Neyman-Pearson Outline: test for simple binary hypotheses, receiver operating characteristic (ROC). An introduction to classical composite hypothesis testing. Reading: Chapter 3 in Kay-II, (part of) Chapter 5 in Levy.
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1
False-alarm and Detection Probabilities for Binary Hypothesis Tests: A Reminder (see handout # 5)
In binary hypothesis testing, we wish to identify which hypothesis is true (i.e. make the appropriate decision): H0 : sp(0) null hypothesis versus
H1 : sp(1) alternative hypothesis where sp(0) sp(1) = sp, sp(0) sp(1) = . Recall that a binary decision rule (x) maps 1, decide H1, 0, decide H0. X
data space
to {0, 1}:
(x) =
which partitions the data space X [i.e. the support of fX | (x | )] into two regions: X0 = {x : (x) = 0}
EE 527, Detection and Estimation Theory, # 5c
and X1 = {x : (x) = 1}.
2
Recall the probabilities of false alarm and miss: PFA((X), ) = E X | [(X) | ] =
X1
fX | (x | ) dx
for sp (0)
(1)
PM((X), ) = E X | [1 - (X) | ] = 1-
X1 PD ((X ),)
fX | (x | ) dx
=
X0
fX | (x | ) dx
for in sp (1)
(2)
and the probability of detection (correctly deciding H1): PD((X), ) = E X | [(X) | ] =
X1
fX | (x | ) dx
for in sp (1).
For simple hypotheses, sp(0) = {0}, sp(1) = {1}, and sp = {0, 1}, the above expressions simplify, as shown in the following.
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3
Probabilities of False Alarm (P FA) and Detection (P D) for Simple Hypotheses
PFA((X), 0) =
X1
fX | (x | 0) dx (3)
= PrX | {test statistic (ts) > | 0}
X X1
PD((X), 1) =
X1
fX | (x | 1) dx (4)
= PrX | {ts > | 1}.
X X1
Comments: (i) As the region X1 shrinks (i.e.
EE 527, Detection and Estimation Theory, # 5c
), both of the above
4
probabilities shrink towards zero. (ii) As the region X1 grows (i.e. grow towards unity. 0), both probabilities
(iii) Observations (i) and (ii) do not imply equality between PFA and PD; in most cases, as X1 grows, PD grows more rapidly than PFA (i.e. we better be right more often than we are wrong). (iv) However, the perfect case where our rule is always right and never wrong (PD = 1 and PFA = 0) cannot occur when the conditional pdfs/pmfs fX | (x | 0) and fX | (x | 1) overlap. (v) Thus, to increase the detection probability PD, we must also allow for the false-alarm probability PFA to increase. This behavior represents the fundamental tradeoff in hypothesis testing and detection theory and motivates us to introduce a (classical) approach to testing simple hypotheses, pioneered by Neyman and Pearson, to be discussed next. Receiver Operating Characteristic (ROC) allows us to visualize the realm of achievable PFA((X), 0) and PD((X), 1).
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A point (PFA, PD) is in the shaded region if we can find a rule (X) such that PFA((X), 0) = PFA and PD((X), 1) = PD.
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6
Neyman-Pearson Test for Simple Hypotheses
Bayesian tests are criticized because they require specification of prior distribution (pmf or, in the composite-testing case, pdf) and the cost-function parameters L(i | j). An alternative classical solution for simple hypotheses is developed by Neyman and Pearson.
Select the decision rule (X) that maximizes PD((X), 1) while ensuring that the probability of false alarm PFA((X), 0) is less than or equal to a specified level . Setup: Simple hypothesis testing: H0 : = 0 H1 : = 1. Parametric data models fX | (x | 0), fX | (x | 1). No prior pdf/pmf on is available.
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versus
Define the set of all rules (X) whose probability of false alarm is less than or equal to a specified level : D = (X) PFA((X), 0) } see also (3). A Neyman-Pearson test NP(x) solves the constrained optimization problem: NP(x) = arg max PD((x), 1).
(x)D
We apply Lagrange multipliers to solve this optimization problem; consider the Lagrangian: L((x), ) = PD((x), 1) + [ - PFA((x), 0)] with 0. A decision rule (x) will be optimal if it maximizes L((x), ) and satisfies the Karush-Kuhn-Tucker (KKT) condition: [ - PFA((x), 0)] = 0. Upon using (3) and (4), the Lagrangian can be written as L((x), ) = +
X1
EE 527, Detection and Estimation Theory, # 5c 8
(5)
[fX | (x | 1) - fX | (x | 0)] dx.
