## Re: question 13.9

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• 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
Message 1 of 3 , Nov 13, 2004
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 ...
Note that,
P(X,Y|e) = P(X|Y,e) P(Y|e)
also,
P(X,Y|e) = P(Y|X,e) P(X|e)

Can u finish it off from here?
-- AI

--- In aima-talk@yahoogroups.com, "brandon_pitt_47"
<brandon_pitt_47@y...> wrote:
>
>
> 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
product rule.
>
> P(X,Y|e) = P(X|Y,e) P(Y|e)
>
> part b: prove the conditionalized version of Bayes' Rule in
Equation (13.10)
>
> P(Y|X,e) = P(X|Y,e) P(Y|e)
> ------------
> P(X)
>
> ANY HELP WOULD BE GREAT.
>
> thanks,
> brandon
• yes, i can. thanks again for your help. brandon __________________________________ Do you Yahoo!? Check out the new Yahoo! Front Page. www.yahoo.com
Message 2 of 3 , Nov 14, 2004
yes, i can. thanks again for your help.

brandon

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