Timeline for Does a univariate random variable's mean always equal the integral of its quantile function?

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Jun 19, 2014 at 19:05	comment	added	Avraham		If you like diagrams, you may be interested in this 1988 paper by Lee: The Mathematics of Excess of Loss Coverages and Retrospective Rating-A Graphical Approach.
Nov 16, 2011 at 8:01	comment	added	Henry		@whuber: If you want to extend to negative $t$, you get $\int_{t=0}^\infty (1-F(t)) \; dt - \int_{t=-\infty}^0 F(t) \; dt$. Note that if this converges for a distribution symmetric about $0$, i.e. $F(t)=1-F(-t)$ then it is easy to see that the expectation is zero. Taking a sum rather than a difference $\int_{t=0}^\infty (1-F(t)) \; dt + \int_{t=-\infty}^0 F(t) \; dt$ gives the average absolute deviation about $0$.
Nov 16, 2011 at 4:14	comment	added	whuber♦		I like pictures, and instinctively feel there's a great idea lurking here--I love the idea--, but I don't understand these particular ones. Explanations would be helpful. One thing that stops me in my tracks is the thought of trying to extend the integral of $(1-F(t))dt$ to $-\infty$: it has to diverge.
Nov 15, 2011 at 23:45	history	answered	Henry	CC BY-SA 3.0