# Probability that one beta random variable is greater than other beta random variables

Suppose we have n independent beta random variables $p_1 \sim beta(\alpha_1, \beta_1), p_2 \sim beta(\alpha_2, \beta_2)...$, the parameters are all known. I want to calculate the probability that $p_i$ is greater than all the others. \begin{align} {\rm Pr}(p_i \geq max(p_1, p_2,...,p_n)) &= \int_0^1\int_0^{p_i}...\int_0^{p_i}f_{p_1}f_{p_2}...f_{p_n}dp_1dp_2...dp_ndp_i\\ &=\int_0^1I_{p_i}(\alpha_1, \beta_1)I_{p_i}(\alpha_2, \beta_2)...I_{p_i}(\alpha_n, \beta_n)f_{p_i}dp_i \end{align} This is what I got so far. Is there any closed-form solution for this question? If not, is there an efficient way to solve the integral? It can be assumed that the $\alpha, \beta$ parameters are very large if necessary, but it would be great if it can be solved without this assumption. Thank you.

If the integral can not be efficiently solved, can there be any approximation?

note: parameters $\alpha, \beta$ are different for different random variables.

• If the $\alpha_i$'s and $\beta_i$'s were the same for all $p_i$'s , then one could find the joint distribution of $(p_i,\max(p_1,\cdots,p_n))$ to calculate the probability. Not sure if that would have made things any easier though. – StubbornAtom Jun 11 '18 at 19:19
• If $\alpha_i, \beta_i$ are very large, $p_i$ is highly concentrated near its mean $\alpha_i/(\alpha_i+\beta_i)$. In that case, if there is one $p_i$ whose mean is greater than the others, the probability that it is the greatest is close to $1$. – Robert Israel Jun 11 '18 at 19:42

Let $M=max(p_1,\dots p_{i-1},p_{i+1},\dots,p_n)$ $$\begin{split} P(p_i\geq y, y\geq M) &= P(M\leq y)-P(p_i\leq y, y\geq M)\\ & = P(M\leq y)-I_i(\alpha_i,\beta_i)(y)P(M\leq y) \end{split}$$
$$$$\begin{split}P(M\leq y) & = P(p_1 \leq y, \dots , p_n \leq y)\\ & =\prod_{k=1, k\neq i}^{n}P(p_k \leq y)\\& = \prod_{k=1,k\neq i}^{n}I_i(\alpha_k,\beta_k)(y) \end{split}$$$$ This happens $\forall y$. So you would need to integrate this over $(0,1)$. I do not think you can do better than this.
In the simple case where the $\alpha_i$'s and $\beta_i$'s are the same across $p_1, \ldots, p_n$, each $p_i$ has equal probability of being the maximum. Thus, $P(p_i \textrm{ is the max}) = \frac{1}{n}$, $\forall i$. Otherwise, I don't think you can solve the probability explicitly.
• If the $\alpha_i$ and $\beta_i$ are integers, the PDF's and CDF's are polynomials, so you can certainly do the integration explicitly. – Robert Israel Jun 11 '18 at 19:45