ELEC 303 – Random Signals (PowerPoint) by shuifanglj


									ELEC 303 – Random Signals

Lecture 17 – Hypothesis testing 2
        Dr. Farinaz Koushanfar
       ECE Dept., Rice University
              Nov 2, 2009
• Reading: 8.2,9.3
• Bayesian Hypothesis testing
• Likelihood Hypothesis testing
      Four versions of MAP rule
•  discrete, X discrete
                             p  () p X| ( x | )

•  discrete, X continuous
                             p  ()f X| ( x | )

•  continuous, X discrete
                             f  () p X| ( x | )
•  continuous, X continuous
                              f  (  ) f X | ( x |  )
          Example – spam filter
• Email may be spam or legitimate
• Parameter , taking values 1,2, corresponding to
  spam/legitimate, prob p(1), P(2) given
• Let 1,…, n be a collection of special words,
  whose appearance suggests a spam
• For each i, let Xi be the Bernoulli RV that denotes
  the appearance of i in the message
• Assume that the conditional prob are known
• Use the MAP rule to decide if spam or not.
      Bayesian Hypothesis testing
• Binary hypothesis: two cases
• Once the value x of X is observed, Use the Bayes
  rule to calculate the posterior P|X(|x)
• Select the hypothesis with the larger posterior
• If gMAP(x) is the selected hypothesis, the correct
  decision’s probability is       P(= gMAP(x)|X=x)
• If Si is set of all x in the MAP, the overall probability
  of correct decision is
               P(= gMAP(x))=i P(=i,XSi)
• The probability of error is: i P(i,XSi)
Multiple hypothesis
 Example – biased coin, single toss
• Two biased coins, with head prob. p1 and p2
• Randomly select a coin and infer its identity
  based on a single toss
• =1 (Hypothesis 1), =2 (Hypothesis 2)
• X=0 (tail), X=1(head)
• MAP compares P(1)PX|(x|1) ? P(2)PX|(x|2)
• Compare PX|(x|1) and PX|(x|2) (WHY?)
• E.g., p1=.46 and p2 =.52, and the outcome tail
Example – biased coin, multiple tosses
• Assume that we toss the selected coin n times
• Let X be the number of heads obtained
• ?
    Example – signal detection and
           matched filter
• A transmitter sending two messages =1,=2
• Massages expanded:
  – If =1, S=(a1,a2,…,an), if =2, S=(b1,b2,…,bn)
• The receiver observes the signal with
  corrupted noise: Xi=Si+Wi, i=1,…,n
• Assume WiN(0,1)
Likelihood Approach to Binary
       Hypothesis Testing
BHT and Associated Error
Likelihood Approach to BHT (Cont’d)
Likelihood Approach to BHT (Cont’d)
       Binary hypothesis testing
• H0: null hypothesis, H1: alternative hypothesis
• Observation vector X=(X1,…,Xn)
• The distribution of the elements of X depend
  on the hypothesis
• P(XA;Hj) denotes the probability that X
  belongs to a set A, when Hj is true
• A decision rule:
  – A partition of the set of all possible values of the
    observation vector in two subsets: “rejection
    region” and “acceptance region”
• 2 possible errors for a rejection region:
  – Type I error (false rejection): Reject H0, even
    though H0 is true
  – Type II error (false acceptance): Accept H0, even
    though H0 is false
          Probability of regions
• False rejection:
  – Happens with probability
           (R) = P(XR; H0)
• False acceptance:
  – Happens with probability
           (R) = P(XR; H1)
            Analogy with Bayesian
• Assume that we have two hypothesis =0 and
  =1, with priors p(0) and p(1)
• The overall probability of error is minimized using
  the MAP rule:
   –   Given observations x of X, =1 is true if
   –   p(0) pX|(x|0) < p(1) pX|(x|1)
   –   Define:         = p(0) / p(1)
   –   L(x) = pX|(x|1) / pX|(x|0)
• =1 is true if the observed values of x satisfy the
  inequality: L(x)> 
               More on testing
• Motivated by the MAP rule, the rejection
  region has the form R={x|L(x)>}
• The likelihood ratio test
  – Discrete:
           L(x)= pX(x;H1) / pX(x;H0)
  – Continuous:
           L(x) = fX(x;H1) / fX(x;H0)
• Six sided die
• Two hypothesis

• Find the likelihood ratio test (LRT) and
  probability of error
      Error probabilities for LRT
• Choosing  trade-offs between the two error
  types, as  increases, the rejection region
  becomes smaller
  – The false rejection probability (R) decreases
  – The false acceptance probability (R) increases
• Start with a target value  for the false
  rejection probability
• Choose a value  such that the false rejection
  probability is equal to :
                  P(L(X) > ; H0) = 
• Once the value x of X is observed, reject H0 if
  L(x) > 
• The choices for  are 0.1, 0.05, and 0.01
          Requirements for LRT
• Ability to compute L(x) for observations X
• Compare the L(x) with the critical value 
• Either use the closed form for L(x) (or log L(x))
  or use simulations to approximate
• A camera checking a certain area
• Recording the detection signal
• X=W, and X=1+W depending on the presence
  of the intruders (hypothesis H0 and H1)
• Assume W~N(0,)
• Find the LRT and acceptance/rejection region
Example (Cont’d)
Example (Cont’d)
Error Probabilities
Example: Binary Channel
Example: Binary Channel
Example: Binary Channel
Example: More on BHT
Example: More on BHT
Example: More on BHT
Example: More on BHT

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