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Topic: On the correct way to estimate p given that x events were observ
Replies: 0

 Luis A. Afonso Posts: 4,758 From: LIsbon (Portugal) Registered: 2/16/05
On the correct way to estimate p given that x events were observ
Posted: Oct 4, 2013 2:17 PM

On the correct way to estimate a proportion given that x events were observed in n trials.

We intend to discuss the method to estimate the proportion through p= (r+1)/(n+1) as shown in the paper:
A Note on the Calculation of Empirical P Values from Monte Carlo ... B. V. North, D. Curtis and P. C. Shame
Am. J. Hum. Genet 71-439-441, 2002

http://www.ncbi.nlm.nih.gov/pmc/articles/PMC379178/

___1____Why we add 1 to the Numerator?

The reason is that we expect that the next occurrence is a success, a forecast that is always hazardous mainly when r is very large compared with x. In terms of expectation, E(r), one have
_p´= (r+E(r))/(n+1) = (r+r/n)/(n+1) = r* (1+1/n)/(n+1) = r/n

Therefore, following this criterion, p=r/n needs not at all to be *corrected* and is independent of how many trials are performed. The current formula is neatly an estimate by excess of the future proportion. . .

An humorist note
I do not know at what extent the AA. have enjoyed to prove numerically trough simulations (!) that (see *)
____________ r/n <= (r+1)/(n+1)
A High School pupil did do : Since (n+1)/n <= (r+1)/r or more evident (1+1/n) <= (1+1/r), 1/n<=1/r, n>=r and the *problem* is solved . . .
* Apparently with *inverted* results : 12´103 against 10´106.

More profitable they were involved in a golf game.

Luis A. Afonso