Consider maximizing L((x), ) with respect to (x) for a given . Then, (x) needs to satisfy 1, (x) > 0 or 1, (x) = (x) = 0, (x) < where
(6)
fX | (x | 1) (x) = fX | (x | 0) is the likelihood ratio. The values x that satisfy (x) = can be allocated to either X1 or X0. To completely specify the optimal test, we need to select a such that the KKT condition (5) holds and an allocation rule for those x that satisfy (x) = . Now, consider two versions of (6) for a fixed threshold : 1, (x) > 1, (x) = U,(x) = 0, (x) < and 1, (x) > 0, (x) = . L,(x) = 0, (x) <
9
EE 527, Detection and Estimation Theory, # 5c
In the first case, all observations x for which (x) = are allocated to X1; in the second case, these observations are allocated to X0. Consider the cumulative distribution function (cdf) of (X) = under H0: F | (l | 0) = Pr | { l | 0}. Define f0 = F | (0 | 0) = Pr | { 0 | 0}. Recall that cdf F | (l | 0) must be nondecreasing and rightcontinuous, but may have discontinuities.
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10
Consider three cases, depending on : (i) When 1 - < f0 i.e. 1 - f0 < we select the threshold = 0 and apply the rule L,0(x) = 1, (x) > 0 . 0, (x) = 0 (7)
(8)
In this case, KKT condition (5) holds and, therefore, the test (8) is optimal; its probability of false alarm is PFA(L,0(x), 0) = 1 - f0 < see (7).
An example of this case corresponds to 1 = 0 and 1 - 1 in the above figure. (ii) Suppose that 1 - f0 i.e. 1 - f0 (9)
and there exists a such that F | ( | 0) = 1 - . Then, by selecting this as the threshold and using L,(x) =
EE 527, Detection and Estimation Theory, # 5c
(10)
1, (x) > 0, (x)
(11)
11
we obtain a test with false-alarm probability PFA(L,(x), 0) = 1 - F | ( | 0) = see (9)
the KKT condition (5) holds, and the test (10) is optimal. An example of this case corresponds to 2 and 1 - 2 in the above figure. (iii) Suppose that 1 - f0 i.e. 1 - f0
as in (ii), but cdf F | (l | 0) has a discontinuity point > 0 such that F | (- | 0) < 1 - < F | (+ | 0) where F | (- | 0) and F | (+ | 0) denote the left and right limits of F | ( | 0) at l = . If this case happens in practice, we can try to avoid the problem by changing our specified , which is anyway not God-given, but chosen rather arbitrarily. We should pick a value of that satisfies the KKT condition. Suppose that we are not allowed to change ; this gives us a chance to practice some basic probability. First, note that
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L,(x) has false-alarm probability PFA(L,(x), 0) = 1 - F | (+ | 0) < , U(x, ) has false-alarm probability PFA{U,(x), 0} = 1 - F | (- | 0) > and KKT optimality condition (5) requires that PFA((x), 0) = . We focus on the tests of the form (6) and construct the optimal test via randomization.
Define the probability p = - PFA(L,(x), 0) PFA(U,(x), 0) - PFA(L,(x), 0)
which clearly satisfies 0 < p < 1.
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Select U,(x) with probability p and L,(x) with probability 1 - p. This test indeed has the form (6); its probability of false alarm is PFA((x), 0) = PFA(L,(x), 0) + p [PFA(U,(x), 0) - PFA(L,(x), 0)] = . Since KKT condition (5) is satisfied, the randomized test is optimal. 1, (x) > and 0 w.p. 1 - p, (x) = 0, (x) <
(x) =
1 w.p. p
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14
ROC Properties when Likelihood Ratio is a Continuous Random Variable Given
Based on the Neyman-Pearson theory, if we set PFA = , then the test that maximizes PD must be a likelihood-ratio test of the form (6). Thus, the ROC curve separating achievable and non-achievable pairs (PFA, PD) corresponds to the family of likelihood-ratio tests.
