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For two mutually exclusive hypotheses 1 brh1m or

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Unformatted text preview: e is exr1 a constant that represents the barrier height f h1. For two mutually exclusive hypotheses 1 Br(h1|m) or, equivalently, Pr(h0) P ackground: Banburismus1at Bletchley d assuming the weight problem-decision rule of evidence is exFormalize the bans, this rearranges to: Pr(h1|m) 1 1 . B 10 Park P(h1|m) (4) indicates that the posterior probability of h1 B nly on the value of the barrier, B, and not on h1 is true How much the evidence favors h1 over h0 lar samples of evidence, m, encountered. In Decision rule: setting B to achieve an expected level of performance (confidence that h1 s, as ilong as the weight of evidence reaches s true). e probability that h1 is correct is a fixed value. Speed-Accuracy trade off: more evidence needs to be accumulated in order to increase Banburismus, setting thethe correctof the barrier height answer. the the probability of getting mined the weight of evidence to accumulate mmitting to a decision was equivalent to setpected level of performance. For example, Implementation: Banburismus Performed by Neurons (in 2 AFC choice) • How does the activity of sensory neurons compute the “weight of evidence”? • How does the brain make use of the sensory evidence? • What is the decision rule based on this evidence? 992;(and assume opponency, giving an ,analogous expres- , Salzman Pr(x y|h1) 2 2 in these areas h0): 2 √1 sion under that the brain Banburismus Performed by Neurons Q where e2 is was most Sensory xSignals-Turing’s Weight of Evidence , (5) Pr( ,y|h1) 2 22 2 Neuron (y 2 (x 1 2 (x √1 1) 0) 1)(y 302 Q 2 2 1 Using the difference (x-y) in spike rates from two where • (5) ) 0 , neurons (one 2 the common variance, other favors Down favors Up motion, the and are the means of 2 is cisions about (x 2 1 ) (y ) 2 (x )(y )0 1 1 0 1 0 motion) x and y, respectively, under h1, and , is the covariance Qlike Tu- 2 tities 2 1 between x and y. Solving for the weight of evidence (in ically related h1: Up motion ( x is larger) 2 is the common variance, 1 and 0...
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