# If all trimmed means are equal does this imply equal distributions?

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.

• Think about the definition of $F_S^{-1}.$ – BruceET Jan 22 '19 at 20:22
• 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
• 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
• 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
• 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

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.
• 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
• 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