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Consider choosing $\theta^*$ that minimizes the expected absolute loss:

\begin{align} \tag{1} \int_{\Theta}|\theta-\theta^*|\pi(\theta|\mathbf{x})d\theta= \int_{-\infty}^{\theta^*}(\theta^*-\theta)\pi( \theta|\mathbf{x})d\theta+\int_{\theta^*}^{\infty}(\theta-\theta^*)\pi(\theta|\mathbf{x})d\theta \end{align}

Differentiating with respect to $\theta^*$ and equating to zero yields: \begin{align} \tag{2} \int_{-\infty}^{\widehat{\theta^*}}\pi(\theta|\mathbf{x})d\theta = \int_{\widehat{\theta^*}}^{\infty}\pi(\theta|\mathbf{x})d\theta \end{align}

EDITED:

I'm confused about the step taken from (1) to (2), which can be found on p. 12 (94) in online notes (Note: I replaced $\delta(\mathbf{x})$ by $\theta^*$) . A more detailed insight will be highly appreciated. My understanding is that it makes use of Leibniz rule. Here $\widehat{\theta^*}$ is the Bayes estimate (which turns out to be the median of $\pi(\theta|\mathbf{x})$), an argument that minimizes (1).

AlexMe
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    Could you explain what kind of mathematical object you mean by "$\delta$"? If it's not a number (the notation suggests it might be a *function*), could you further tell us what you mean by "differentiating" with respect to it would be? – whuber Nov 07 '18 at 17:39
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    Please explain how one differentiates with respect to an estimator! The only way I can think of doing this in general is explained at https://stats.stackexchange.com/questions/369933, but it's unclear whether that's what you have in mind. – whuber Nov 07 '18 at 17:50
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    Yes it's a well-known result. But it is derived using actual mathematics rather than fanciful operations like those you seem to be using! (That's not to denigrate such work--non-rigorous or even nonsensical forms of mathematics can provide some intuition. But that's not what you seem to be asking about.) Until we know more about what you mean by differentiating with respect to a function, none of your work will be in the least clear. – whuber Nov 07 '18 at 18:19
  • @whuber thanks, you're right, I will reformulate the question. This is exactly why I found that proof confusing. – AlexMe Nov 07 '18 at 18:21
  • @whuber I was wondering if the question is more clearly formulated. If the edited question is clear, then we could delete additional comments so that the subsequent readers do not get distracted by them (and also this place will look "cleaner"). Thanks :) – AlexMe Nov 07 '18 at 19:25
  • @Xi'an what if the support of $\pi(\theta|\mathbf{x)}$ is $(-\infty, \infty)$? Does the result in https://stats.stackexchange.com/questions/333521/integral-of-a-cdf still hold? – AlexMe Nov 07 '18 at 20:11
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    @AlexMe You can avoid the Leibniz rule if you write $\int_{-\infty}^t(t-x)\pi(x)\,dx=t\int_{-\infty}^t \pi(x)\,dx -\int_{-\infty}^t x\,\pi(x)\,dx$. Then differentiate the first piece using the product rule, and the second piece using FTC. – grand_chat Nov 07 '18 at 22:25
  • Many thanks @grand_chat! One of the proofs that I have seen online made use of Leibniz rule, and hence I went down that road. Apparently it was much simpler than that. Wonder why that article did not consider this simpler proof!? – AlexMe Nov 08 '18 at 19:17

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