bonferroni   13

Simultaneous confidence intervals for multinomial parameters, for small samples, many classes? - Cross Validated
- "Bonferroni approach" is just union bound
- so Pr(|hat p_i - p_i| > ε for any i) <= 2k e^{-ε^2 n} = δ
- ε = sqrt(ln(2k/δ)/n)
- Bonferroni approach should work for case of any dependent Bernoulli r.v.s
q-n-a  overflow  stats  moments  distribution  acm  hypothesis-testing  nibble  confidence  concentration-of-measure  bonferroni  parametric  synchrony
february 2017 by nhaliday
bounds - What is the variance of the maximum of a sample? - Cross Validated
- sum of variances is always a bound
- can't do better even for iid Bernoulli
- looks like nice argument from well-known probabilist (using E[(X-Y)^2] = 2Var X), but not clear to me how he gets to sum_i instead of sum_{i,j} in the union bound?
edit: argument is that, for j = argmax_k Y_k, we have r < X_i - Y_j <= X_i - Y_i for all i, including i = argmax_k X_k
- different proof here (later pages): http://www.ism.ac.jp/editsec/aism/pdf/047_1_0185.pdf
Var(X_n:n) <= sum Var(X_k:n) + 2 sum_{i < j} Cov(X_i:n, X_j:n) = Var(sum X_k:n) = Var(sum X_k) = nσ^2
why are the covariances nonnegative? (are they?). intuitively seems true.
- for that, see https://pinboard.in/u:nhaliday/b:ed4466204bb1
- note that this proof shows more generally that sum Var(X_k:n) <= sum Var(X_k)
- apparently that holds for dependent X_k too? http://mathoverflow.net/a/96943/20644
q-n-a  overflow  stats  acm  distribution  tails  bias-variance  moments  estimate  magnitude  probability  iidness  tidbits  concentration-of-measure  multi  orders  levers  extrema  nibble  bonferroni  coarse-fine  expert  symmetry  s:*  expert-experience  proofs
february 2017 by nhaliday
probability - How to prove Bonferroni inequalities? - Mathematics Stack Exchange
- integrated version of inequalities for alternating sums of (N choose j), where r.v. N = # of events occuring
- inequalities for alternating binomial coefficients follow from general property of unimodal (increasing then decreasing) sequences, which can be gotten w/ two cases for increasing and decreasing resp.
- the final alternating zero sum property follows for binomial coefficients from expanding (1 - 1)^N = 0
- The idea of proving inequality by integrating simpler inequality of r.v.s is nice. Proof from CS 150 was more brute force from what I remember.
q-n-a  overflow  math  probability  tcs  probabilistic-method  estimate  proofs  levers  yoga  multi  tidbits  metabuch  monotonicity  calculation  nibble  bonferroni  tricki  binomial  s:null  elegance
january 2017 by nhaliday

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