For simplicity, we focus here on the case where the likelihood ratio is a continuous random variable given . First, note that,
EE 527, Detection and Estimation Theory, # 5c 15
for the likelihood-ratio test, PFA( ) =
X1 +
fX | (x | 0) dx f | (l | 0) dl
= PrX | {(X) > | 0} = PD( ) =
X1 +
(12)
fX | (x | 1) dx f | (l | 1) dl
= PrX | {(X) > | 1} =
(13)
where denotes the threshold. Under the continuity assumption for the likelihood ratio, as we vary between 0 and +, the point PFA((X), 0), PD((X), 1) moves continuously along the ROC curve. If we set = 0, we always select H1 and, therefore, PFA(0) = PD(0) = 1. Conversely, if we set = +, we always select H0 and, therefore, PFA(+) = PD(+) = 0. In summary, ROC Property 1. If the likelihood ratio is a continuous random variable given , the points (0, 0) and (1, 1) belong to ROC.
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Now, differentiate (12) and (13) with respect to : dPD( ) d dPD( ) d implying f | ( | 1) dPD( ) = = . dPFA( ) f | ( | 0) In summary, ROC Property 2. If the likelihood ratio is a continuous random variable given , the slope of ROC at point (PFA( ), PD( )) is equal to the threshold of the corresponding likelihood-ratio test. In particular, this result implies that the slope of ROC is = + at (0, 0) and = 0 at (1, 1). ROC Property 3. The domain of achievable pairs (PFA, PD) is convex and the ROC curve is concave. This property holds in general, including the case where the likelihood ratio is a mixed or discrete random variable given . HW: Prove ROC Property 3.
EE 527, Detection and Estimation Theory, # 5c 17
= -f | ( | 1) = -f | ( | 0)
ROC Property 4. All points on ROC curve satisfy PD( ) PFA( ). This property holds in general, including the case where the likelihood ratio is a mixed or discrete random variable given .
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Example: Simple Hypotheses, Coherent Detection in Gaussian Noise with Known Covariance Matrix
Simple hypotheses: the space of the parameter and its partitions are sp = {0, 1}, sp(0) = {0}, sp(1) = {1}. The measurement vector X given is modeled using fX | (x | ) = N (x | , C) = 1 |2 C|
1 exp[- 2 (x - )T C -1 (x - )]
where C is a known positive definite covariance matrix. Our likelihood-ratio test is (x)
likelihood ratio
fX | (x | 1) = fX | (x | 0)
H1
1 exp[- 2 (x - 1)T C -1 (x - 1)] = exp[- 1 (x - 0)T C -1 (x - 0) 2
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.
19
Therefore,
H1
-1 2 i.e.
(x - 1) C
T
-1
1 (x - 1) + 2
(x - 0) C
T
-1
(x - 0)
ln
(1 - 0) C and, finally,
H1
T
-1
[x - (0 + 1)]
1 2
H1
ln .
T (x) = s C
T
-1
x
ln + 1 (1 - 0)T C -1 (1 + 0) = 2
where we have defined s = 1 - 0. False-alarm and detection/miss probabilities. Given , T (x) is a linear combination of Gaussian random variables, implying that it is also Gaussian, with mean and variance: E X | [T (X) | ] = sT C -1 varX | [T (X) | ] = sT C -1s
EE 527, Detection and Estimation Theory, # 5c
(not a function of ).
20
Now, PFA = PrX | {T (X) > | 0}
standard normal random variable
= PrX |
T (X) - sT C -10 sT C -1 s
- sT C -10 > 0 T C -1 s s (14)
- sT C -10 = Q sT C -1s and PD = 1 - PM = PrX | {T (X) > | 1}
standard normal random variable
= PrX |
T (X) - sT C -11 sT C -1 s
- sT C -11 > | 1 T C -1 s s
- sT C -11 = Q . T C -1 s s
We use (14) to obtain a that satisfies the specified PFA: sT C -1 0 = Q-1(PFA) + sT C -1 s sT C -1 s
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implying PD = Q Q
-1
(PFA) -
sT C -1 s (15)
= Q Q-1(PFA) - d Here, d= sT C -1 s = (1 - 0)T C -1 (1 - 0)
is the deflection coefficient.
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Decentralized Detection for Simple Hypotheses
Consider a decentralized detection scenario depicted by
Assumptions: The observations X[n], n = 0, 1, . . . , N - 1 made at N spatially distributed sensors (nodes) follow the same marginal probabilistic model: fX | (x[n] | ) (16)
and are conditionally independent given = , which may not always be reasonable, but leads to an easy solution.
