But the frequentist answer is only different when the frequentist can't use common sense. If you use min(D) as the frequentist estimator, you would get a very d

But the frequentist answer is only different when the frequentist can't use common sense. If you use min(D) as the frequentist estimator, you would get a very different confidence interval, as it would have the form [min(D) - constant, min(D)]. The CDF of the truncated exponential is F(x) = 1-exp(theta-x), and the CDF of the minimum of three samples is 1-(1-F(x))^3. I get that the frequentist 95% CI is [9.00142, 10], which for all intents and purposes is the same as the credible interval the author computes. I agree that credible intervals and confidence intervals answer different questions. I don't think that it's obvious that the confidence interval approach is wrong, and the example in the blog post is definitely not evidence towards this.