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You're using a shorthand that captures an important point but does not quite dispose of the objection. Let R(X) be an indicator function that's 1 when largest
by waldrews 2y ago
You're using a shorthand that captures an important point but does not quite dispose of the objection. Let R(X) be an indicator function that's 1 when largest root is real and 0 otherwise; then we want E[R(X)]. I think your point is that E[R(aX)]=E[R(X)], so the choice of Uniform(-1,1) rather than Uniform(-LargeNumber, LargeNumber) is a 'without loss of generality' simplification. Likewise we could take any scale mixture E[R(AX)] for A real, almost surely nonzero, and independent of X. That's a broad class of distributions, but it doesn't cover every 'natural' way of generating coefficient vectors; for example, you could sample X from a hypersphere instead of a hypercube.
- scotty79 2y ago> but it doesn't cover every 'natural' way of generating coefficient vectors; But it covers one of the 'natural' ways. It seems like reasonable question to ponder.
- waldrews 2y agoOh yeah, definitely reasonable. I'm a statistician by trade, so if I see R^n, I want to put a multivariate normal measure on it, which means hyperspheres. Hypercubes are cool too. Just as long as we don't treat the limit from growing the hypercube to cover R^n as being equivalent to a uniform measure on R^n. Unlike, say, in probabilistic number theory, where they shamelessly get away with defining the equivalent of a uniform probability measure on infinitely many integers, much to the consternation of every other kind of mathematician.