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:
- The distribution of therapeutic effect \theta_i as i ranges over individuals in the population
- 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.)
What do the Bayesians think?