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We wish to test: H0 : = 0 H1 : = 1. Each node n makes a hard local decision d[n] based on its local observation x[n] and sends it to the headquarters (fusion center), which collects all the local decisions and makes the final global decision H0 versus H1. This structure is clearly suboptimal: it is easy to construct a better decision strategy in which each node sends its (quantized, in practice) likelihood ratio to the fusion center, rather than the decision only. However, such a strategy would have a higher communication (energy) cost. The false-alarm and detection probabilities of each node's local decision rules can be computed using (16). Suppose that we have obtained them for each n: PFA,n, PD,n, n = 0, 1, . . . , N - 1. versus
We now discuss the decentralized detection problem. Note that
dn pD(n) | (dn | 1) = PD,n (1 - PD,n)1-dn
Bernoulli pmf
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and, similarly,
dn pD(n) | (dn | 0) = PFA,n (1 - PFA,n)1-dn
Bernoulli pmf
where PFA,n is the nth sensor's local detection false-alarm probability. Now, pD(n) | (dn | 1) ln (d) = ln pD(n) | (dn | 0) n=1
N N
=
n=1
ln
dn PD,n (1 - PD,n)1-dn dn PFA,n (1
H1
- PFA,n
)1-dn
ln .
To be able to further simplify the above expression, we nof focus on the case where all sensors have identical performance: PD,n = PD, PFA,n = PFA
i.e. all local decision thresholds at the nodes are identical. Define the number of sensors deciding locally to support H1:
N -1
u1 =
n=0
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d[n].
25
Then, the log-likelihood ratio becomes 1 - PD PD + (N - u1) log log (d) = u1 log PFA 1 - PFA or PD (1 - PFA) u1 log PFA (1 - PD)
H1 H1
log
log + N log
1 - PFA . 1 - PD
(17)
Clearly, each node's local decision dn is meaningful only if PD > PFA, which implies PD (1 - PFA) >1 PFA (1 - PD) the logarithm of which is therefore positive, and the decision rule (17) further simplifies to u1
H1
.
The Neyman-Person performance of analysis this detector is easy: the random variable U1 is binomial given (i.e. conditional on the hypothesis) and, therefore, PrU1 | {U1 = u1 | } =
EE 527, Detection and Estimation Theory, # 5c
N pu1 (1 - p)N -u1 u1
26
where p = PFA under H0 and p = PD under H1. Hence, the "global" false-alarm probability is PFA,global = PrU1 | {U1 > | 0}
N
=
u1 =
N u1 PFA (1 - PFA)N -u1 . u1
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27
An Introduction to Classical Composite Hypothesis Testing
First, recall that, in composite testing of two hypotheses, we have sp(0) and sp(1) that form a partition of the parameter space sp: sp(0) sp(1) = sp, sp(0) sp(1) = and that we wish to identify which of the two hypotheses is true: H0 : sp(0) null hypothesis versus
H1 : sp(1) alternative hypothesis. Here, we adopt the classical Neyman-Pearson approach: given an upper bound on the false-alarm probability, maximize the detection probability. The fact that H0 is composite means that the false-alarm probability for a rule (X) is a function of : PFA((X), )) where sp(0). Therefore, to satisfy the upper bound , we
EE 527, Detection and Estimation Theory, # 5c 28
consider all tests (X) such that
sp (0)
max PFA((X), )) .
(18)
In this context,
sp (0)
max PFA((X), )
(19)
is typically referred to as the size of the test (X). Therefore, the condition (18) states that we focus on tests whose size is upper-bounded by . Definition. Among all tests (X) whose size is upperbounded by [i.e. (18) holds], we say that UMP(X) is a uniformly most powerful (UMP) test if it satisfies PD(UMP(X), ) PD((X), ) for all sp(1). This is a very strong statement and very few hypothesis-testing problems have UMP tests. Note that Neyman-Pearson tests for simple hypotheses are UMP. Hence, to find an UMP test for composite hypotheses, we need to first write a likelihood ratio for the simple hypothesis test with sp(0) = {0}, sp(1) = {1}, and sp = {0, 1} and then transform this likelihood ratio in such a way that unknown quantities (e.g. 0 and 1) disappear from the test statistic.
EE 527, Detection and Estimation Theory, # 5c 29
(1) If such a transformation can be found, there is hope that a UMP test exists. (2) However, we still need to figure out how to set a decision threshold ( , say) such that the upper bound (18) is satisfied.
