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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis TestingIn deriving Bayes decision rule, we assumed that we know both the a priori

probabilities 0 and 1 as well as the likelihoods )|( 0Hyp and )|( 1Hyp . Thismeans that we have both the knowledge of the mechanism generating the state of the

nature and the mechanism affecting our observations (our measurements about the

state of nature). It is, however, possible that we may not have access to all the

information. For example, we may not know the a prioriprobabilities. In such a case,

the Bayes decision rule is not a good rule since it can only be derived for a given a

prioriprobability.An alternative to the Byes hypothesis testing, in this case, is the Minimax

Hypothesis testing. The minimax decision rule minimizes the maximum possible risk,

i.e., it minimizes,

)(),(max 10 RR over all .

Lets look at),( 0 r , i.e., the overall risk for a decision rule when the a priori

probability is]1,0[0 . It can be written as:

)()1()(),( 10000 RRr

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis Testing

Note that, for a given decision rule , as 0 varies from 0 to 1, ),( 0 r goes

linearly from ),0()(1 rR to ),1()(0 rR . Therefore, for a given decision rule

the maximum value of ),( 0 r as 0 varies over the interval ]1,0[ occurs either at

00 or 10 and it is )}(),(max{ 10 RR .

So minimizing )}(),(max{ 10 RR is equivalent to minimizing ),(max 010 0

r

.

Thus, the minimax decision rule is:),(maxmin 0

10 0

r

Let0

denote the optimum (Bayes) decision rule for the a priori probability 0 .

Denote the corresponding minimum Bayes risk by )( 0V , i.e., ),()( 00 rV .

It is easy to show that )( 0V is a continuous concave function of 0 for

]1,[0 o and has the end points 11)0( CV and 00)1( CV .

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis Testing

The following figure shows a typical graph of )( 0V and),( 0 r . Lets draw a tangent to )( 0V parallel to ),( 0 r .

Denote this line by ),(00

r .

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis TestingSince ),(

00 r lies entirely below the line ),( 0 r , it has a lower maximum

compared to ),( 0 r . Also note that since it touches )( 0V at 00 then 0 is the

minimum risk (Bayes) rule for a prioriprobability 0 . Since for any ]1,0[0 we can

draw a tangent to )( 0V and find the minimum risk rule as a Bayes rule, it is clear that

the minimax decision rule is the Bayes rule for the value of 0 that maximizes )( 0V .

Denoting this point by L , we note that point,

)()()}(),(max{ 1010 LLLL RRRR

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis Testing

Proposition: The Minimax Test

Let L be the a prioriprobability that maximizes )( 0V and such that either 0L , or 1L , or

),()( 10 LL RR

thenL

is a minimax rule.

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis Testing

Proof

Let ),()( 10 LL RR then for any 0 we have,

),,(),(),(minmax 0010 0

LL rrr L

So, we have

),,(maxmin),(max),(minmax 010

010

010 000

rrrL

Also, for each we have).,(minmax),(max 0

100

10 00

rr

This implies that,

).,(minmax),(maxmin 010

010 00

rr

Combining the two inequalities, we get,

.),(minmax),(maxmin 010

010 00

rr

Therefore,

.),(maxmin),( 010 0

rr LL

That is,L

is the minimax rule.

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis TestingDiscussion

By definition: ),()(0

00 rV . So, for every ]1,0[0 , we have )(),( 00

0 Vr and

).(),( 00

0 Vr Since ),(

00 r , as a function of 0 is a straight line, it has to be tangent to

)( 0V at .00 If )( 0V is differentiable at 0 , we have,

).()(/),()(000 10000

RRddrV

Now consider the case that )( 0V has an interior maximum but is not differentiable at that point. In

this case we define two decision rules 00lim LL

and .lim00 LL

The critical regions for these two decision rules are,

),|()()|())(1(|{01000101111

HypCCHypCCyLL

and

),|()()|())(1(|{01000101111

HypCCHypCCyLL

Take a number ]1,0[q and devise a decision rule~

L that uses the decision rule

L with probability

q and uses L

with probability q1 . It means that it decides1

H if 1

y , decides0

H if cy )(1

and

decides 1H with probability q ify is on the boundary of 1 .

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ELEC6111: Detection and Estimation Theory

Minimax Hypothesis TestingDiscussion

Note that the Bayes risk is not a function ofq , so )(),(

~

LL Vr L but the conditional risksdepend on q ,

).()1()()(~

LLL jjj

RqqRR

To achieve )()(~

1

~

0 LLRR , we need to choose,

.

)()()()(

)()(

1010

10

LLLL

LL

RRRR

RRq

Note that )()()( 00 LL

RRV L , so we have:

.)()(

)(

LL

L

VV

Vq

This is called a randomized decision rule.

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ELEC6111: Detection and Estimation Theory

Example: Measurement with Gaussian Error

Consider the measurement with Gaussian error with unifom costs

The function )( 0V can be written as,

)),(1)(1()()( 100

00

QQV

With

.2

)1

log( 10

0

0

01

2

We can find the rule making conditional risks )(0 R and )(1 R equal by letting,

))(1()( 10

QQ

and solving for .

