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What is the $\mu$ s.t. $$\int_{\mu}^{\infty}1-F(x)dx = \int_{-\infty}^{\mu}F(x)dx?$$

Here $F(x) = P(X\leq x).$

Should $\mu$ be the median of X, i.e. $0.5=F(\mu)$? I think $\mu$ should be the point so that $F(\mu) = 1-F(\mu)$, which is the median of X. But how do I derive it mathematically?

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    $\begingroup$ Why do you integrate CDF? $\endgroup$
    – gunes
    Oct 16 '20 at 18:14
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    $\begingroup$ That $\mu$ to $\infty$ integral is going to be interesting if you don’t mean the PDF (which you do). $\endgroup$
    – Dave
    Oct 16 '20 at 18:17
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    $\begingroup$ Hint: what's the definition of a median for an absolutely continuous CDF $F$? Can you write the definition of the median in terms of $F$? $\endgroup$
    – Sycorax
    Oct 16 '20 at 18:22
  • $\begingroup$ I was using $E(X)=\int_{0}^{\infty}F(x)dx$, where X is non-negative random variable, to get $E|X-\mu]$. The last step I got was $\int_{\mu}^{\infty}1-F(x)dx + \int_{-\infty}^{\mu}F(x)dx$. But there should be a $\mu$ so that the last two parts of the RHS equals. I wonder what is the $\mu$? $\endgroup$
    – Tan
    Oct 16 '20 at 18:22
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    $\begingroup$ What is "integrability from X"? $\endgroup$
    – Tan
    Oct 17 '20 at 0:35
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The mean of a variable $X$ can be computed as

$$\mu_X = \int_{0}^{\infty}1-F(x)dx - \int_{-\infty}^{0} F(x)dx $$

The mean of a shifted variable $X-\mu_X$ (which equals zero) is computed as

$$0 = \int_{0}^{\infty}1-F(x+\mu_X)dx - \int_{-\infty}^{0} F(x+\mu_X)dx $$

Or

$$ 0 = \int_{\mu_X}^{\infty}1-F(x)dx -\int_{-\infty}^{\mu_X} F(x)dx$$

Which is equivalent to your equation.

Therefore the mean $\mu_X$ in these computations is the same as the parameter $\mu$ in your question.

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