3 ms·

Also, the median can only be 1stddev away from the mean in any distribution, so this can further winnow the search.

• mjburgess 4y ago
  ...this isn't true: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,10^30
  • zeristor 4y ago
    Reminds me of the distribution in salaries. I read …, 1, 10, 30 but …, 1, 10^30 is far more dramatic.
  • Chirono 4y ago
    I’d double check your working there because it certainly is true. In your example the standard deviation (2e29) is far bigger than the difference between the median and mean (5e28).
  • hhmc 4y ago
    Someone has already explained why your example is wrong, but for a general proof, for dist X, let u = mean, m = median | u - m | = |E(X - m)| <= E(|X - m|) <= E(|X - u|) <= E(sqrt((X - u)^2)) <= stddev mostly steps via Jensen's inequality
    • mjburgess 4y ago
      oops... I interpreted the claim, admittedly bizarrely, as something like: fit N(u, s) to {(x, y)...} then m will lie u+-s rather than, def. s^2 = E[X^2] - E[X]^2