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Bayes rule in its original, probability, form says that
Bayes rule in its original, probability, form says that


::''p''(''F'' | ''E'') = ''p''(''E'' | ''F'')p(''F'')/p(''E'')
::''p''(''F'' | ''E'') = ''p''(''E'' | ''F'') ''p''(''F'') / ''p''(''E'')


providing p(''E'') is not zero…  Suppose that your probability for ''F'' were zero, then since multiplication of zero by any number always gives the same reulst, zero, the right-hand, and hence also the left-hand, sides will always
providing p(''E'') is not zero…  Suppose that your probability for ''F'' were zero, then since multiplication of zero by any number always gives the same reulst, zero, the right-hand, and hence also the left-hand, sides will always

Revision as of 18:55, 13 January 2014

Dennis Lindley

Leading British statistician, Dennis Lindley, dies
StatsLife.org, 16 December 2013

Provides link to 32 minute video of an interview with Lindley.

In Lindley's book Understanding Uncertainty (Wiley, 2005), Section 6.8 (pp. 90-91) a devoted to an discussion of Cromwell's Rule, which was referenced in Quotations above. Quoting from Lindley's description:

Bayes rule in its original, probability, form says that

p(F | E) = p(E | F) p(F) / p(E)

providing p(E) is not zero… Suppose that your probability for F were zero, then since multiplication of zero by any number always gives the same reulst, zero, the right-hand, and hence also the left-hand, sides will always be zero whatever be the evidence E. In other words, if you have probability zero for something, F, you will always have probability zero for it, whatever evidence E you receive.

Submitted by Paul Alper