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Bayesian Inference a) [3 credits) Verify that p(6) = 3092(1 9)2 is a valid probability distribution on [0, 1] (i.e that it is always nonnegative
Bayesian Inference
a) [3 credits) Verify that p(6) = 3092(1 9)2 is a valid probability distribution on [0, 1] (i.e that it is always nonnegative and that it is normalised.) We ip the coin a number of times After each coin ip, we update the probability distribution for 6 to reect our new belief of the distribution on 6, based on evidence. Suppose we ip the coin four times? and obtain the sequence of coin ips El $14 = 0101. For its two subsequences 01 and 0101. denoted by 131.2, 371.4 (and for the case before any coins are ipped), complete the following questions. b) (15 credits) Compute their probability distribution functions after observing the two subsequences 531.2 and 331:4, respectively. c) (3 credits) Compute their expectation values p of 9 before any evidence as well as after observing the two subsequences mm and 331:4, respectively. (1) (3 credits) Compute their variances 02 of 6 before any evidence as well as after observing the two subsequences 221:2 and 531:4, respectively. e) (5 credits) Compute their maximum a posterior? estimations GMAP of 9 before any evidence as well as after observing the two subsequences 531:2 and 231:4, respectively. Present your results in a table like as shown below. Posterior 13(9) p(9|$1;2 = 01) p(9|$1;4 = 0101) f) (5 credits) Plot each of the probability distributions p(6),p[6|53132 = 01),p(6|:c1:4 = 0101) over the interval 0 g 6' g 1 on the same graph to compare them. g) (6 credits) What behaviour would you expect of the posterior distribution p(6|:c1:n) if we updated on a very long sequence of alternating coin ips 331:\" = 01010101 . . .? What would you expect 11, 0'2, 6).,pr to look like for large n? Sketch/ draw an estimate of what p(6|$1:n) would approximately look like against the other distri- butions. Let X be a random variable representing the outcome of a biased coin with possible outcomes & = {0, 1}, x E X. The bias of the coin is itself controlled by a random variable O, with outcomes ] 0 E 0, where 0 = 10 ER : 0Step by Step Solution
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