Page 221 - Introduction to Statistical Pattern Recognition

5 Parameter Estimation 203

-~~Ah(X)rjwhcX)p”(X)dwdX jAh(X)G(h(X))i(X)dX
1
=
2K
= Ah(X)i(X)dX = 0 , (5.62)
I1 (X )=o

where we used the fact that the inverse Fourier transform of 1 is 6(.). Equation
(5.62) becomes zero because F(X) = 0 at h (X) = 0.

The second step involves showing that the second order variation of
(5.61) is positive regardless of Ah(X).
-
1
=
~~~Ah’(X)eiwhcx)p(X)d~dX -jAh2(X)- dF(h) p(X)dX . (5.63)
2n 2 dh
The derivative of the unit impulse, d6(h)/dh, is zero except in the region very
close to h (X) = 0, where d6(h)/dh > 0 for h < 0 and dS(h)/dh < 0 for h > 0.
On the other hand, i(X) > 0 for h < 0 and i(X) c 0 for h > 0. Since
Ah2(X) > 0 regardless of Ah(X), (5.63) is always positive.

Bias: The expected value of 2, E, with respect to the design samples is

=&+E. (5.64)

Then, the bias, G, may be approximated by

jo
-jjEd [ Ah(X) + -Ah2(X)}ejwh(X)i(X)dodX . (5.65)
1
2n 2
When h is a function of q parameters, y ,, . . . ,yq, Ah is, from (5.1)

(5.66)