Question

I am currently going through the following article: http://www.nature.com/nbt/journal/v26/n8/full/nbt1406.html

In this article, how did they arrive at the values in the Estimation step (Figure 1 Step b)? Specifically, how did they come up with values such as (0.45, 0.55), (0.80, 0.20), etc.? Any explanation is much appreciated.

Answer 1

These numbers are the normalized likelihoods that the results given in the 10 toss vector are obtained from the current distributions the coin A (or respectively B).

I'll work out the first two rows for illustration:

The guessed Bernoulli parameter for type A is 0.6 and for type B is 0.5. According to the binomial distribution formula, the unnormalized likelihood for obtaining 5H 5T are From A:

L_A = C(10,5)(0.6)^5(0.4)^5

where C(10,5) is the binomial coefficient 10!/5!5!

Similarly from B we obtain:

L_B = C(10,5)(0.5)^5(0.5)^5

The normalized likelihoods are obtained as

For A: L_A/(L_A+L_B) = 0.4491

For B: L_B/(L_A+L_B) = 0.5509

For the second case 9H 1T

L_A = C(10,9)(0.6)^9(0.4)^1

L_B = C(10,9)(0.5)^9(0.5)^9

The normalized likelihoods:

For A: L_A/(L_A+L_B) = 0.8050

For B: L_B/(L_A+L_B) = 0.1950