Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.stat.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
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 71439441, 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



