elec 303 – random signals lecture 17 – hypothesis testing 2 dr. farinaz koushanfar ece dept.,...
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Four versions of MAP rule discrete, X discrete discrete, X continuous continuous, X discrete continuous, X continuousTRANSCRIPT
ELEC 303 – Random Signals
Lecture 17 – Hypothesis testing 2Dr. Farinaz Koushanfar
ECE Dept., Rice UniversityNov 2, 2009
outline
• Reading: 8.2,9.3• Bayesian Hypothesis testing• Likelihood Hypothesis testing
Four versions of MAP rule
• discrete, X discrete
• discrete, X continuous
• continuous, X discrete
• continuous, X continuous
)|x(p)(p |X
)|x(f)(p |X
)|x(p)(f |X
)|x(f)(f |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,XSi)
• The probability of error is: i P(i,XSi)
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 WiN(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(XA;Hj) denotes the probability that X
belongs to a set A, when Hj is true
Rejection/acceptance
• 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(XR; H0)• False acceptance:– Happens with probability
(R) = P(XR; 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)
Example
• 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
LRT
• 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
Example
• 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
Example
Example
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