1 Introduction

The radio spectrum is one of the most important resources for communica-tions. So spectrum detection is essential for wireless communication, whichis the key issue in Cognitive Radio.

1.1 What is cognitive radio?

Cognitive Radio is generic term used to describe a radio that is aware of thesurrounding environment and can accordingly adapts it transmission. More-over Cognitive radio is a flexible system as it can change the communicationparameter according to the condition.

1.2 Purpose

The purpose of cognitive radio Spectrum sensing plays a main role in CRbecause it’s important to avoid interference with PU and guarantee a goodquality of service of the PU. It’s predicted that CR identifies portions ofunused spectrum to transmits in it to avoid interferences with others users.The cognitive radio basically detects the unused spectrum and transmits theinformation to the fusion center whether the user is present is or not. Nowour aim is to check whether the primary user is present or not.So for that wewill perform hypothesis testing :

1. Simple hypothesis testing

2. Composite hypothesis testing.

For simple hypothesis testing there is :

1. Binary hypothesis testing.

2. Bayesian hypothesis testing.

Let us study about Bayesian hypothesis testing in detail.

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2 Bayesian Binary Hypothesis Testing.

In Bayesian Binary Hypothesis Testing we just select from two hypothesis,H0 and H1 based on the observation on random vector Y . Here we considerthe domain Y . From this we consider,

H0 = Y ∼ f(y/H0) & H1 = Y ∼ f(y/H1) (1)

If we consider as a discrete case then,

H0 = P (y/H0)

H1 = P (y/H1)

So from given Y we will decide whether H0 or H1 is true.This can beaccomplished using a function called δ(y) which is also called a decisionfunction which takes values (0,1). Hence,

δ(y) = 1; ifH1holds and

δ(y) = 0; ifH0holds.

On the basis of decision function we have a disjoint set which is given byY0& Y1 where,

Y0 = {y : δ(y) = 0}

Y1 = {y : δ(y) = 1}

The Hypothesis has prior probabilities which are :

π0 = P (H0) & π1 = P (H1).

with

π0 + π1 = 1.

Now there exist a Cost function which can be given by Cij i.e it statesthat cost incurred on selecting Hi when Hj holds. On the basis of the costfunction, the Baye’s Risk can be given as follows :

R(δ/Hj) =1∑i=0

CijP [Yi/Hj] (2)

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Therefore the average Baye’s Risk is

R(δ) =1∑i=0

R(δ/Hj)πj (3)

Hence the optimum Baye’s Decision rule δB is obtained by reducing the riskgiven over here i.e,

R(δ) =n∑i=0

n∑j=0

CijP [Yi/Hj]πj (4)

Now as the Y′ and Y∞are disjoint sets,we can write as

P (Y1/Hj) + P (Y0)/Hj) = 1 (5)

so for j=0,1 we can get

R(δ) =1∑j=0

C0jπj +1∑j=0

(C1j − C0j)P [Y1/Hj] (6)

R(δ) =1∑j=0

C0jπj +

∫Y1

1∑j=0

(C1j − C0j)f(y/Hj), dy (7)

Hence by reducing the Y1 we can minimize the Risk.

Y1 = {yεRn :1∑j=0

(C1j − C0j)f(y/Hj)} (8)

Now for j=1,we can see that taking H1 and H1 holds is lest costly then theselecting H0 H1 holds i.e.(C11 ≺ C01) In order to make a correct a decisionwe introduce a likelihood ratio which takes the values of vector Y,given as

L(y) =f(y/H1)

f(y/H0)(9)

This can be given as if L(y) ≤ τ then H0 is present and if L(y) ≥τthenH1is present.

τ =(C10 − C00)

(C01 − C11)(10)

Then under the condition C11 < C01as optimum bayesian decision rule canbe written as ,

δB =

{1 if L(y)≥ τ0 if L(y)≤ τ

(11)

From this we can derive equation for Minimum Probability of error .

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2.1 Minimum Probability of Error

Taking Cij=0 for i=j and Cij=1 for i 6= jThe Cost is incurred only when the error occurs i.e when H1 is true whenH0holds and H0 is true when H1holds.

