Problem 331. Compute Area from Fixed Sum Cumulative Probability
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4 Comments
A really nice problem : concrete origin, careful formulation, then a good walk in 3D geometry, analysis, integration and at the end a single formula. The critical value cuts the triangle in four pieces : my son said it was Zelda !
Raphael: I agree completely!
Perhaps you could have a precision requirement a bit more liberal in order to allow numerical approximation algorithms or other approaches as well? just my two cents, the problem looks great
Tip: the probability grows from the vertices of simplex (triforce/triangle) toward its center (which has the highest frequency).
Solution Comments
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2 Comments
This is not a solution, and it should be deleted. And If it is possible, then more test cases should be added.
The solution has been rescored, and such solutions are no longer allowed on this question.
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1 Comment
I am pleased that you solved this problem, David. Congratulations! I didn't find any particularly easier way of solving it. The crucial step is showing that the probability density is proportional to your 1/y^3 for points within the corresponding "kite-shaped region". I used the Jacobian between two coordinate systems to show that. After dividing that region into two halves everything falls into place, though in my dotage I had to make heavy use of the Symbolic Toolbox to check for errors. (I hope this problem will serve as a warning to people who recommend this method of producing random numbers with a predetermined sum.) R. Stafford
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1 Comment
It is inherent in the definition of P here that the density, dP/dA, must increase as P increases and therefore dA/dP must decrease. In your proposed solution you have dA/dP increasing as P increases. R. Stafford
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