4 ms·
Yes, for a normal, which is known to have extremely thin tails (decaying with a squared exponential!). Chebyshev's bound holds for any possible distribution (wi
by FabHK 5y ago
Yes, for a normal, which is known to have extremely thin tails (decaying with a squared exponential!). Chebyshev's bound holds for any possible distribution (with finite variance).
- FabHK 5y agoNot squared exponential (that's just an exponential), but exponential of a square, even. Anyway.