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Following this answer, it is claimed that we can solve the problem in the following way:

$$\displaystyle \int_{c}^{\infty} (x-c) dF(x) = \lim_{y \rightarrow \infty} (y-c) F(y) - \displaystyle \int_{c}^{\infty} F(x) dx.$$

where $F$ is the cumulative distribution function of a given random variable.

Does this limit exist though? In my understanding, since distribution function saturates at 1, the first limit should converge to infinity? What am I missing?

Did
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runr
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  • @CalvinKhor how is it so, if both $y$ and $P(y)$ are nondecreasing? I will try to wrap my head around it – runr Dec 18 '18 at 14:13
  • Oh I may have misunderstood, is P the cumulative distribution function? – Calvin Khor Dec 18 '18 at 14:16
  • @CalvinKhor it should be, yes. I agree, the notation is bad for a CDF, copied it from the linked answer. Will edit to avoid confusion, thanks! – runr Dec 18 '18 at 14:18
  • If F is indeed a cumulative distribution function then the second integral is similarly infinite and should cancel the divergence of thr first term ie as in the answer below – Calvin Khor Dec 18 '18 at 14:21
  • A wrong argument in an answer with 7 upvotes? Just an ordinary day on mse... :-) – Did Dec 18 '18 at 15:03
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    The correct argument is $$ \int_c^\infty (x-c) dF(x) = \lim_{y \to\infty} -(y-c) (1-F(y)) +\int_c^\infty (1-F(x)) dx=\int_c^\infty (1-F(x)) dx$$ – Did Dec 18 '18 at 15:06
  • @Did good point. One upvote is mine, since the first part before the edit seems OK (didn't check, but arrived to the answer), though may have to reconsider..:) As for your second comment -- how did you arrive to $(1-F(y))$ part? My quick guess would be $\int (x-c)dF(x) = - \int(x-c)d(1-F(x))$? If so, this seems as a useful trick! Either way, Thanks! – runr Dec 18 '18 at 16:46
  • @Nutle Yes, this is simply $dF=-d(1-F)$ and the notion that $1-F$ is better than $F$ here since with limit zero at +oo. – Did Dec 18 '18 at 16:50
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    @Did finally, sorry if this is a dumb question: $(y-c) \rightarrow \infty$ while $(1-F(y)) \rightarrow 0$. Unless $1-F(y)$ decays exponentially (or with $>1$ power?), how can we be sure that it reaches $0$ faster than $\infty$? Or is$\infty \cdot 0 = 0$? Though on the other hand, I can't imagine such $F$ to contradict that result. – runr Dec 18 '18 at 20:41
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    The question is not dumb in the least. It happens that, by Lebesgue dominated convergence theorem, $(y-c)(1-F(y))\to0$ as soon as $X$ is integrable (actually, $X^+$ integrable). – Did Dec 18 '18 at 21:19

1 Answers1

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Shouldn't the formula be,

$\displaystyle \int_c^\infty dP(x) = \lim_{y\to\infty} \left\{ (y-c) P(y) - \int_c^y P(x) dx \right\} $

Then for large $ x $, $ P(x) $ approaches 1, and you end up with both terms approaching $\infty$.

Not very helpful in finding the answer but perhaps explains the confusion.

WA Don
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  • Thanks. You're right, missed that the second term would approach $-\infty$ too. Still, it indeed does not seem helpful. Unless trying simplifying the formula you used for fixed $y$, and the limit might look nicer if some terms could get cancelled out? – runr Dec 18 '18 at 13:00