# Why is a frequentist confidence interval equivalent to a credible interval with flat priors?

It's a commonly quoted result that frequentist confidence intervals are equivalent to a bayesian credible interval assuming a flat prior. Ignoring for now questions about invariance under reparameterization or reasonability of a flat prior over the real line in practice, why is this true mathematically?

Let $$X$$ be random vector representing our data. $$f_\theta$$ the distribution of $$X$$ conditional on some value of $$\theta$$. Suppose $$a(X), b(X)$$ are functions such that

$$\int_{X \mid a(X) < \theta < b(X)} f_{\theta}(X) dX = 1 - \alpha$$

Then, for some realization of the data $$X = x$$, the interval $$[a(x), b(x)]$$ is a confidence interval of level $$\alpha$$.

I need to go from that to $$\Pr(a(x) < \theta < b(x) \mid X = x) = 1 - \alpha$$ and have no idea how. Bayes theorem gives proportionality, but seemingly no more. I suspect I’m not understanding something about the rigorous treatment of flat priors.

• – Glen_b Aug 7 '19 at 23:58