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On the denominator of equation 9.75, should not be the position of the productory and sumamtory exchanged?

Posterior probability

1 Answers1

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I think the equation is correct, but you also have a point, in the sense that in this particular instance you can swap the sum and product:

$$ \sum_\mathbf{Z} \prod_{n=1}^N p(\mathbf{x}_n, \mathbf{z}_n | \mathbf{\theta}) = \prod_{n=1}^N \sum_{\mathbf{z}_n} p(\mathbf{x}_n, \mathbf{z}_n | \mathbf{\theta}). $$

Here is a proof sketch:

$$ \begin{aligned} \sum_\mathbf{Z} \prod_{n=1}^N p(\mathbf{x}_n, \mathbf{z}_n | \mathbf{\theta}) &= \sum_{\mathbf{z_1}} \cdots \sum_{\mathbf{z_N}} \prod_{n=1}^N p(\mathbf{x}_n, \mathbf{z}_n | \mathbf{\theta}) \\ &= \sum_{\mathbf{z_1}} \cdots \sum_{\mathbf{z_N}} p(\mathbf{x_1}, \mathbf{z_1} | \mathbf{\theta}) \cdots p(\mathbf{x_N}, \mathbf{z_N} | \mathbf{\theta}) \\ &= \sum_{\mathbf{z_1}} p(\mathbf{x_1}, \mathbf{z_1} | \mathbf{\theta}) \cdots \sum_{\mathbf{z_N}} p(\mathbf{x_N}, \mathbf{z_N} | \mathbf{\theta}) \\ &= \prod_{n=1}^N \sum_{\mathbf{z}_n} p(\mathbf{x}_n, \mathbf{z}_n | \mathbf{\theta}). \end{aligned} $$