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30
Example 1: Detecting a Positive DC Level in AWGN (versus zero DC level)
Consider the following composite hypothesis-testing problem: H0 : H1 : = 0 i.e. sp(0) = {0} >0 versus
i.e. sp(1) = (0, +)
where the measurements X[0], X[1], . . . , X[N - 1] are conditionally independent, identically distributed (i.i.d.) given = , modeled as {X[n] | = } = + W [n] n = 0, 1, . . . , N - 1 with W [n] a zero-mean white Gaussian noise with known variance 2, i.e. W [n] N (0, 2) implying fX | (x | ) = 1 exp - 2 2 (2 2)N 1
N -1
(x[n]-)2 (20)
n=0
where x = [x[0], x[1], . . . , x[N - 1]]T . A sufficient statistic for is N 1 x= x[n]. N n=1
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Now, find the pdf of x given = : fX | (x | ) = N (x | , 2/N ). (21)
We start by writing the classical Neyman-Pearson test for the simple hypotheses with spsimple(0) = {0} and spsimple(1) = {1}, 1 (0, +):
1 2 -1/2 exp[- 2 2/N (x - 1)2] fX | (x | 1) (2 /N ) = 1 fX | (x | 0) (2 2/N )-1/2 exp[- 2 2/N (x)2] H1
.
Taking log etc. leads to 1 x
H1
.
Since we know that 1 > 0, we can divide both sides of the above expression by 1 and accept H1 if (x) : x
H1
.
Hence, we transformed our likelihood ratio in such a way that 1 disappears from the test statistic, i.e. we accomplished (1) above. Now, on to (2). How to determine the threshold such that the upper bound (18) is satisfied? Based on (25), we know: fX | (x | 0) = N (x | 0, 2/N )
EE 527, Detection and Estimation Theory, # 5c 32
and, therefore, PFA((X), 0) = PrX | {X > 0} = PrX | X -0 2/N
standard normal random var.
>
2/N
0
= Q
2/N
.
Note that max PFA((X), ) = PFA((X), 0) = Q 2/N =
sp (0)
see (18) and (19). The most powerful test is achieved if the upper bound in (18) is reached by equality: 2 Q-1(). N
=
(22)
Hence, we have accomplished (2), since this yields exactly size for our test (X). To study the performance of the above test, we substitute
EE 527, Detection and Estimation Theory, # 5c 33
(22) into the power function: PrX | {X > | } = PrX | X - 2/N
standard normal random var.
>
- 2/N
= Q
- 2/N
= Q Q-1() -
2/N
.
(23)
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34
Example 2: Detecting a Positive DC Level in AWGN (versus nonnegative DC level)
Consider the following composite hypothesis-testing problem: H0 : H1 : 0 >0 i.e. sp(0) = (-, 0] versus i.e. sp(1) = (0, +) are
where the measurements X[0], X[1], . . . , X[N - 1] conditionally i.i.d. given = , modeled as {X[n] | = } = + W [n] n = 0, 1, . . . , N - 1
with W [n] a zero-mean white Gaussian noise with known variance 2, i.e. W [n] N (0, 2) implying fX | (x | ) = 1 exp - 2 2 )N 2 (2 1
N -1
(x[n]-)2 (24)
n=0
where x = [x[0], x[1], . . . , x[N - 1]]T . A sufficient statistic for is N 1 x= x[n]. N n=1
EE 527, Detection and Estimation Theory, # 5c 35
and (25) fX | (x | ) = N (x | , 2/N ). We start by writing the classical Neyman-Pearson test for the simple hypotheses with spsimple(0) = {0} and spsimple(1) = {1}, where 0 (-, 0] and 1 (0, +), implying
1 2 -1/2 exp[- 2 2/N (x - 1)2] fX | (x | 1) (2 /N ) = 1 fX | (x | 0) (2 2/N )-1/2 exp[- 2 2/N (x - 0)2] H1
and 0 < 1. Taking log etc. leads to (1 - 0) x and, since 0 < 1, to (x) : x
H1 H1
.
Hence, we transformed our likelihood ratio in such a way that 0 and 1 disappear from the test statistic, i.e. we accomplished (1) above. The power function of this test is PrX | {X > | } = PrX |
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X - - - > =Q / N / N / N
36
which is an increasing function of . Recall the definition (19) of test size:
sp (0)
max PFA((X), ) = =
sp (0)
max PrX | {X > | } max Q - / N =Q .
(-,0]
/ N
The most powerful test is achieved if the upper bound in (18) is reached by equality: = Q-1(). N Hence, we have accomplished (2), since this yields exactly size for our test (X).
