1
$\begingroup$

I've been reading the Baum-Welch algorithm and somewhere it mentioned this statistical property: $$ P(X \mid Y,Z) =\frac{ P(X,Y \mid Z)}{P(Y \mid Z)} $$ being based on Bayes' Theorem. I do understand the basic Bayes' Theorem, which is, $$ P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}, $$ but I can't seem to figure out $P(X \mid Y,Z) =\frac{ P(X,Y \mid Z)}{P(Y \mid Z)}$.

$\endgroup$
1
$\begingroup$

The definition of a conditional probability is

$$ \mathbb P( A \mid B) = \frac{\mathbb P(A , B) }{\mathbb P(B) } $$

From that, for three events $X,Y$ and $Z$, we have: \begin{align} \mathbb P( X \mid Y,Z ) = \frac{\mathbb P(X, Y,Z)}{\mathbb P(Y,Z)} \qquad (*) \end{align} Morevoer, $$ \mathbb P(X,Y \mid Z) = \frac{\mathbb P(X, Y,Z)}{\mathbb P(Z)} $$ thus, $$ \mathbb P(X, Y,Z) = \mathbb P(X,Y \mid Z) \mathbb P(Z). $$ If we use this equality in $(*)$ we get,

\begin{align*} \mathbb P(X \mid Y, Z) &= \frac{ \mathbb P(X,Y \mid Z) \mathbb P(Z)}{\mathbb P(Y,Z)} \\ &= \frac{ \mathbb P(X,Y \mid Z) }{\mathbb P(Y \mid Z)}. \end{align*} The last line comes from the fact that $$ \frac{\mathbb P(Z)}{\mathbb P(Y,Z)} = \frac{1}{\mathbb P(Y \mid Z)} $$ So finally,

$$ \mathbb P(X \mid Y, Z) = \frac{ \mathbb P(X,Y \mid Z) }{\mathbb P(Y \mid Z)}. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.