## question 13.9

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• 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
Message 1 of 3 , Nov 2, 2004
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
• 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 2 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 3 of 3 , Nov 14, 2004
yes, i can. thanks again for your help.

brandon

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