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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?
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.