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37
Example 3: Detecting a Completely Unknown DC Level in AWGN
Consider now the composite hypothesis-testing problem: H0 : H1 : = 0 i.e. sp(0) = {0} versus
= 0 i.e. sp(1) = (-, +)\{0} are
where the measurements X[0], X[1], . . . , X[N - 1] conditionally i.i.d. given = , following fX | (x | ) = 1 (2 2)N 1 exp - 2 2
N -1
(x[n] - )2
n=0
and x = [x[0], x[1], . . . , x[N - 1]]T . A sufficient statistic for N 1 is x = N n=1 x[n] and the pdf of x given = is fX | (x | ) = N (x | , 2/N ). (26)
We start by writing the classical Neyman-Pearson test for the simple hypotheses with sp(0) = {0} and sp(1) = {1 = 0}: 1 x > . We cannot accomplish (1), since 1 cannot be removed from the test statistic; therefore, UMP test does not exist for the above problem.
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Monotone Likelihood-ratio Criterion
Consider a scalar parameter . We say that fX | (x | ) belongs to the monotone likelihood ratio (MLR) family if the pdfs (or pmfs) from this family satisfy the identifiability condition for (i.e. these pdfs are distinct for different values of ) and there is a scalar statistic T (x) such that, for 0 < 1, the likelihood ratio fX | (x | 1) (x ; 0, 1) = fX | (x | 0) is a monotonically increasing function of T (x). If fX | (x | ) belongs to the MLR family, then use the following test: 1, for T (x) , (x) = 0, for T (x) <
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and set = PFA((X), 0) = PrX | {T (X) | 0} e.g. use this condition to find the threshold . This test has the following properties: (i) With given by (27), (x) is UMP test of size for testing H0 : > 0 H1 : 0. (ii) For each , the power function PrX | {T (X) | } is a monotonically increasing function of . Note: Consider the one-parameter exponential family fX | (x | ) = h(x) exp[() T (x) - B()]. (29) (28) versus (27)
Then, if () is a monotonically increasing function of , the class of pdfs (pmfs) (29) satisfies the MLR conditions.
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Example: Detection for Exponential Random Variables
Consider conditionally i.i.d. measurements X[0], X[1], . . . , X[N - 1] given the parameter > 0, following the exponential pdf: fX | (x[n] | ) = Expon(x[n] | 1/) = 1 exp(--1 x[n]) i(0,+)(x[n]).
The likelihood function of for all observations x = [x[0], x[1], . . . , x[N - 1]]T is 1 fX | (x | ) = N exp[--1 T (x)] where
N -1 N -1
i(0,+)(x[n])
n=0
T (x) =
n=0
x[n].
Since fX | (x | ) belongs to the one-parameter exponential family (29) and () = --1 is a monotonically increasing function of . Therefore, the test (x) = 1, for T (x) , 0, for T (x) <
41
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is UMP for testing H0 : > 0 H1 : 0. The sum of i.i.d. exponential random variables follows the Erlang pdf (which is a special case of the gamma pdf): fT | (T | ) = 1 T N -1 exp(-T /) i(0,+)(T ) N (N - 1)! versus
= Gamma(T | N, -1). Therefore, the size of the test can be written as = PrX | {T (X) | 0} = = 1 N 0 1+
+
tN -1 exp(-t/0) dt (N - 1)!
N -1
1 + + 0 (N - 1)! 0
exp(-/0)
where the integral is evaluated using integration by parts. For N = 1, we have = 0 ln(1/).
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Generalized Likelihood Ratio (GLR) Test
Recall again that, in composite testing of two hypotheses, we have sp(0) and sp(1) that form a partition of the parameter space sp: sp(0) sp(1) = sp, sp(0) sp(1) = and that we wish to identify which of the two hypotheses is true: H0 : sp(0) null hypothesis versus
H1 : sp(1) alternative hypothesis. In GLR tests, we replace the unknown parameters by their maximum-likelihood (ML) estimates under the two hypotheses. Hence, accept H1 if maxsp(1) fX | (x | ) GLR(x) = > . maxsp(0) fX | (x | ) This test has no UMP optimality properties, but often works well in practice.
