# Which $\mu$ hold so that integral of CDF (from $\mu$ to $\infty$) equals to integral of 1-CDF (from $-\infty$ to $\mu$)?

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?

• Why do you integrate CDF? Oct 16, 2020 at 18:14
• That $\mu$ to $\infty$ integral is going to be interesting if you don’t mean the PDF (which you do).
– Dave
Oct 16, 2020 at 18:17
• 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 $F$?
– Sycorax
Oct 16, 2020 at 18:22
• 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 $\mu$?
– Tan
Oct 16, 2020 at 18:22
• What is "integrability from X"?
– Tan
Oct 17, 2020 at 0:35

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.