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I am trying to prove the following:

Given that $\forall \alpha\in [0,1]$:

$$\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,dx = \int_{F_0^{-1}(\alpha)}^{\infty}yf_0(y)\,dy$$

where $F_S^{-1}(\alpha)$ and $F_0^{-1}(\alpha)$ are the $\alpha$th quantiles of the distribution. I wand to show that $f_S(x)$ is identically distributed to $f_0(x)$.

It seems like it should be true, but I haven't been able to prove it definitively. Any advice or directions to explore would be appreciated.

What I've tried so far:

integration by parts led me to the following:

$yF_{S}(y)\big|_{F_{S}^{-1}(\alpha)}^{\infty}-\int_{F_{S}^{-1}(\alpha)}^{\infty}F_{S}(y)dy=yF_{0}(y)\big|_{F_{0}^{-1}(\alpha)}^{\infty}-\int_{F_{0}^{-1}(\alpha)}^{\infty}F_{0}(y)dy$

But I can't substitute infinity for y in the first term without both sides going to infinity. To get around this I considered the fact that for some $\epsilon>0$:

$$\int_{F_S^{-1}(\alpha)}^{F_S^{-1}(\alpha+\epsilon)}xf_S(x)\,dx = \int_{F_0^{-1}(\alpha)}^{F_0^{-1}(\alpha+\epsilon)}yf_0(y)\,dy$$

which then applying integration by parts leads to:

$(\alpha+\epsilon)F_{S}^{-1}(\alpha+\epsilon)-\alpha F_{S}^{-1}(\alpha)-\int_{F_{S}^{-1}(\alpha)}^{F_{S}^{-1}(\alpha+\epsilon)}F_{S}(y)dy = (\alpha+\epsilon)F_{0}^{-1}(\alpha+\epsilon)-\alpha F_{0}^{-1}(\alpha)-\int_{F_{0}^{-1}(\alpha)}^{F_{0}^{-1}(\alpha+\epsilon)}F_{0}(y)dy$

and I feel if I could show that the integral components of this were equal to 0 then I could solve it.

Alex
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    Think about the definition of $F_S^{-1}.$ – BruceET Jan 22 '19 at 20:22
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    This amounts to solving a homework problem: you should thus provide more details on what you attempted so far and add the self-study tag to the question. – Xi'an Jan 22 '19 at 20:27
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    Even simpler: integrating by parts and then choosing an appropriate substitution reduces this to an almost trivial question. (That's a time-honored mathematical technique: reduce a theorem to a matter of definition by means of a succession of simplifications.) – whuber Jan 22 '19 at 21:18
  • @whuber I tried integration by parts, but get:

    $$yF_{S}(y)\big|{F{S}^{-1}(\alpha)}^{\infty}-\int_{F_{S}^{-1}(\alpha)}^{\infty}F_{S}(y)dy=yF_{0}(y)\big|{F{0}^{-1}(\alpha)}^{\infty}-\int_{F_{0}^{-1}(\alpha)}^{\infty}F_{0}(y)dy$$ which takes both sides of the equation to infinity because of the first

     – Alex Jan 23 '19 at 04:41
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    Good try. You can fix the problem by using $-(1-F_S(y))$ as the antiderivative of $f_S(y).$ In doing so I believe you'll discover that for this result to be true, an additional assumption about the distribution is needed: namely, that it has an expectation and the expectation is finite. – whuber Jan 23 '19 at 13:39
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    Just to be clear: consider the power-law distributions given by $$F_\lambda(y)=1-y^{-\lambda},, y \ge 1$$ (and equal to $0$ otherwise) for $0 \lt \lambda \le 1.$ All the integrals involved in this question are infinite (and therefore equal), but different values of $\lambda$ give different distributions, thereby yielding counterexamples. – whuber Jan 23 '19 at 14:47
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    using your suggestion $-(1-F_S(y)$ as the antiderivative of $f_S(y)$ I am still left with an integral that I cannot make sense of i.e.

