were viewed as LSQ hypotheses.
 c  (x)   c  x
i
i
i
i1
x i 2 ...x i k
Probabilistic assumptions:
+ Guiding feature selection but also
- Not allowing the use of more general features.
A unified approach: a lot of classifiers, probabilistic and others can be
viewed as linear classier over an appropriate feature space.
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
What’s Next?
(1) If probabilistic hypotheses are actually like other linear
functions, can we interpret the outcome of other linear learning
algorithms probabilistically?

Yes
(2) If probabilistic hypotheses are actually like other linear
functions, can you actually train them similarly (that is,
discriminatively)?



Yes.
Classification: Logistics regression/Max Entropy
HMM: can be learned as a linear model, e.g., with a version of
Perceptron (Structured Models class)
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Recall: Naïve Bayes, Two Classes
In the case of two classes we have:
log
but since
P(v j  1 | x )
P(v j  0 | x )
P(v
j


i
w ixi  b
 1 | x )  1 - P(v
j
 0 |x )
We get (plug in (2) in (1); some algebra):
P(v
j
 1|x ) 
1
1  exp(-

i
w i x i  b)
Which is simply the logistic (sigmoid) function used in the
neural network representation.
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Conditional Probabilities
(1) If probabilistic hypotheses are actually like other linear
functions, can we interpret the outcome of other linear
learning algorithms probabilistically?

Yes
General recipe

Train a classifier f using your favorite algorithm (Perceptron, SVM,
Winnow, etc). Then:

Use Sigmoid1/1+exp{-(AwTx + B)} to get an estimate for P(y | x)
A, B can be tuned using a held out that was not used for training.
Done in LBJava, for example


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Logistic Regression
(2) If probabilistic hypotheses are actually like other linear
functions, can you actually train them similarly (that is,
discriminatively)?
The logistic regression model assumes the following model:
P(y= +/-1 | x,w)= [1+exp(-y(wTx + b)]-1
This is the same model we derived for naïve Bayes, only that
now we will not assume any independence assumption. We
will directly find the best w.
How?
Therefore training will be more difficult. However, the weight
vector derived will be more expressive.
It can be shown that the naïve Bayes algorithm cannot represent all
linear threshold functions.
 On the other hand, NB converges to its performance faster.
14
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
Logistic Regression (2)
Given the model:
P(y= +/-1 | x,w)= [1+exp(-y(wTx + b)]-1
The goal is to find the (w, b) that maximizes the log likelihood of the data:
{x1,x2… xm}.
We are looking for (w,b) that minimizes the negative log-likelihood
minw,b 1m log P(y= +/-1 | x,w)= minw,b 1m log[1+exp(-yi(wTxi + b)]
This optimization problem is called Logistics Regression
Logistic Regression is sometimes called the Maximum Entropy model in the
NLP community (since the resulting distribution is the one that has the
largest entropy among all those that activate the same features).
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Logistic Regression (3)
Using the standard mapping to linear separators through the origin, we
would like to minimize:
minw 1m log P(y= +/-1 | x,w)= minw, 1m log[1+exp(-yi(wTxi)]
To get good generalization, it is common to add a regularization term, and
the regularized logistics regression then becomes:
minw f(w) = ½ wTw + C 1m log[1+exp(-yi(wTxi)],
Empirical loss
Regularization term
Where C is a user selected parameter that balances the two terms.
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Comments on discriminative Learning
minw f(w) = ½ wTw + C 1m log[1+exp(-yiwTxi)],
Regularization term
Empirical loss
Where C is a user selected parameter that balances the two terms.
Since the second term is the loss function
Therefore, regularized logistic regression can be related to other learning
methods, e.g., SVMs.
L1 SVM solves the following optimization problem:
minw f1(w) = ½ wTw + C 1m max(0,1-yi(wTxi)
L2 SVM solves the following optimization problem:
minw f2(w) = ½ wTw + C 1m (max(0,1-yiwTxi))2
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Optimization: How to Solve
All methods are iterative methods, that generate a sequence wk that
converges to the optimal solution of the optimization problem above.
Many options within this category:


Iterative scaling: Low cost per iteration, slow convergence, updates
each w component at a time
Newton methods: High cost per iteration, faster convergence
 non-linear conjugate gradient; quasi-Newton methods; truncated Newton
methods; trust-region newton method.
 Currently: Limited memory BFGS is very popular

Stochastic Gradient Decent methods
 The runtime does not depend on n=#(examples); advantage when n is very large.
 Stopping criteria is a problem: method tends to be too aggressive at the beginning
and reaches a moderate accuracy quite fast, but it’s convergence becomes slow if
we are interested in more accurate solutions.
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Summary
(1) If probabilistic hypotheses are actually like other linear
functions, can we interpret the outcome of other linear
learning algorithms probabilistically?

Yes
(2) If probabilistic hypotheses are actually like other linear
functions, can you actually train them similarly (that is,
discriminatively)?

Yes.
 Classification: Logistic regression/Max Entropy
 HMM: can be trained via Perceptron (Structured Learning Class: Spring
2016)
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Conditional Probabilities
Data: Two class (Open/NotOpen Classifier)
0.1
The plot shows a
(normalized) histogram
of examples as a
function of the dot
product
act = (wTx + b)
and a couple other
functions of it.
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
In particular, we plot
the positive Sigmoid:
0
0.2
0.4
act
[1+exp(-(wTx
0.6
sigmoid(act)
0.8
1
e^act
-1
P(y= +1 | x,w)=
+ b)]
Is this really a probability distribution?
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Conditional Probabilities
M apping Clas s ifie r's activation to Conditional
Probability
Plotting: For example z:
y=Prob(label=1 | f(z)=x)
(Histogram: for 0.8, # (of examples
with f(z)<0.8))
Claim: Yes; If Prob(label=1 | f(z)=x) = x
Then f(z) = f(z) is a probability dist.
That is, yes, if the graph is linear.
Theorem: Let X be a RV with
distribution F.
(1) F(X) is uniformly distributed in (0,1).
(2) If U is uniform(0,1), F-1(U) is
distributed F, where F-1(x) is the value
of y s.t. F(y) =x.
Alternatively:
1
0.8
0.6
0.4
0.2
0
0
0.2
act
f(z) is a probability if: ProbU {z| Prob[(f(z)=1 · y]} = y
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0.4
0.6
sigmoid(act)
0.8
1
e^act
Plotted for SNoW (Winnow).
Similarly, perceptron; more tuning
is required for SVMs.
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