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edited Apr 6 2010 at 22:33

The numbers in question are the conditional probabilities that each data set was generated using coin A or coin B given the number of heads that occured in the data set and the current estimate of each coin's bias. We can To compute these numbers, we calculate likelihoods of head counts given coin assignments , and biases, then use Bayes' theorem relates to relate these likelihoods to the desired conditional probabilitesprobabilities.

Then (for example) the first red number in Figure 1(b), step 2 is $\mathbb{P}(z_1 = A | X_1 = x_1, \theta).$

Apply Bayes' theroem to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) \mathbb{P}(z_1 = A | \theta)}{\mathbb{P}(X_1 = x_1 | \theta)}$.

Marginalize the denominator over $z_1$ and use the fact that $\mathbf{z}$ is assumed a priori to be uniformly distributed and independent of $\theta$ to cancel terms of the form $\mathbb{P}(z_1 | \theta)$ to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta)}{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) + \mathbb{P}(X_1 = x_1 | z_1 = B, \theta)} = \frac{\theta_A^{x_1} (1 - \theta_A)^{n - x_1}}{\theta_A^{x_1} (1 - \theta_A)^{10 - x_1} + \theta_B^{x_1} (1 - \theta_B)^{10 - x_1}}$.

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edited Apr 6 2010 at 22:27

Jeff Hussmann
71●4

The numbers in question are the conditional probabilities that each data set was generated using coin A or coin B given the number of heads that occured in the data set and the current estimate of each coin's bias. We can compute likelihoods of head counts given coin assignments, and Bayes' theorem relates these likelihoods to the desired conditional probabilites.

Then (for example) the $i$'th first red number in Figure 1(b), step 2 is $\mathbb{P}(z_1 = A | X_1 = x_1, \theta).$

Apply Bayes' theroem to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) \mathbb{P}(z_1 = A | \theta)}{\mathbb{P}(X_1 = x_1 | \theta)}$.

Marginalize the denominator over $z_1$ and use the fact that $\mathbf{z}$ is assumed a priori to be uniformly distributed and independent of $\theta$ to cancel terms of the form $\mathbb{P}(z_1 | \theta)$, giving theta)$ to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta)}{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) + \mathbb{P}(X_1 = x_1 | z_1 = B, \theta)} = \frac{\theta_A^{x_1} (1 - \theta_A)^{n - x_1}}{\theta_A^{x_1} (1 - \theta_A)^{n theta_A)^{10 - x_1} + \theta_B^{x_1} (1 - \theta_B)^{n theta_B)^{10 - x_1}}$.

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answered Apr 6 2010 at 22:18

Jeff Hussmann
71●4

As in the paper, let $\mathbf{z}$ be the current vector of (random variables representing) assignments of coin A or coin B to each data set, let $\mathbf{X}$ be the vector of head counts in each data set, and let $\theta = (\theta_A, \theta_B)$ be the current estimate of the coins' biases. Then the $i$'th red number in Figure 1(b), step 2 is $\mathbb{P}(z_1 = A | X_1 = x_1, \theta).$ Apply Bayes' theroem to give $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) \mathbb{P}(z_1 = A | \theta)}{\mathbb{P}(X_1 = x_1 | \theta)}$. Marginalize the denominator over $z_1$ and use the fact that $\mathbf{z}$ is assumed a priori to be uniformly distributed and independent of $\theta$ to cancel terms of the form $\mathbb{P}(z_1 | \theta)$, giving $\mathbb{P}(z_1 = A | X_1 = x_1, \theta) = \frac{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta)}{\mathbb{P}(X_1 = x_1 | z_1 = A, \theta) + \mathbb{P}(X_1 = x_1 | z_1 = B, \theta)} = \frac{\theta_A^{x_1} (1 - \theta_A)^{n - x_1}}{\theta_A^{x_1} (1 - \theta_A)^{n - x_1} + \theta_B^{x_1} (1 - \theta_B)^{n - x_1}}$.