Page 285 - Schaum's Outlines - Probability, Random Variables And Random Processes

DECISION THEORY [CHAP. 8

(b) Let

Then by Eqs. (4.108) and (4.1 12), and the result of Prob. 5.60, we see that Y is a normal r.v. with zero
mean and variance lln under H, , and is a normal r.v. with mean 1 and variance l/n under H ,. Thus

Note that PI = PI,. Using Table A (Appendix A), we have
PI = PI, = 1 - @(l.ll8) = 0.1318 for n = 5
PI = PI, = 1 - O(1.581) = 0.057 for n = 10

8.16. In the binary communication system of Prob. 8.6, suppose that s,(t) and sl(t) are arbitrary
signals and that n observations of the received signal x(t) are made. Let n samples of s,(t) and
sl(t) be represented, respectively, by

So=C~Ol,S02,.-.,~onl~ and s1=Cs11,s12,...,s,,3T
where T denotes "transpose of." Determine the MAP test.
we
For each Xi, can write
1
f (xi H,) = - exp[ 4 (xi - s,,)']
(
Jz;;

Since the noise components are independent, we have

and the likelihood ratio is given by

Thus, by Eq. (8.1 5), the MAP test is given by

Taking the natural logarithm of both sides of the above expression yields