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It's impossible to have a uniform distribution over (-inf, +inf). Since the distribution is uniform, each digit has an equal chance of being zero through nine,
by jprete 2y ago
It's impossible to have a uniform distribution over (-inf, +inf). Since the distribution is uniform, each digit has an equal chance of being zero through nine, but that's true of all digits or else it's no longer uniform. At that point 100% of the values drawn in any sampling of the distribution are infinite.
If scaling the coefficients down to [-1, +1] gives a uniform distribution, then the original distribution must have been uniform in [-x, +x] where x is the largest coefficient in the original polynomial. That, too, is a contradiction! It means that either -x or +x are guaranteed to be one of the random coefficients, which is very much not a uniform distribution.
(That said, I think the (-1, +1) version of the problem still has interest!)
- zeroonetwothree 2y agoTypically you would do a uniform (-x, x) and then take the limit. This isn’t necessarily the same as the (-1, 1) version though
- eigenket 2y agoIts the same because the uniform distribution over (-x,x) can be obtained by taking a random variable uniformly distributed over (-1,1) and multiplying it by x. This doesn't change anything for the question about roots because the roots of a polynomial don't change when you multiply it by x.
- thaumasiotes 2y ago> Typically you would do a uniform (-x, x) and then take the limit. But this is completely uninformative for the interval (-inf, +inf). Taking limits is a tool you can use, and it works well when the infinite behavior matches the limit of the finite behavior. The opposite is the case here and the infinite behavior is radically unlike the finite behavior. You might as well try to determine the value of an impulse function at zero by seeing what the limit is as you approach zero.