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112 CHAPTER 12. PROBLEM SOLUTIONS
Because the covariance matrices are not equal, it follows from Eq. (12.2-26) that
d 1 ( x ) = −
ln(0.25) −
x T
x
ln(0.25) − (x 21 + x 22 )
and
d 2 ( x ) = −
ln(4.00) −
x T
x
ln(4.00) −
(x 21 + x 22 )
where the term lnP( ω j ) was not included because it is the same for both deci- sion functions in this case. The equation of the Bayes decision boundary is
d ( x ) = d 1 ( x ) − d 2 ( x ) = 1.39 −
(x 12 + x 22 ) = 0.
(b) Figure P12.8 shows a plot of the boundary.
Problem 12.
From basic probability theory,
p (c ) =
x
p (c / x )p ( x ).
For any pattern belonging to class ω j , p (c / x ) = p ( ω j / x ). Therefore,
p (c ) =
x
p ( ω j / x )p ( x ).
Substituting into this equation the formula p ( ω j / x ) = p ( x /ω j )p ( ω j ) / p ( x ) gives
p (c ) =
x
p ( x /ω j )p ( ω j ).
Because the argument of the summation is positive, p (c ) is maximized by maxi- mizing p ( x /ω j )p ( ω j ) for each j. That is, if for each x we compute p ( x /ω j )p ( ω j ) for j = 1, 2, ..., W , and use the largest value each time as the basis for selecting the class from which x came, then p (c ) will be maximized. Since p (e ) = 1 − p (c ), the probability of error is minimized by this procedure.
Problem 12.
We start by taking the partial derivative of J with respect to w :
∂ J ∂ w
y sgn( w T^ y ) − y
where, by definition, sgn( w T^ y ) = 1 if w T^ y > 0, and sgn( w T^ y ) = −1 otherwise. Substituting the partial derivative into the general expression given in the prob- lem statement gives
w (k + 1 ) = w (k ) + c 2
y ( k ) − y ( k )sgn w ( k )T^ y ( k )
where y (k ) is the training pattern being considered at the k th iterative step. Sub- stituting the definition of the sgn function into this result yields
w (k + 1 ) = w (k ) + c
0 if w ( k )T^ y ( k ) y (k ) otherwise
where c > 0 and w ( 1 ) is arbitrary. This expression agrees with the formulation given in the problem statement.
Problem 12.
The single decision function that implements a minimum distance classifier for two classes is of the form
d (^) i j ( x ) = x T^ ( m i − m j ) −
( m Ti m i − m Tj m j ).
Thus, for a particular pattern vector x , when d (^) i j ( x ) > 0, x is assigned to class ω 1 and, when d (^) i j ( x ) < 0, x is assigned to class ω 2. Values of x for which d (^) i j ( x ) = 0 are on the boundary (hyperplane) separating the two classes. By letting w = ( m i − m j ) and w (^) n + 1 = − 12 ( m Ti m i − m Tj m j ), we can express the above decision function in the form d ( x ) = w T^ x − w (^) n + 1.
This is recognized as a linear decision function in n dimensions, which is imple- mented by a single layer neural network with coefficients
w (^) k = (m (^) i k − m (^) j k ) k = 1, 2,... , n
and
θ = w (^) n + 1 = −
( m Ti m i − m Tj m j ).
Figure P12.
as we move further down the tree). We prove that k (^) a c ≥ min[k (^) ab , k (^) b c ] by con- tradiction. For k (^) a c ≤ min[k (^) ab , k (^) b c ] to hold, shape A has to separate from shape C before (1) shape A separates from shape B, and (2) before shape B separates from shape C , otherwise k (^) ab ≤ k (^) a c or k (^) b c ≤ k (^) a c , which automatically vio- lates the condition k (^) a c < min[k (^) ab , k (^) b c ]. But, if (1) has to hold, then Fig. P12. shows the only way that A can separate from C before separating from B. This, however, violates (2), which means that the condition k (^) a c < min[k (^) ab , k (^) b c ] is vi- olated (we can also see this in the figure by noting that k (^) a c = k (^) b c which, since k (^) b c < k (^) ab , violates the condition). We use a similar argument to show that if (2) holds then (1) is violated. Thus, we conclude that it is impossible for the condition k (^) a c < min[k (^) ab , k (^) b c ] to hold, thus proving that k (^) a c ≥ min[k (^) ab , k (^) b c ] or, equivalently, that D(A,C ) ≤ max[D(A, B ), D(B,C )].
