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Unformatted text preview: University of Illinois Fall 2009 ECE 313: Problem Set 7: Solutions DecisionMaking Under Uncertainty 1. [A warmup exercise] (a) The likelihood matrix L is as shown below and the maximumlikelihood decision rule is indicated shading. Hypothesis X = 0 X = 1 X = 2 X = 3 H 1 . 4 . 3 . 2 . 1 H . 1 . 2 . 3 . 4 It is easy to get P FA = sum of unshaded entries on H row = 0 . 1 + 0 . 2 = 0 . 3 and P MD = sum of unshaded entries on H 1 row = 0 . 1 + 0 . 2 = 0 . 3 also. (b) By the law of total probability, P ( E ) = P FA + 1 P MD = 0 . 7 . 3 + 0 . 3 . 3 = 0 . 3. (c) The joint probability matrix J = LD where D = diag[ 1 , ] is as shown below, and the Bayesian decision rule is as indicated by the shading. Hypothesis X = 0 X = 1 X = 2 X = 3 H 1 . 12 . 09 . 06 . 03 H . 07 . 14 . 21 . 28 P ( E ) = sum of all the unshaded entries in the J matrix = 0 . 25 is smaller than the average error probability for the ML rule, as it should be. (d) In general, the J matrix can be written as Hypothesis X = 0 X = 1 X = 2 X = 3 H 1 . 4 1 . 3 1 . 2 1 . 1 1 H . 1 . 2 . 3 . 4 Now, if 0 . 1 > . 4 1 = 0 . 4(1 ), that is, > . 8, then the Bayesian decision is in favor of H whenever X = 0. But, if > . 8, then the same decision (favoring H ) holds whenever X equals 1, 2, or 3 also. (Work it out!) Hence, if > . 8, the Bayesian decision always is in favor of H . From the symmetry of the problem, it should be obvious that if 1 > . 8, the Bayesian decision always is in favor of H 1 . The incredulous are invited to work out the details for themselves. Alternatively, the likelihood ratio takes on values, 4, 3 2 , 2 3 and 1 4 , and hence always exceeds 1 = 1 if < . 2, that is, if 1 > . 8, and can never exceed 1 = 1 if > . 8. The result to be proved follows from this....
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This note was uploaded on 10/25/2010 for the course ECE 313 taught by Professor Milenkovic,o during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Milenkovic,O

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