    $\int_{F_S(y)}^{\infty} xf_S(x) dx = F_S^{-1}(\alpha)(1-\alpha) - \int_{F_S(y)}^{\infty} xf_S(x) dx$

    which you were right gets rid of the infinity issue but the integral is still troubling me.

    Also thank you very much for including the assumption that the distribution has an expectation and the expectation is finite.

     – Alex Jan 23 '19 at 17:48
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    Sorry the integral above after integration by parts should be:

    $\int_{F_S(x)}^\infty -(1 - F_S(x))dx$

     – Alex Jan 24 '19 at 01:23
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    The plots at https://stats.stackexchange.com/a/18449/919 help us see that each integral, as a function of $\alpha,$ sweeps out the area within a region lying beneath a moving horizontal line (located at $\alpha$). The equality of the integrals expresses the equality of these area functions, which implies the congruence of the corresponding regions, which means nothing other than that the two distribution functions are the same. – whuber Jan 24 '19 at 15:59

1 Answers1

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Recall that $$\frac{\text{d}}{\text{d}\theta} \int_\theta^\infty f(x)\,\text{d}x=-f(\theta)$$ and hence that $$\frac{\text{d}}{\text{d}\theta} \int_{g(\theta)}^\infty f(x)\,\text{d}x=-(f\circ g)(\theta)\times g'(\theta)$$ Hence, assuming the expectations of $F_S$ and $F_0$ are well-defined and finite, $$\frac{\text{d}}{\text{d}\alpha}\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,\text{d}x = - F_S^{-1}(\alpha)\times(f_S\circ F_S^{-1})(\alpha)\times\frac{\text{d}}{\text{d}\alpha}F_S^{-1}(\alpha)$$ Recall further that $$\frac{\text{d}}{\text{d}\alpha}F_S^{-1}(\alpha)=\frac{1}{[\frac{\text{d}}{\text{d}x}F_S](F_S^{-1}(\alpha))}=\frac{1}{f_S(F_S^{-1}(\alpha))}=\frac{1}{(f_S\circ F_S^{-1})(\alpha)}$$ and you should directly deduce the result.

This resolution is in fact equivalent to a change of variable in the integral$$\int_{F_S^{-1}(\alpha)}^{\infty}xf_S(x)\,\text{d}x$$ since expressing $x$ as $x=F_S^{-1}(\beta)$ in this integral leads to $$\int_{\alpha}^{1}F_S^{-1}(\beta)\,\text{d}\beta$$thanks to the same cancellation of $(f_S\circ F_S^{-1})(\alpha)$ as above. Hence $$\int_{\alpha}^{1}F_S^{-1}(\beta)\,\text{d}\beta=\int_{\alpha}^{1}F_0^{-1}(\beta)\,\text{d}\beta$$ for all $0<\alpha<1$ and taking the derivative in $\alpha$: $$-F_S^{-1}(\alpha)=F_S^{-1}(\alpha)$$ for all $0<\alpha<1$, as expected.

Xi'an
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    One basic check is needed at the outset: does the integral of $x f_S(x)\mathrm{d}x$ even exist? :-) – whuber Jan 23 '19 at 13:40
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    So starting at the beginning:

    $\int_{F_{S}^{-1}(\alpha)}^{\infty}xf_{S}(x)dx=\int_{F_{0}^{-1}(\alpha)}^{\infty}yf_{0}(y)dy$

    Then applying your suggested U-substitution:

    $int_{\alpha}^{1}F_{S}^{-1}(\beta)d\beta=\int_{\alpha}^{1}F_{0}^{-1}(\beta)d\beta$

    Then we can combine the integrals:

    $\int_{\alpha}^{1}F_{S}^{-1}(\beta)-F_{0}^{-1}(\beta)d\beta=0$

    Does this then imply that $F_{S}^{-1}(\alpha)=F_{0}^{-1}(\alpha)\forall;\alpha\in[0,1]$? I believe it does, which would complete the proof. Very elegant solution.

     – Alex Jan 23 '19 at 17:53