
\\
&amp; = &amp; \mathsf{normal}(y_n \mid \mu_k, \sigma_k) \cdot \lambda_k.
\end{array}
\]</span></p>
<p>The normalization can be done via summation, because <span class="math inline">\(z_n \in 1{:}K\)</span> only
takes on finitely many values. In detail,
<span class="math display">\[
p(z_n = k \mid y_n, \mu, \sigma, \lambda)
\ = \
\frac{p(y_n \mid z_n = k, \mu, \sigma) \cdot p(z_n = k \mid \lambda)}
     {\sum_{k&#39; = 1}^K p(y_n \mid z_n = k&#39;, \mu, \sigma)
                    \cdot p(z_n = k&#39; \m