Sent from my iPhone
On Sep 24, 2008, at 5:16 PM, Jason House <[EMAIL PROTECTED]>
wrote:
On Sep 24, 2008, at 2:40 PM, Jacques Basaldúa <[EMAIL PROTECTED]> wr
ote:
Therefore, the variance of the normal that best approximates the
distribution of both RAVE and
wins/(wins + losses) is the same n·p·(1-p)
See above, it's slightly different.
If this is true, the variance you are measuring from the samples
does not contain any information
about the precision of the estimators. If someone understands this
better, please explain it to
the list.
This will get covered in my next revision. A proper discussion is
too much to type with my thumb...
My paper writing time is less than I hadhiped, so here's a quick and
dirty answer.
For a fixed win rate, the probabilities of a specific number of wins
and losses follows the binomial distribution. That distribution keeps
p (probability of winning) constant and the number of observed wins
and losses variable.
When trying to reverse this process, the wins and losses are kept
constant and p varies. Essentially prob(p=x) is proportional to
(x^wins)(1-x)^losses.
This is a Beta distribution with known mean, mode, variance, etc...
It's these values which should be used for approximating the win rate
estimator as a normal distion. Does that makes sense?
I've glossed over a very important detail when "reversing". Bayes
Theorem requires some extra a priori information. My preferred
handling alters the reversed equation's exponents a bit but the basic
conclusion (of a beta distribution) is the same._______________________________________________
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