{
  "$type": "site.standard.document",
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  "path": "/t/where-are-the-exceptional-responders/17961#post_10",
  "publishedAt": "2026-03-25T07:38:14.000Z",
  "site": "https://discourse.datamethods.org",
  "tags": [
    "here"
  ],
  "textContent": "It occurred to me yesterday that I might have gained at least some _formal_ advantage by proceeding directly to a Bayesian _expectation_ :\n\n\\begin{aligned} \\text{E}[\\text{P}(\\theta_i < \\theta \\mid n)] &= \\int_0^\\infty \\theta^\\lambda\\,p(\\lambda\\mid n) \\,\\text{d}\\lambda \\\\\\ \\\\\\ &= \\int_0^\\infty \\theta^\\lambda \\frac{\\beta_n\\,^\\alpha}{\\Gamma(\\alpha)} \\lambda^{\\alpha-1} e^{-\\beta_n\\lambda}\\,\\text{d}\\lambda \\\\\\ \\\\\\ &= \\left(\\frac{\\beta_n}{\\beta_n-\\ln\\theta}\\right)^\\alpha \\int_0^\\infty \\frac{(\\beta_n-\\ln\\theta)^\\alpha}{\\Gamma(\\alpha)} \\lambda^{\\alpha-1} e^{-(\\beta_n-\\ln\\theta)\\lambda}\\,\\text{d}\\lambda \\\\\\ \\\\\\ &= \\left(\\frac{\\beta_n}{\\beta_n-\\ln\\theta}\\right)^\\alpha \\int_0^\\infty \\text{d}\\,\\text{Gamma}(\\alpha, \\beta_n-\\ln\\theta) \\\\\\ \\\\\\ &= \\left(\\frac{\\beta_n}{\\beta_n-\\ln\\theta}\\right)^\\alpha. \\end{aligned}\n\nNow this would _seem_ to let us solve directly for \\theta_p — _if only_ we can validly equate\n\np = \\text{E}[\\text{P}(\\theta_i < \\theta_p \\mid n)]. \\tag{$\\star$}\n\nThis is because, taking logs on both sides of\n\np = \\left(\\frac{\\beta_n}{\\beta_n-\\ln\\theta_p}\\right)^\\alpha = \\left(1 - \\frac{\\ln\\theta_p}{\\beta_n}\\right)^{-\\alpha},\n\nwe would obtain\n\n-\\frac{\\ln p}{\\alpha} = \\ln\\left(1 - \\frac{\\ln\\theta_p}{\\beta_n}\\right) < -\\frac{\\ln\\theta_p}{\\beta_n},\n\nfrom which we get\n\n\\ln\\theta_p < \\frac{\\beta_n}{\\alpha}\\,\\ln p \\implies \\theta_p < p^{\\beta_n/\\alpha} = p^{(\\beta+n\\ln(1/\\theta_c))/\\alpha} < p^{(n/\\alpha)\\ln(1/\\theta_c)}.\n\nSubstituting the same p=0.9 and \\theta_c=0.8 as above, we would obtain\n\n\\theta_{0.9} < 0.9^{(200/2)\\ln(1/0.8)} \\doteq 0.9^{22.3} \\doteq 0.095.\n\nBUT: I have serious doubts about Eq (\\star), which seems to conflate two distinct _kinds_ of distribution:\n\n  1. The distribution of therapeutic effect \\theta_i as i ranges over individuals in the population\n  2. The Bayesian’s subjective probability over a family of these distributions, indexed by \\lambda.\n\n\n\n(Indeed, I’m reminded of the quote I’d previously referenced here.)\n\n**What do the Bayesians think?**",
  "title": "Where are the exceptional responders?"
}