R(δ/H1) = P [E/H0]

R(δ/H0) = P [E/H1]

Hence,P [E] = P [E/H0] + P [E/H1]

and consequently the τ = π0π1

3 Neyman-Pearson Test

In Bayesian hypothesis Testing it would require the knowledge of cost func-tions and prior probabilities π0&π1. In Neyman-Pearson Testing the aim is todesign the decision function δ in such a way the it maximizes the probabilityof Detection PD by bounding the Probability of false alarm PF to α.

Dα = {δ : PF (δ) ≤ α} (12)

δNP = arg maxδ∈Dα

PD(δ) (13)

The optimization problem can be solved by using Lagrangian Optimization,

L(δ, λ) = PD(δ) + λ(α− PF (δ)) (14)

A Test δ will be optimal if it maximizes L(δ, λ)& satisfies KKT condition

λ(α− PF (δ)) = 0 (15)

Now

PD(δ) =

∫Y1

f(y/H1) dy

PF (δ) =

∫Y1

f(y/H0) dy

The Lagrangian can be written as,

L(δ, λ) =

∫Y1

f(y/H1) dy + λ(α−∫Y1

f(y/H0) dy)

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L(δ, λ) =

∫Y1

(f(y/H1)− λf(y/H0)) dy + λα (16)

To maximize L(δ, λ),δ(y) should be chosen such that

f(y/H1)− λf(y/H0) ≥ 0

f(y/H1)

f(y/H0)≤ λ

L(y) ≤ λ

Thus the δ(y) can be written as

δ(y) =

1 if L(y) > λ0 or 1 if L(Y)=λ0 if L(y) < λ

(17)

4 Bayesian Sequential Hypothesis Testing

The standard Hypothesis problem involves fixed number ofobservation.On the other hand in sequential Hypothesis testing Problem, thenumber of observation are not fixed .Depending the observed samples decisionmay be taken after just few samples or large number of samples are observedif the decision is not concluded.We consider infinite number of I.I.D (inde-pendant identically Distributed ) observations { YK :K≥ 1} are available.Byusing this a sequential decision rule can be formed by a pair (φ, δ),whereφ = {φn, n ∈ N} a sampling plan or stopping rule and δ = {δn, n ∈ N}denotes the terminal decision rule.The function φn(Y1, Y2, ......Yn) maps Yn into {0,1}.After observing YK (for1 ≤ K ≥ n) we have, φn(Y1, Y2, ....., Yn) = 0indicates we should take one moresample till the decision is made ,and if at the same we have φn(Y1, Y2, ....., Yn) =1 one should stop sampling and make a decision.The terminal decision function δn(Y1, Y2, ....., Yn) takes the values in 0,1 whereδn(Y1, Y2, ....., Yn)= 0 or 1 depending on whether H0 or H1 holds,more overδn(Y1, Y2, ....., Yn) can be defined only if we have decided to stop sampling.

φn(Y1, Y2, ....., Yn) =

{1 for n=N0 for n 6= N

(18)

δB =

{undefined for n 6= Nδn(Y1, Y2, ....., Yn) for n=N

(19)

Now we associate cost for decision in order to determine the sequential deci-sion rule(φ, δ) in the Bayesian setting.To compute Baye’s Risk for a sequential

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decision rule (φ, δ) as R(φ, δ) one needs to first compute Conditional Baye’srisk for each of the hypothesis.R(φ, δ/H0) = conditional baye’s risk of (φ, δ) under H0 R(φ, δ/H1) = con-ditional baye’s risk of (φ, δ) under H1 Proir probabilities are given by π0 =P[H0] & π1 = P[H1]. Let CM=the cost corresponding to a miss,i.e. choosingH0 when H1 is true. and CF=the cost corresponding to a false alarm,i.e.choosing H1 when H0 is true. Unlike in the case of fixed sample size hy-pothesis testing problem ,here we also assume that each observation incursa cost denoted by D,i.e. strictly positive (D>0). In the absence of this costfor each observation D, one can collect an infinite number of observations,which would ensure that the decision is error-free.With this one can express the conditional Baye’s Risk for H0 as

R(φ, δ/H0) = CFP [δN(Y1, Y2, ....., YN) = 1/H0] +DE[N/H0] (20)

where ”N” denotes the random stopping time. The Conditional Baye’s riskfor H1 given by ,

R(φ, δ/H1) = CMP [δN(Y1, Y2, ....., YN) = 0/H1] +DE[N/H0] (21)

