An E step

Topic: Mixtures and the EM algorithm · more questions on this topic
‹ Mean and variance of a mixture An M step ›

For the mixture of the previous exercise, compute \tau_1(x) at x=1 and at x=3.

Worked answer (click to show)

A responsibility is a normalised weighted density. Pass the standard deviation, rather than the variance, when a normal density routine is parameterised by the standard deviation. Component 1 is \operatorname{N}(0,1) and component 2 is \operatorname{N}(4,4), so the second standard deviation is 2.

At x=1: f_1=\varphi(1)=0.2420 and f_2=\varphi(-1.5)/2=0.1295/2=0.0648. The weighted values are 0.3\times 0.2420=0.0726 and 0.7\times 0.0648=0.0453, so

\tau_1(1)=\frac{0.0726}{0.1179}=0.6156.

At x=3: f_1=\varphi(3)=0.00443 and f_2=\varphi(-0.5)/2=0.3521/2=0.1760. The weighted values are 0.3\times 0.00443=0.00133 and 0.7\times 0.1760=0.1232, so

\tau_1(3)=\frac{0.00133}{0.1246}=0.0107.