Proof that posterior median is the Bayes estimate of absolute loss?

Question

It is always argued that the posterior median is the Bayes estimate associated to the absolute loss function. The proofs I have come across rely on differentiating the conditional Bayesian risk and equating this to zero. However, this shows that this is an inflection point, but not that it is a minimum. How can one prove that the posterior median indeed minimizes the Bayesian risk?

Most, if not all, textbooks just go next level and just leave it as an exercise to the poor reader.

Answer 1

Second derivative yields

$$2 \pi (\delta | x) \geq 0$$

So the original function is convex and hence the median corresponds to a minimum not an inflection point

Answer 2

Adding to Xiaomi answer, here is the derivation, using Leibnitz rule:

$\require{cancel} \frac{\partial \rho}{\partial \delta} = \int_{-\infty}^{\delta}\pi(\theta|x)d\theta - \int_{\delta}^{\infty}\pi(\theta|x)d\theta \\ \frac{\partial^2 \rho}{\partial \delta^2} = \pi(\delta|x)\cdot1 - \cancel{\pi(\delta|x)\cdot0} + \cancel{\int_{-\infty}^{\delta}0d\theta} - [\cancel{\pi(\delta|x)\cdot0} - \pi(\delta|x)\cdot1 + \cancel{\int_{\delta}^{\infty}0d\theta}] = 2\pi(\delta|x) $