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Example: Detecting a Completely Unknown DC Level in AWGN
Consider again the composite hypothesis-testing problem from p. 38: H0 : H1 : = 0 i.e. sp(0) = {0} versus
= 0 i.e. sp(1) = (-, +)\{0} are
where the measurements X[0], X[1], . . . , X[N - 1] conditionally i.i.d. given = , following fX | (x | ) = 1 (2 2)N 1 exp - 2 2
N -1
(x[n] - )2
n=0
and x = [x[0], x[1], . . . , x[N - 1]]T . A sufficient statistic for N 1 is x = N n=1 x[n] and the pdf of x given = is fX | (x | ) = N (x | , 2/N ). Our GLR test accepts H1 if maxsp(1) fX | (x | ) GLR(x) = > . fX | (x | 0) Now, x = arg max fX | (x | )
sp (1)
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and fX | (x | 0) = N (x | 0, 2/N ) x2 = exp - 1 2 2 /N 2 2/N 1 fX | (x | x) = N (x | 0, 2/N ) = 2 2/N 1 yielding N x2 ln GLR(x) = . 2 2 Therefore, we accept H1 if (x)2 > or |x| > . We compare this detector with the (not realizable, also called clairvoyant) UMP detector that assumes the knowledge of the sign of under H1. Assuming that the sign of under H1 is known, we can construct the UMP detector, whose ROC curve is given by PD = Q(Q-1(PFA) - d) where d = N 2/ 2 and is the value of the parameter under H1; see (23) for the case where > 0 under H1. All other detectors have PD below this upper bound.
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GLR test: Decide H1 if |x| > . To make sure that the GLR test is implementable, we must be able to specify a threshold so that the false-alarm probability is upper-bounded by a given size . This is possible in our example:
PFA((x), 0) = PrX | {|X| > | 0}
symmetry
see (26) 2/N ) see (26)
=
2 PrX | {X > | 0} = 2 Q(/
PD((x), ) = PrX | {|X| > | } = = = - 2/N
-1
PrX | {X > | } + PrX | {X < - | } Q +Q + 2/N 2/N 2/N .
Q Q
(/2) - (/2) +
+Q Q
-1
In this case, GLR test is only slightly worse than the clairvoyant detector (Figure 6.4 in Kay-II):
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Example: DC level in WGN with A and 2 both unknown. Recall that 2 is called a nuisance parameter since we care exclusively about . Here, the GLR test for
H0 : H1 :
= 0 i.e. sp(0) = {0}
versus
= 0 i.e. sp(1) = (-, +)\{0}
accepts H1 if
max,2 fX | ,2 (x | , 2) GLR(x) = > max2 fX | ,2 (x | 0, 2)
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where fX | ,2 (x | , 2) = Here, max fX | ,2 (x | , ) =
, 2 2
1 exp - 2 2 (2 2)N
1
N -1
(x[n]-)2 .
n=0
(30)
max fX | ,2 (x | 0, 2) =
2
1 e-N/2 2 [2 1 (x)]N/2 1 e-N/2 2 [2 0 (x)]N/2
where
2 0 (x) =
1 N 1 N
N
x2[n]
n=1 N
2 1 (x) =
(x[n] - x)2.
n=1
Hence, GLR(x) =
2 0 (x) 2 1 (x) N/2
i.e. GLR test fits data with the "best" DC-level signal ML = x, 2 finds the residual variance estimate 1 , and compares this 2 estimate with the variance estimate 0 under the null case (i.e.
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2 for = 0). When sufficiently strong signal is present, 1 and GLR(x) 1.
2 0
Note that
2 1 (x) =
1 N 1 N
N
(x - x[n])2
n=1 N
=
(x2[n] - 2 x x[n] + x2)
n=1 N
=
1 N
x2[n] - 2 x2 + x2
n=1
2 = 0 (x) - x2.
Hence,
2 1 0 (x) . 2 ln GLR(x) = N ln 2 2 = N ln 2 2 (x) 0 (x) - x 1 - x /0
x2 1 0 2 0 (x) and ln[1/(1 - z)] is monotonically increasing on z (0, 1). Therefore, an equivalent test can be constructed as follows: x2 > . T (x) = 2 0 (x)
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Note that
The pdf of T (X) given = 0 does not depend on 2 and, therefore, GLR test can be implemented, i.e. it is CFAR. Definition. A test is constant false alarm rate (CFAR) if we can find a threshold that yields a test whose size is equal to . In other words, we should be able to set the threshold independently of the unknown parameters, i.e. the distribution of the test statistic under H0 does not depend on the unknown parameters.
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