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Detection, Estimation modulation theory part 1 - Vantress H.

Vantress H. Detection, Estimation modulation theory part 1 - Wiley & sons , 2001. - 710 p.
ISBN 0-471-09517-6 Previous << 1 .. 15 16 17 18 19 20 < 21 > 22 23 24 25 26 27 .. 254 >> Next LRT ó ł PF Pd
0 0 1 1
1 1 1 — e~mo 1 — e~ml
2 1 5
1 ......1 ¦ 1 Î 1 1 O | 1 u 1 |D /Ň
a 4 3 /
_ 5 ok / -
? X /
6 \ /
_ \ / -
7 î /
7 2 /
_ / -
? /
8 /
--- / -
î /
3 /
- / -
/
/
-. / ---
4 /
/
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/
\ î mo = 2, mi = 4 _
? mo = 4, mi = 10

.. r
\
\
\
1
/ 1 1 1 1 1 1 ł 1 l .
0 0.2 0.4 0.6 0.8 1.0
Pf------>
Fig. 2.11 (a) Receiver operating characteristic, Poisson problem.
Performance: Receiver Operating Characteristic 43
Pf ^
Fig. 2.11 (6) Receiver operating characteristic with randomized decision rule.
To get the desired value of PF we use LRT No. 0 with probability ł and LRT No. 1 with probability i. The test is
If n = 0, say Hi with probability i,
say Ho with probability 1, n > 1 say H\.
This procedure, in which we mix two likelihood ratio tests in some probabilistic manner, is called a randomized decision rule. The resulting PD is simply a weighted combination of detection probabilities for the two tests.
PD = 0.5(1) + 0.5(1 - e~mi) = (1 - 0.5e"mi). (86)
We see that the ROC for randomized tests consists of straight lines which connect the points in Fig. 2.11a, as shown in Fig. 2.116. The reason that we encounter a randomized test is that the observed random variables are discrete. Therefore A(R) is a discrete random variable and, using an ordinary likelihood ratio test, only certain values of PF are possible.
44 2.2 Simple Binary Hypothesis Tests
Looking at the expression for PF in (56) and denoting the threshold by 77, we have
If PF(rj) is a continuous function of 77, we can achieve a desired value from 0 to 1 by a suitable choice of 77 and a randomized test will never be needed. This is the only case of interest to us in the sequel (see Prob. 2.2.12).
With these examples as a background, we now derive a few general properties of receiver operating characteristics. We confine our discussion to continuous likelihood ratio tests.
Two properties of all ROC’s follow immediately from this example.
Property 1. All continuous likelihood ratio tests have ROC’s that are concave downward. If they were not, a randomized test would be better. This would contradict our proof that a LRT is optimum (see Prob. 2.2.12).
Property 2. All continuous likelihood ratio tests have ROC’s that are above the PD — PF line. This is just a special case of Property 1 because the points (PF = 0, PD = 0) and (PF = 1, PD = 1) are contained on all ROC’s.
Property 3. The slope of a curve in a ROC at a particular point is equal to the value of the threshold 77 required to achieve the PD and PF of that point.
Proof.
Differentiating both expressions with respect to 77 and writing the results as a quotient, we have
(88)
dPp/dr) = ~Pa\h1(v\^i) = dPD dPp/diq ~Pa\h0(t]\Hq) dP f
(89)
We now show that
Pa\h1(7}\^i) __ Đë\í0(ó\Íî) ^
(90)
Let
(91)
Then
PM A pr {A(R) > rjlHJ = Pr^m dR
Performance: Receiver Operating Characteristic 45
where the last equality follows from the definition of the likelihood ratio. Using the definition of Q(rj), we can rewrite the last integral
PDfo)=f A(R)Pr,Ho(R\H0)dR= Ă XpA\Ho(X\H0)dX. (93)
Jr\
Differentiating (93) with respect to 77, we obtain
= — VPa\h0(i\Ho)- (94)
Equating the expression for dPD(n)\dr\ in the numerator of (89) to the right side of (94) gives the desired result.
We see that this result is consistent with Example 1. In Fig. 2.9˙, the curves for nonzero d have zero slope at PF = PD = 1 (77 = 0) and infinite slope at PF = PD = 0 (77 = 00). Previous << 1 .. 15 16 17 18 19 20 < 21 > 22 23 24 25 26 27 .. 254 >> Next 