# Where are the exceptional responders?

**URL:** <https://discourse.datamethods.org/t/where-are-the-exceptional-responders/17961>\
**Category:** research methods\
**Created:** [July 24, 2024, 10:38pm UTC](https://discourse.datamethods.org/t/where-are-the-exceptional-responders/17961 "2024-07-24T22:38:09Z")\
**Posts on this page:** 1\
**Showing post:** 10

<div class="post-metadata">

**Author:** ![davidcnorrismd](https://discourse.datamethods.org/user_avatar/discourse.datamethods.org/davidcnorrismd/32/3502_2.png) [@davidcnorrismd](https://discourse.datamethods.org/u/davidcnorrismd)\
**Post date:** [March 25, 2026, 7:38am UTC](https://discourse.datamethods.org/t/where-are-the-exceptional-responders/17961/10 "2026-03-25T07:38:14Z")

</div>

It occurred to me yesterday that I might have gained at least some _formal_ advantage by proceeding directly to a Bayesian _expectation_:

\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}

Now this would _seem_ to let us solve directly for \theta\_p — _if only_ we can validly equate

p = \text{E}[\text{P}(\theta\_i \< \theta\_p \mid n)]. \tag{$\star$}

This is because, taking logs on both sides of

p = \left(\frac{\beta\_n}{\beta\_n-\ln\theta\_p}\right)^\alpha = \left(1 - \frac{\ln\theta\_p}{\beta\_n}\right)^{-\alpha},

we would obtain

-\frac{\ln p}{\alpha} = \ln\left(1 - \frac{\ln\theta\_p}{\beta\_n}\right) \< -\frac{\ln\theta\_p}{\beta\_n},

from which we get

\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)}.

Substituting the same p=0.9 and \theta\_c=0.8 as above, we would obtain

\theta\_{0.9} \< 0.9^{(200/2)\ln(1/0.8)} \doteq 0.9^{22.3} \doteq 0.095.

BUT: I have serious doubts about Eq (\star), which seems to conflate two distinct _kinds_ of distribution:

1. The distribution of therapeutic effect \theta\_i as i ranges over individuals in the population
2. The Bayesian’s subjective probability over a family of these distributions, indexed by \lambda.

(Indeed, I’m reminded of the quote I’d previously referenced [here](https://discourse.datamethods.org/t/good-ways-to-verify-coverage-of-cis-for-probabilities-in-variational-bayes-binary-classification/28450/2).)

**What do the Bayesians think?**

---

_[View the full topic](https://discourse.datamethods.org/t/where-are-the-exceptional-responders/17961)._