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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?

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

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  • $\begingroup$ this is not an answer, but rather a request. could you please share the reference from where the image was taken? Thanks a lot! $\endgroup$
    – user390869
    Commented Jun 21, 2023 at 20:49
  • $\begingroup$ Relevant (a proof that does not use differentiation): stats.stackexchange.com/questions/600225/… $\endgroup$
    – Zhanxiong
    Commented Jun 21, 2023 at 21:23

2 Answers 2

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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

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  • $\begingroup$ How do you get Zero? Are you using $d/dx \int_a^x f(y)dy =f(x)$ and $. - \int_x^a f(y)dy = \int_a^x f(y)dy $? $\endgroup$
    – Xiaomi
    Commented Nov 14, 2018 at 1:58
  • $\begingroup$ The second derivative as a function of $\delta$ is not zero everywhere. The first derivative is $ 2 \int^\delta \pi(y)dy - 1$. Taking the derivative a second time gives the above result which is clearly not zero everywhere $\endgroup$
    – Xiaomi
    Commented Nov 14, 2018 at 2:13
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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) $

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