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I need to prove this Bayes' rule in multi-variable case:

In the probability theorem, if $t_{i}=(y_{i},\delta_{i})$, how could I prove that

$p(t_{i}|\theta) = p(y_{i},\delta_{i}|\theta) = p(y_{i}|\delta_{i},\theta)p(\delta_{i}|\theta)$


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Just apply the definition of conditional probability: $$ P(\color{Red}{A} \mid \color{Blue}{B}) = \frac{P(\color{Red}{A} \cap \color{Blue}{B})}{P(\color{Blue}{B})} = \frac{P(\color{Red}{A} \color{Blue}{B})}{P(\color{Blue}{B})} ,$$ assuming $P(\color{Blue}{B}) \gt 0$. (I will use $P$ instead of $p$ everywhere.)

$$ \begin{align*} P(\color{Red}{y_i, \delta_i} \mid \color{Blue}{\theta}) &= \frac{P(\color{Red}{y_i, \delta_i}, \color{Blue}{\theta})}{P(\color{Blue}{\theta})} \\ &= \frac{P(y_i, \delta_i, \theta)}{P(\delta_i, \theta)} \frac{P(\delta_i, \theta)}{P(\theta)} \\ &= \frac{P(\color{Red}{y_i}, \color{Blue}{\delta_i, \theta})}{P(\color{Blue}{\delta_i, \theta})} \frac{P(\color{Red}{\delta_i}, \color{Blue}{\theta})}{P(\color{Blue}{\theta})} \qquad \qquad \text{added color coding} \\ &= P(\color{Red}{y_i} \mid \color{Blue}{\delta_i, \theta}) \cdot P(\color{Red}{\delta_i} \mid \color{Blue}{\theta}) \end{align*} $$

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thanks, that is answer I am after. – xuesong Nov 26 '11 at 3:42

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