Quote:
Originally Posted by Ether
Note that this problem can be easily solved using all three methods discussed in this thread:
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Yes, apart from the different piecing together the different lengths, every term in the integrals for this problem appeared in Aren's solution.
Quote:
Originally Posted by Ether
Standard deviation is not so meaningful when the distribution looks this this. |
This is a non-sequitur. Z.B. was referring to the standard deviation across groups of samples, not standard deviation across line lengths. However, the graph of the population also suggests that a nominal MonteCarlo solution might do well to treat these as three separate populations, as Ether has done with the integration cases.
- The "same side" population ramps up from zero, follows a third order polynomial, and goes back to zero at 1.0.
- The "adjacent side" population ramps up linearly from zero to one, has a kink (continuous value, discontinuous slope), and decays back to zero at the square root of two. As Ether has indicated implicitly, note that this side has to be weighted double the other two, because given the first point, the second point can be on only one "same" side, one "opposite" side, or two "adjacent" sides.
- The "opposite side" set has no population with length less than unity, 13+% of its values between 1.00 and 1.01, and 39+% of its lengths between 1.0 and 1.1, and the population continues to decay down to the square root of two.
Edit: By decay, I mean that the population curve is concave up as it approaches zero, not necessarily an exponential decay.