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Not a homework question, but a problem I need to solve in real life. Seems easy at first, but somehow I cannot prove tha:

A,B,C are independent random variables under each of the two probability measures $P_1$ and $P_2$, In addition, we know that $P_1(A>B) > P_2(A>B)$ and $P_1(A>C) > P_2(A>C)$. Prove that $P_1(A>B, A>C) > P_2(A>B, A>C)$.

Thanks!

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    What you mean by bonus? You are ordering others to solve or request? Because it doesn't seem to be request. – 2017-01-24
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    @JeanMarie, this will not work since A>B does not have to be independent of A>C. – 2017-01-24
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    @cruvadom You are right. I delete this silly explanation. – 2017-01-24
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    @MeesdeVries, I try to decompose it somehow using the independecy, but nothing leads to the desired outcome. – 2017-01-24
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    What do you call "real life", where one would need to solve this? – 2017-01-26

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The result doesn't hold. Under different probabilities, random variables $A,B$ and $C$ have different distributions. Here's a counterexample for the question.

Let $A,B$ and $C$ be independent random variables with distributions:

  • Under measure $P_1$: $$ A:\left(\genfrac{}{}{0pt}{}{0}{0.8}\,\genfrac{}{}{0pt}{}{2}{0.2}\right),\quad B,C : \left(\genfrac{}{}{0pt}{}{-1}{0.75}\,\genfrac{}{}{0pt}{}{1}{0.25}\right) $$
  • Under measure $P_2$: $$ A:\left(\genfrac{}{}{0pt}{}{0}{0.3}\,\genfrac{}{}{0pt}{}{2}{0.7}\right),\quad B,C : \left(\genfrac{}{}{0pt}{}{-1}{0.2}\,\genfrac{}{}{0pt}{}{1}{0.8}\right)$$

Then, easy calculation gives: $$P_1(A>B)=0.8>0.76=P_2(A>B)\ \\P_1(A>C)=0.8>0.76=P_2(A>C),$$ but, on the other hand: $$ P_1(A>B,A>C) = 0.65 < 0.712 = P_2(A>B,A>C). $$