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Pr (e) A p0pF + ðãðì. (69a)
t The function that is usually tabulated is erf (X) = V2/it exp (—y2) dy, which is
related to (65) in an obvious way.
38 2.2 Simple Binary Hypothesis Tests
Fig. 2.9 (a) Receiver operating characteristic: Gaussian variables with unequal means.
The threshold for this criterion was given in (40). For the special case in which Bo — Pi the threshold tj equals one and
Pr (ñ) = ÊPF + PM). (69b)
Using (67) and (68) in (69), we have
Pr(t) = Jl^exp K) * =erfc* (+f)' (70)
It is obvious from (70) that we could also obtain the Pr (º) from the ROC. However, if this is the only threshold setting of interest, it is generally easier to calculate the Pr (c) directly.
Before calculating the performance of the other two examples, it is worthwhile to point out two simple bounds on erfc* (X). They will enable
Performance: Receiver Operating Characteristic 39 us to discuss its approximate behavior analytically. For X > 0
Tjh (‘ - ã») Hr) ‘erfc*m < ('?)• (71)
This can be derived by integrating by parts. (See Problem 2.2.15 or Feller .) A second bound is
erfc# (X) < ³ exp * > 0, (72)
Fig. 2.9 (b) detection probability versus </.
40 2.2 Simple Binary Hypothesis Tests
which can also be derived easily (see Problem 2.2.16). The four curves are plotted in Fig. 2.10. We note that erfc* (ËÃ) decreases exponentially.
The receiver operating characteristics for the other two examples are also of interest.
Fig. 2.10 Plot of erfc* (X) and related functions.
Performance: Receiver Operating Characteristic 41 Example 2. In this case the test is
/(R) = ² R* f (in v- N In 2-°) = y, (ai > a0). (73)
1 = 1 H0 Vl — <*0 \
The performance calculation for arbitrary N is somewhat tedious, so we defer it until Section 2.6. A particularly simple case appearing frequently in practice is N = 2. Under H0 the r* are independent zero-mean Gaussian variables with variances equal to a02:
PF = Pr (/ > y|tf0) = Pr (ri2 4- r22 > y|tf0). (74)
To evaluate the expression on the right, we change to polar coordinates:
T\ — z cos â, z = Vri2 4- r22
r2 = ã sin 0, â = tan'1 — **i
Prb> ã èàä - <7«
Integrating with respect to 0, we have
We observe that /, the sufficient statistic, equals z2. Changing variables, we have
(Note that the probability density of the sufficient statistic is exponential.)
/.D = exp(__2_). (79)
To construct the ROC we can combine (78) and (79) to eliminate the threshold y. This gives
Po = (PFyo*i°i2. (80)
In terms of logarithms
1ïË> = ^1ïÐê. (81)
As expected, the performance improves monotonically as the ratio <ri2/a02 increases.
We shall study this case and its generalizations in more detail in Section 2.6.
The two Poisson distributions are the third example.
Example 3, From (38), the likelihood ratio test is
_In òó 4- mi - m0 _ . ^ ___ 4 /M4
n < In -fTTZ— = Ó’ (mi > m0)•
Ho In m³ — In mo
Because n takes on only integer values, it is more convenient to rewrite (82) as
42 2.2 Simple Binary Hypothesis Tests
where yi takes on only integer values. Using (35),
PD = 1 - e-1 2 Ó³ = î, 1, 2,..., (84)
n = 0 '
and from (36)
Ó² “ 1 (/*jn)n
PF = 1 - e~mo 2 Ó² = 0, 1, 2............................. (85)
ÿ = Î ï'
The resulting ROC is plotted in Fig. 2.11a for some representative values of m0 and mi.
We see that it consists of a series of points and that PF goes from 1 to 1 — e~mo when the threshold is changed from 0 to 1. Now suppose we wanted PF to have an intermediate value, say 1 — ie~mo. To achieve this performance we proceed in the following manner. Denoting the LRT with yt = 0 as LRT No. 0 and the LRT with ó ³ = 1 as LRT No. 1, we have the following table: