ECE 407: PATTERN RECOGNITION
PROFESSOR: Dr. J. Ben-Arie
KL PROBLEM SOLUTION
Problem a
The mean and covariance.
mx = [2 1]T ;
Cx =
1
4
10 3
3 2
(1)
Problem b
1 = 2.75 ;
e1 = 1 [3 1]T
10
2 = 0.25 ;
e2 = 1 [1 3]T
10
3 other combinations of solutions with the
ECE 407: Pattern Recognition I
Prof. J. Ben-Arie
HW 9 Solutions
32. We seek to evaluate f (I) = 22:01 air, at a point :t' Where the n coefcients al
are given.
_ (a) A straightforward @(nz) algorithm is:
Algorithm 0 (Basic. polynomial evaluation)
I beg
ECE 407: Pattern Recognition I
Prof. J. Ben-Arie
HW 4 Solutions
31. Recall that from the Bayes rule, to minimize the probability of error, the decision
boundary is given according to:
Choose wl if P(w1|:z:) > P(w211:); otherwise choose 012.
(a) In this
Chapter 11, Solution 3.
I
+
90 F
C
1600
R
1
1
j 5.5556
6
j C j 90 x10 x 2 x103
I = 160/60 = 2.667A
The average power delivered to the load is the same as the average power absorbed by
the resistor which is
P avg = 0.5|I|260 = 213.4 W.
Chapter 11, Solutio
Solutions to Selected Problems In:
Pattern Classication
by Duda, Hart, Stork
John L. Weatherwax
February 24, 2008
Problem Solutions
Chapter 2 (Bayesian Decision Theory)
Problem 11 (randomized rules)
Part (a): Let R(x) be the average risk given measurement