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$G_{1}$ and $G_{2}$ are CDFs of some non-degenerate distributions specified on $\mathbb{R}$ and $\widetilde{G}_{1}$, $\widetilde{G}_{2}$ are corresponding generalized inverses ($\widetilde{f}(u)=\inf \lbrace t: f(t) \geq u \rbrace$). Prove the existance of $u_{1}<u_{2}$, such that: $$ \widetilde{G}_{1}(u_{1}) < \widetilde{G}_{1}(u_{2}) \\ \widetilde{G}_{2}(u_{1}) < \widetilde{G}_{2}(u_{2}) $$ and $ u_{1},u_{2} \in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2}) $, where $ \mathbb{T}(f) = \lbrace x \in \mathbb{R}$ such that $f$ is continuous in x and $f(x)$ is finite $\rbrace $

I know that: $\widetilde{G}_{1},\widetilde{G}_{2}$ are non-decreasing and set of discontinuity points of any monotonic function might be at least countable. Thus, it's easy to show that $ u_{1},u_{2} \in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2}) $, but I don't know how to prove the two remaining inequalities are satisfied. Thanks for any help.

user2280549
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  • The generalized inverses are actually continuous: where $G_1,G_2$ had jumps, the inverses have flat segments. And $G_1,G_2$ can only have jumps, no other types of discontinuities, because of the monotonicity. So you basically have two nondecreasing continuous functions and you want to show that their difference is somewhere decreasing. – Ian Feb 22 '16 at 21:01
  • Actually, they might not be continuous. The problem lies, where $G_{1}$, $G_{2}$ are constant – user2280549 Feb 22 '16 at 21:02
  • Oh, yes, you're right, the flat segments in $G_1,G_2$ create problems. My mistake. – Ian Feb 22 '16 at 21:03
  • Since $G_i$ are non-degenerate, there exist $u_{i,j}$, s.t. $0<u_{i,1}<u_{i,2}<1$ and $\widetilde{G}i(u{i,1})<\widetilde{G}i(u{i,2})$. Now choose $u_1\in(0,\min(u_{i,1}))$ and $u_2\in(\max(u_{i,2}),1)$ which you can do because the sets excluded by $u_{1},u_{2} \in \mathbb{T}(\widetilde{G}{1}) \cap \mathbb{T}(\widetilde{G}{2})$ of measure zero. – A.S. Mar 01 '16 at 07:48
  • what do you mean by saying $\min(u_{i,1})$ – user2280549 Mar 01 '16 at 11:23

1 Answers1

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$G$ is a monotonic increasing(non decreasing) function. So we know$$ a\le b \Leftrightarrow G(a) \le G(b)$$(Strict inequality if continuous inceasing over the period containing $a,b$).What if we put $a=\widetilde {G}(u_1)$ and $b$ similarly by choosing any $u_1\lt u_2$ in $\mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2})$ ? Consider the fact that if $u\in \mathbb{T}(\widetilde{G}_{1}) \cap \mathbb{T}(\widetilde{G}_{2})$ then at $u$ ,$G$ doesn't have any flat region (i.e. is one one) ,so we can write $u=G(\widetilde{G} (u))$

Qwerty
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  • Imho, it's not so easy. If $u$ is located on the interval, where $\widetilde{G}{1}$/$\widetilde{G}{2}$ is constant, your equation is invalid. To visualize it please see corresponding plot here: http://math.stackexchange.com/questions/1517134/convergence-of-generalized-inverses – user2280549 Mar 01 '16 at 11:12
  • You have to prove there exists at least 1 existence of $u_1 $ and $u_2$.So what if we chose our $u_1$ and $u_2$ appropriately? – Qwerty Mar 01 '16 at 13:19
  • What does it mean "appropriately"? Maybe I don't understand your hint. I understand that for each inverse CDF separately, I am able to find such pair of points, but what I don't know is how to proof that one particular pair is good for both of them. There might exist "many" pairs $(u_{1},u_{2})$, that are good e.g. for $ \wildtilde{G){1} $ but $ \wildtilde{G{2})(u){2} = \wildtilde{G){2}(u_{2}) $ – user2280549 Mar 01 '16 at 13:37