Both follow the normal proof methods ....
Note that P(X,Y|e) = P(X,Y,e)/P(e)
and P(X|Y,e) = P(X,Y,e)/P(Y,e)
and P(Y|e) = P(Y,e)/P(e)
The first part is immediate from this.
The second part is almost trivial after this ...
P(X,Y|e) = P(X|Y,e) P(Y|e)
P(X,Y|e) = P(Y|X,e) P(X|e)
Can u finish it off from here?
--- In firstname.lastname@example.org, "brandon_pitt_47"
> Hope someone can help me with this one. completely stuck. any
solutions out there? it's
> on page 490 about proving the conditionalized version of the general
> P(X,Y|e) = P(X|Y,e) P(Y|e)
> part b: prove the conditionalized version of Bayes' Rule in
> P(Y|X,e) = P(X|Y,e) P(Y|e)
> ANY HELP WOULD BE GREAT.