6

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?

kjetil b halvorsen
  • 63,378
  • 26
  • 142
  • 467
Tan
  • 1,349
  • 1
  • 13
  • 2
    Why do you integrate CDF? – gunes Oct 16 '20 at 18:14
  • 1
    That $\mu$ to $\infty$ integral is going to be interesting if you don’t mean the PDF (which you do). – Dave Oct 16 '20 at 18:17
  • 1
    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 '20 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 '20 at 18:22
  • Sorry, my mistake. I mean $E(X) = \int_{0}^{\infty} 1-F(x)dx$ – Tan Oct 16 '20 at 18:30
  • Integrating both sides by parts sheds some light on the question. – whuber Oct 16 '20 at 19:02
  • Can you please illustrate a bit more? @whuber F(x) may not be differentiable on R. – Tan Oct 16 '20 at 19:06
  • That's a good point, but the resulting integral nevertheless is well-defined as a Riemann-Stieltjes (or Lebesgue-Stieltjes) integral. See https://stats.stackexchange.com/a/222497/919. – whuber Oct 16 '20 at 19:07
  • Doesn't this require integrability from $X$ to hold? – Freguglia Oct 16 '20 at 23:55
  • 1
    What is "integrability from X"? – Tan Oct 17 '20 at 00:35

1 Answers1

6

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.

Sextus Empiricus
  • 43,080
  • 1
  • 72
  • 161