classify x = [0.3, ∗]t . Answer: i )P (ωi ) (a) p(ωi |x) = p(x|ωp(x) . Since P (ω1 ) = P (ω2 ) = P (ω3 ) = 1/3, we just need to compare the likelihood p(x|ωi ). 1 1 exp − [0.3, 0.3][0.3, 0.3]t = 0.1455 2π 2 1 1 p(x|ω2 ) = exp − [−0.7, −0.7][−0.7, −0.7]t = 0.0975 2π 2 p(x|ω1 ) = 2 1 1 exp − [−0.2, −0.2][−0.2, −0.2]t 4π 2 1 1 + exp − [0.8, −0.2][0.8, −0.2]t 4π 2 = 0.1331 p(x|ω3 ) = Since p(x|ω1 ) is the largest likelihood, x…
Words 648 - Pages 3