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I have problems analyzing this: Let $F$ and $G$ be distribution functions, both on $\mathbb{R}$. Now define, for $t\in[0,1]$ $$H_t(x,y) = \begin{cases} F(x)^{1-t}G(y) & F(x)\geq G(y) \\ F(x)G(y)^{1-t} & F(x)\leq G(y) \end{cases} $$

Is $H_{t}(x,y)$ a distibution function?

I think that the limit (When both tends to infinite is 1 and when just one tends to $-\infty$ is 0) property is true, but... The 2-increasing property and the continuity property I just don't find a counterexample or don't know how to prove that.

BlueRedem1
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    $H_t(x,y)$ looks to me to be a weakly increasing function of $x$ and $y$ because $F(x)$ and $G(y)$ are, and as left-continuous as $F(x)$ and $G(y)$ – Henry Mar 09 '20 at 01:57
  • @Henry So you think that the 2-increasing property fails? I was trying with that but since we don't know how the functions are, I find kind of difficult to prove that – BlueRedem1 Mar 09 '20 at 02:01
  • @EduardoCuéllar - no, I think the increasing property is present – Henry Mar 09 '20 at 02:03
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    @pegasus - is there any reason to believe the derivative has to exist? – Henry Mar 09 '20 at 02:04
  • I don't know if the 1-t affects continuity or 2-increasing property, so that's kind of difficult for me to give a proof. Just checked the limits property but I'm stuck now – BlueRedem1 Mar 09 '20 at 02:09
  • @MichaelHardy sorry, you are right. That was a typo – BlueRedem1 Mar 09 '20 at 02:36
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    @Henry It is obvious that $F(x,y)$ is increasing and right continuous, but the hard part is showing that it is 2-increasing. – Mike Earnest Mar 11 '20 at 23:52
  • @MikeEarnest Right. That's why I posted this because I really don't know how to even start – BlueRedem1 Mar 12 '20 at 04:44

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