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ELEC6111: Detection and Estimation Theory

Example: Measurement with Gaussian Error

We can solve this by inspection and get:

.2

10

L

So, the minimax decision rule is:

.2/)(0

2/)(1)(

10

10

yif

yify

L

Conditional risks for measurement with Gaussian error

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Hypothesis Testing

In Bayes hypothesis testing as well as minimax, we are concerned with the average risk, i.e., the

conditional risk averaged over the two hypotheses. Neyman-Pearson test, on the other hand, recognizes theasymmetry between the two hypotheses. It tries to minimize one of the two conditional risks with the other

conditional risk fixed (or bounded).

In testing the two hypotheses 0H and 1H , the following situations may arise:

0H is true but 1H is decided. This is called a type I error or afalse alarm. This comes from radarapplication where 0H represents no target and 1H is the case of target present. The probability

of this event is calledfalse alarm probability orfalse alarm rate and is denoted as )(F

P

1H is true but 0H is decided. This is called a type II error or a miss. The probability of this event iscalled miss probability and is denoted as )(MP

0H is true and 0H is decided. Probability of this event is )(1 FP . 1H is true and 1H is decided. This case represents a detection. The detection probabilityis

)(1)( MD PP .

In testing 0H versus 1H , one has to tradeoff between the probabilities of two types of errors. Neyman-

Pearson criterion makes this tradeoff by bounding the probability of false alarm and minimizing miss

probability subject to this constraint, i.e., the Neyman-Pearson test is,

)(max

DP subject to ,)( FP

where is the bound on false alarm rate. It is called the levelof the test.

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Hypothesis Testing

For obtaining a general solution to the Neyman-Pearson test, we need to define a randomized decisionrule. We define the randomized test,

L

L

L

yLif

yLifq

yLif

yL

)(0

)(

)(1

)(~

where L is the threshold corresponding to L .

While in a non-randomized rule, )(y gives the decision, in a randomized rule, )(

~

yL gives the probability ofdecision.

Then we have,

,)|()()}({)( 0~~

0

~

dyHypyYEPF

where {.}0E is expectation under hypothesis 0H . Also,

.)|()()}({)(1

~~

1

~

dyHypyYEPD

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Lemma

Consider a hypothesis pair 0H and 1H :00 ~: PYH

Versus

11 ~: PYH

where jP has density )|()( jj Hypyp for 1,0j . For 0 , the following statements are true:

1. Optimality: Let ~be any decision rule satisfying .)( ~ FP Let ~be any decision rule of the form)(

),|()|(0

)|()|()()|()|(1

01

01

01~

A

HypHypif

HypHypifyHypHypif

where 0 and 1)(0 y are such that .)(~

FP Then ).()(~~

DD PP

This means that any size- decision rule of form (A) is Neyman-Pearson rule.

2. Existence: For any )1,0( there is a decision rule, NP~ , of form (A) with 0)( y for which.)(

~

NPFP

3. Uniqueness: Suppose that is any Neyman-Pearson rule of size- for 0H versus 1H . Then must be ofthe form (A).

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Lemma (Proof)

1.Not that, by definition, we always have 0)]|()|()][()([ 01

~~

HypHypyy (why?)So, we have,

.0)]|()|()][()([ 01

~~

dyHypHypyy Expanding the above expression, we get,

.)|()()|()()|()()|()( 0

~

0

~

1

~

1

~

dyHypydyHypydyHypydyHypy

Applying the expressions for the detection probabilityandfalse alarm rate, we have:

.0)]([)]()([)()(~~~~~ FFFDD PPPPP

2. Let 0 be the smallest number such that (look at the Figure in next slide): )]|()|([ 0010 HYpHYpP .

Then if )]|()|([ 0010 HYpHYpP , choose,

)]|()|([

)]|()|([

0010

00100

HYpHYpP

HYpHYpP

Otherwise, choose 0 arbitrarily. Consider a Neyman-Pearson decision rule, NP~

, with 0 and 0)( y . For this

decision rule, the false alarm arte is,

)]|()|([)]|()|([}{)( 001000010~

0

~

HYpHYpPHYpHYpPEP NPNPF .

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Lemma (Proof)

.

2. See the text.

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Lemma (Example): Measurement with Gaussian Error

For this problem, we have,

),(1)()())([)]|()|([ 00010

QYPyLPHYpHYpP

where .2

)log( 10

01

2

Any value of can be achieved by choosing,

.)1()( 01

0

1

0

Q

Since 0)( 0 YP , the choice of 0 is arbitrary and we can choose 10 . So, we have

.0

1)(

0

0~

yif

yifyNP

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ELEC6111: Detection and Estimation TheoryNeyman-Pearson Lemma (Example): Measurement with Gaussian Error

The detection probability for NP

~

is dQQQQQYPYEP NPNPD

)()()()()}({)( 10111001

~

1

~