Question

Suppose that the number of defects on a roll of magnetic recording tape has a Poisson distribution for which the mean λ is known to be either 1 or 5. Suppose the prior pmf for λ is: P[λ = 1] = 0.4 P[λ = 5] = 0.6 Suppose that we examine five tapes, and find that the number of defects are x1 = 3,x2 = 1,x3 = 4,x4 = 6 and x5 = 2. Show that the posterior distribution for λ is: P[λ = 1|¯ x] = 0.012 P[λ = 5|¯ x] = 0.988

Answer 1

la <- sum(log(dpois(c(3, 1, 4, 6, 2), 1)))
lb <- sum(log(dpois(c(3, 1, 4, 6, 2), 1.5)))
.4*exp(la) / (.4*exp(la) + .6*exp(lb))

.6*exp(lb) / (.4*exp(la) + .6*exp(lb))

> la <- sum(log (dpois (c(3, 1, 4, 6, 2), 1))) > lb <- sum (log(dpois (c(3, 1, 4, 6, 2), 1.5))) > .4*exp(la) / (.4*exp(la) +.6*exp(lb)) [1] 0.01221372 > .6*exp(lb) / (.4*exp(la) +.6*exp(lb)) [1] 0.9877863