Therefore the average Baye’s Risk for the sequential decision rule(φ, δ) isgiven by ,

R(φ, δ) =1∑j=0

R(φ, δ/Hj)πj (22)

Our aim is to choose a decision rule (φ, δ),so that the Baye’s Risk can beminimized i.e if (φB, δB)denotes the optimum Bayesian Sequential decisionrule , then

R(φB, δB) = min(φ,δ)

R(φ, δ) = V (π0) (23)

Next we want to divide the set of all sequential decision rules into the fol-lowing two categories:

S = {(φ, δ) : φ0 = 0}and{(φ0 = 1, δ0 = 1) ∪ (φ0 = 1, δ0 = 0)} (24)

”S” here corresponds to the case of a sequential decision rule set,where onesample is taken to decide between H0 and H1. While {(φ0 = 1, δ0 = 1)∪(φ0 =1, δ0 = 0)} corresponds to the case when one doesnt take further more sample.Let

J(π0) = min(φ,δ)∈S

R(φ, δ) (25)

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Note that since N≥ 1 for all rules in S therefore we have E[N/H0] or E[N/H1]isgreater than or equal to 1. Thus we have ,

R(φ, δ) ≥ D

⇒ min(φ,δ)∈S

R(φ, δ) ≥ D

⇒ J(π0) ≥ D

Also note that for π0=1 and π0=0,the

P [δN(Y1, Y2, ......YN) = 1/H0]and[P [δN(Y1, Y2, ......YN) = 0/H1]are equal to zero

Therefore, for π0 = 1, R(φ, δ = DE[N/H0] = D. Thus J(1)=D.Similarly forπ0 = 0,J(0)=D.Next we compute the Baye’s risk for the following two cases of sequentialdecision rules, when no sample is taken implying φ0 = 1.When (φ0 = 1, δ0 = 1) since no samples is taken during this , we haveE[N/H0]=0.This implies

R(φ, δ/H0) = CFP [δN(Y1, Y2, ......YN) = 1/H0] = CF

and R(φ, δ/H1) = 0 as P [δN(Y1, Y2, ......YN) = 1/H1] = 0

⇒ R(φ, δ) = CFπ0. (26)

Similarly for (φ0 = 1, δ0 = 0),this implies

R(φ, δ) = CM(1− π0) (27)

.Therefore the minimum Baye’s risk for sequential decision rule that do

not take any sample which correspondsto ( the rules (φ = 1, δ = 1)&(φ =1, δ = 0) is therefore given by the piecewise linear function,

T (π0) = min{CFπ0, CM(1− π0)} (28)

T (π0) =

{CFπ0 forπ0 <

CM(CF+CM )

CF (1− π0) forπ0 >CM

(CF+CM )

(29)

Since the Baye’s risk obtained by any of other strategies,should lie betweenJ(π0) and T (π0),therefore,we have

V (π0) = min(T{π0), J(π0)} (30)

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The sketch of V (π0) vs π0 is shown in the figure given below,Depending on the fact that whether fact whether J(π0) always remain

above T (π0) or whether it intersects T (π0)at 2 pts as shown in figure abovetwo cases are possible, In the First case, when J(π0) remains above T (π0),the optimum decision rule is to make a decision immediately i.e. φ0=1 andδ0 is given as follows :

δ0 =

{1 ifπ0 ≤ CM

(CF+CM )

0 ifπ0 ≥ CM(CF+CM )

(31)

In the second case the optimum decision is

φ0 = 1, δ0 = 1forπ0 ≤ πL

φ0 = 1, δ0 = 0forπ0 ≥ πU

andφ0 = 0 for πL < π0 < πU

Thus sequential decision rule can be summarized as follows:

• Decide H1 or H0 if π0 falls below a low threshold πL or above highthreshold πU repectively.

• We keep on taking one sample when πL < π0 < πU .

Now next we want to get what one should do in case when πL < π0 < πU forthat what is to be done by taking one sample Y1 = y1.It is the same situationwhat we are having at first stage that we have CF and CM and D which aresame.However with knowledge of observation, y1 one can update one’s priorprobability π0 as follows using Baye’s rule:

π0(y1) = Pr[H0/Y1 = y1]

=Pr[Y1 = y1/H0]Pr[H0]∑

j = 01Pr[Y1 = y1/Hj]Pr[Hj]

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