Comments on: Two Mathematical Bugs http://www.algorithm.co.il/blogs/math/two-mathematical-bugs/ Algorithms, for the heck of it Tue, 21 Jun 2011 21:07:08 +0000 hourly 1 http://wordpress.org/?v=3.1.3 By: lorg http://www.algorithm.co.il/blogs/math/two-mathematical-bugs/#comment-100 lorg Sun, 09 Mar 2008 23:28:11 +0000 http://www.algorithm.co.il/blogs/index.php/programming/python/two-mathematical-bugs/#comment-100 Thanks for the proof! Just the statement "sampling a random point from the area under the curve" is more elegant and understandable than my original proof. Thanks for the proof!
Just the statement “sampling a random point from the area under the curve” is more elegant and understandable than my original proof.

]]> By: Adam Morrison http://www.algorithm.co.il/blogs/math/two-mathematical-bugs/#comment-99 Adam Morrison Sun, 09 Mar 2008 09:55:49 +0000 http://www.algorithm.co.il/blogs/index.php/programming/python/two-mathematical-bugs/#comment-99 Actually, I think your solution is correct. Essentially you are sampling a random point from the area under the curve. More formally, denote your random variable by z. Then, for x to the left of the curve's symmetry axis, Pr[z <= x] is the integral from P(a) to P(x) of (x - P^{-1}(y)) dy, which if you think about it, is exactly the integral of P(x) over [a,x]. The argument for the other x values is analogous. (All of this assumes that P(x) is itself a proper distribution, otherwise you need to divide by the area under the curve.) I don't see why the fact that the probability for passing the 2nd stage isn't constant is relevant, since these are not independent samples... Actually, I think your solution is correct. Essentially you are sampling a random point from the area under the curve.

More formally, denote your random variable by z. Then, for x to the left of the curve’s symmetry axis, Pr[z <= x] is the integral from P(a) to P(x) of (x – P^{-1}(y)) dy, which if you think about it, is exactly the integral of P(x) over [a,x]. The argument for the other x values is analogous. (All of this assumes that P(x) is itself a proper distribution, otherwise you need to divide by the area under the curve.)

I don’t see why the fact that the probability for passing the 2nd stage isn’t constant is relevant, since these are not independent samples…

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