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Suppose X is a random variable. Is there a way to find $\min\{E|X-c|, c \in \mathbb{R}\}$?

What have I attempted to do so far:$\DeclareMathOperator{\Pr}{Pr}$ $$\Pr(|X - c| \leq t) = \begin{cases} 0 & \text{if } \;t \leq 0\newline\Pr(X \leq t + c) - \Pr(X \leq c - t)& \text{if } t > 0 \end{cases}$$ That results in the probability density function of $|X - c|$ being $$P_{|X - c|}(t) = \begin{cases} 0 & \text{if } t \leq 0,\cr P_{X}(t + c) + P_{X}(c - t)& \text{if } t > 0,\end{cases}$$ where $P_X$ is the probability density function of $X$. That means, that $$E|X-c| = \int_0^\infty t(P(t + c) + P(c - t))\,dt.$$ However this gives me nothing, as finding $\min\{\int_0^\infty t(P_X(t + c) + P_X(c - t))\,dt,\; c \in \mathbb{R}\}$ does not seem to be any easier.

Any help will be appreciated.

Bernard
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Chain Markov
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  • I think there is no general solution for this. Whether it can be solved will depend on the distribution that is involved. Further it might well be that the minimum will be achieved quite often for $c=\mathbb EX$. – drhab Jun 06 '18 at 10:35
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    The minimum is achieved at a median of $X$. See this answer, for instance. – Sangchul Lee Jun 06 '18 at 10:58

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