I have been working with beta-bernoulli posteriors recently. Is it true that if $X,Y$ are independent rvs with $X \sim Beta(a_1+1,b_1+1)$ and $Y \sim Beta(a_2+1,b_2+1)$ then $\mathbb{P}(X>Y)>0.5$ iff $\frac{a_1}{a_1+b_1}>\frac{a_2}{a_2+b_2}$? $a_1,a_2,b_1,b_2$ are assumed to be whole numbers.

This makes intuitive sense as the estimated bernoulli parameter is greater in $X$ than in $Y$. I have observed this for all parameters within the range $0 \leq a_1,b_1,a_2,b_2\leq 6$. However, it does not seem to immediately follow from the definitions.

I have tried writing $\mathbb{P}(X>Y)=\frac{\int_{0}^{1}\int_{0}^{x}y^{a_2}(1-y)^{b_2}x^{a_1}(1-x)^{b_1}dy dx}{B(a_1+1,b_1+1)B(a_2+1,b_2+1)}$, but the numerator doesn't seem that easy to work with. I don't see a similar question on stackexchange. Can this be proved or can a counter example be produced for this?

Update1: It seems the result is false if $a_1,b_1,a_2,b_2$ are just positive real numbers as initially posed. See Henrys answer below. Can a counterexample be produced for whole numbers?

Update2: Henry has produced a whole number counterexample. This answers the question in the negative conclusively.


1 Answer 1


It does not seem to be a correct conjecture.

It seems your condition is that the mode for $X$ is greater than the mode for $Y$. Since in non-symmetric Beta distributions, the mode is not equal to the mean, it should be possible to find a counterexample.

One way is to take a non-symmetric case where $\frac{a_1}{a_1+b_1}=\frac{a_2}{a_2+b_2}$, find which tends to be greater more often, and then adjust the parameters slightly to get a counter-example which works.

So, using R, we might look at

XgreaterY <- function(a1,b1,a2,b2,cases){ 
   mean(rbeta(cases, a1+1, b1+1) > rbeta(cases, a2+1, b2+1))
XgreaterY(3, 1, 30, 10, 10^7)
# 0.390946  
XgreaterY(3, 1, 29, 11, 10^7) 
# 0.4385228

suggesting that $a_1=3, b_1=1, a_2=29, b_2=11$ provides a counterexample, since $\frac{a_1}{a_1+b_1} = 0.75 > 0.725 = \frac{a_2}{a_2+b_2}$ but it seems $\mathbb{P}(X>Y) < 0.44 < 0.5$

  • $\begingroup$ Could this hold if $a_1,b_1,a_2,b_2$ are all whole numbers? I had tried it only for whole numbers. $\endgroup$ Jan 12, 2023 at 14:57
  • $\begingroup$ @SushantVijayan No: try $a_1=3,b_1=1,a_2=29,b_2=11$ or XgreaterY(3, 1, 29, 11, 10^7) to get a probability less than $0.44$ $\endgroup$
    – Henry
    Jan 12, 2023 at 15:02
  • $\begingroup$ Thanks. can you update your answer with this example. I will accept your answer. $\endgroup$ Jan 12, 2023 at 15:06
  • $\begingroup$ would this be true if say I had used the means instead of the mode? $\endgroup$ Jan 12, 2023 at 15:13
  • 1
    $\begingroup$ @SushantVijayan I doubt it, and a similar approach should work, perhaps reversing the adjustment. The (mode) example in the answer compares $\text{Beta}(4,2)$ and $\text{Beta}(30,12)$. A (mean) example might compare $\text{Beta}(301,149)$ and $\text{Beta}(4,2)$ $\endgroup$
    – Henry
    Jan 12, 2023 at 15:22

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