2
$\begingroup$

Can you tell me steps how to prove that $\int_{-\infty}^\infty F(x) \text{d}F(x) = 0,5$?

$\endgroup$
  • 4
    $\begingroup$ Welcome to CV. This question is much too brief. You're assuming that the equation is sufficient information to obtain a response. This is not the case. You need to elaborate on your objectives in posing this question, and why you're interested in a proof or risk having the question removed. $\endgroup$ – Mike Hunter Apr 10 '16 at 10:32
  • 1
    $\begingroup$ Is this a question from a course of textbook? Please add the [self-study] tag & read its wiki. Then tell us what you understand thus far, what you've tried & where you're stuck. We'll provide hints to help you get unstuck. Please make these changes as just posting your homework & hoping someone will do it for you is grounds for closing. $\endgroup$ – Silverfish Apr 10 '16 at 11:09
  • $\begingroup$ Did you try integration by parts? $\endgroup$ – kjetil b halvorsen Apr 10 '16 at 11:16
  • 1
    $\begingroup$ I'm voting to keep this closed because the question does not comply with the site's self-study policy. $\endgroup$ – Sycorax says Reinstate Monica Apr 11 '16 at 12:03
  • $\begingroup$ Related, more general, question that asks for the value of $$\int_{-\infty}^\infty G(x)\,\mathrm dF(x)$$ where $G(x)$ and $F(x)$ are CDFs. $\endgroup$ – Dilip Sarwate Apr 11 '16 at 13:47
6
$\begingroup$

A simple (and obvious) substitution solves this immediately.

Simply by looking at the integral, the substitution $u=F(x)$ leaps out from the page and tugs at our locks. The result (the remainder of which is even simpler, and which I leave for you) follows. (Keep in mind this is a definite integral. Pay attention to the limits.)

As JohnK explains in comments, this result will be valid for continuous distributions.

$\endgroup$
  • $\begingroup$ That could have been a comment, no? $\endgroup$ – JohnK Apr 10 '16 at 13:36
  • 2
    $\begingroup$ @JohnK Keeping in mind the guidelines on answering self-study questions, in particular the first two points, and yet the need to actually answer questions rather than leave them unanswered, what would you suggest that would still leave the OP something substantive to do? The guidance in the answer leads to an immediate solution to an essentially trivial question. I did try putting the substitution in anyway, but took it out again as it seems to give away too much. $\endgroup$ – Glen_b -Reinstate Monica Apr 10 '16 at 13:49
  • $\begingroup$ @JohnK on a different sort of question, I definitely agree that it would be more like a comment, and I also agree that as it stands it's less than ideal as an answer. My question in the previous comment was not rhetorical -- I would like to give an answer that better meets the two conflicting goals here and would appreciate any thoughts you have on that. $\endgroup$ – Glen_b -Reinstate Monica Apr 11 '16 at 1:37
  • $\begingroup$ Glen, I understand the conflicting ideas here. On the one hand it is a self study question so you don't want to give it away. On the other, answers need to be more than two-liners. In my opinion, a good answer here would offer the substitution, make a comment about how general this result is-it's for all continuous distributions- and ask the OP to fill in the details. $\endgroup$ – JohnK Apr 11 '16 at 8:29
  • $\begingroup$ @JohnK It's a reasonable position on things; I have done as you suggest. $\endgroup$ – Glen_b -Reinstate Monica Apr 11 '16 at 10:12
2
$\begingroup$

Integration by parts: $$\int_{-\infty}^\infty F(x)f(x) \text{dx} =F^2(x)|^{\infty}_{-\infty}-\int_{-\infty}^\infty f(x)F(x) \text{dx}$$

Adding $\int_{-\infty}^\infty F(x)f(x) \text{dx}$ to both sides of the equation: $$2\int_{-\infty}^\infty F(x)f(x) \text{dx} =F^2(x)|^{\infty}_{-\infty}=1-0$$ Dividing by 2 $$\int_{-\infty}^\infty F(x)f(x) \text{dx}=\frac{1}{2}$$

$\endgroup$
  • 2
    $\begingroup$ Please see the help as it relates to homework-style questions and the guidelines on answering self-study questions, in particular the first two points. I think this is a good, useful answer, but it looks like the OP is doing study-work and as such we probably should lean more toward hints and guidance than explicit complete answers, especially when the OP has not shown an attempt. $\endgroup$ – Glen_b -Reinstate Monica Apr 10 '16 at 13:46

Not the answer you're looking for? Browse other questions tagged or ask your own question.