This is a follow up to this question. Consider again two independent rvs $X\sim Beta(a_1+1,b_1+1)$ and $Y \sim Beta(a_2+1,b_2+1)$. Here $a_1,b_1,a_2,b_2$ are in $\mathbb{N}$.

Is it true that $\mathbb{P}(X>Y)>0.5 \implies \frac{a_1}{(a_1+b_1)}>\frac{a_2}{(a_2+b_2)}$ ? Note that the counterexample presented in here rules out the opposite direction.

So far I have been able to show the following:

  1. $\mathbb{P}(X>Y)>0.5 \implies \frac{a_1+1}{(a_1+b_1+2)}>\frac{1}{2}\frac{a_2}{(a_2+b_2+1)}$.
  2. $a_1 \geq a_2$ $\&$ $b_1 \leq b_2$ or $a_1\geq b_1$ $\&$ $a_2 \leq b_2$ $\implies \mathbb{P}(X>Y)>0.5$.
  • 1
    $\begingroup$ Presumably you are supposing the $a_i$ and $b_i$ are non-negative, right? It's unclear what you mean by "similar structure:" that sounds like you are looking for us to create a question for you. What are you really trying to ask? $\endgroup$
    – whuber
    Commented Jan 16, 2023 at 17:38
  • $\begingroup$ Do some exploration to refine your question. Consider $(a_1,b_1)=(3,14)$ and $(a_2,b_2)=(2,9),$ for instance. $\endgroup$
    – whuber
    Commented Jan 16, 2023 at 20:12
  • $\begingroup$ @whuber In your example I get $\mathbb{P}(X>Y) \approx 0.4554503$. While $\frac{a_1}{(a_1+b_1)}=0.1764706<0.1818182=\frac{a_2}{(a_2+b_2)}$. Isn't this in line with the conjecture? I have clarified that $a_i,b_i \in \mathbb{N}$. I have also removed the "similar structure" part. $\endgroup$ Commented Jan 16, 2023 at 20:26
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    $\begingroup$ Sorry, I made an error shifting between your parameterization and the usual parameterization. Let's check the situation with $(19,3)$ and $(8,1).$ The modes are $19/22\approx 0.864\lt 0.889=8/9.$ But $\Pr(X\gt Y)=0.515\gt0.5.$ $\endgroup$
    – whuber
    Commented Jan 16, 2023 at 22:33
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    $\begingroup$ @whuber Thanks for the counterexample. Let me try and do some computations with the mean instead of the mode. If you'd like you can post this example as an answer and I can accept it. For the mean the example you have given doesn't work since $\frac{20}{24} = 0.8333333 >0.8181818=\frac{9}{11}$. $\endgroup$ Commented Jan 17, 2023 at 6:46


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