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Let $Y = \max_i X_i$ where $X_i \sim \text{Beta}(\alpha_i, \beta_i)$ independently. What is $\operatorname*{E} Y$?

Let $[\cdot]$ denote the CDF of a random variable at $x$. Since $Y \geq 0$, \begin{align} \operatorname*{E} Y &= \int_0^\infty (1 - [Y]) \,\mathrm{d}x \\ &= \int_0^1 (1 - [Y]) \,\mathrm{d}x \\ &= 1 - \int_0^1 [Y] \,\mathrm{d}x \\ &= 1 - \int_0^1 \left[\max_i X_i\right] \,\mathrm{d}x \\ &= 1 - \int_0^1 \prod_i [X_i] \,\mathrm{d}x \\ &= 1 - \int_0^1 \prod_i I_x(\alpha_i, \beta_i) \,\mathrm{d}x \\ \end{align}

where $I_x$ is the regularized incomplete beta function. Is there a closed form for this integral, perhaps in terms of transcendental functions?

user76284
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  • If we call the cdf of your beta distribution $F$, then you get that $\mathbb{E}[Y]=\int_0^1 (1-F(x)^n) dx$ where $n$ is the number of $X_i$. Maybe you can try something from there using a binomial theorem. – Jfischer May 17 '20 at 20:22
  • For a positive RV $Y$ with CDF $G$ we have the identity $\mathbb{E}[Y]=\int_0^1 1-G(x) dx$ (https://en.wikipedia.org/wiki/Expected_value#Alternative_formula_for_expected_value). And $G(x) = \mathbb{P}[Y\leq x] = \mathbb{P}[X_1\leq x,...,X_n\leq x] = \mathbb{P}[X_1\leq x]^n = F(x)^n$. – Jfischer May 17 '20 at 20:27
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    For whatever it's worth, I don't think there is a general symbolic solution. However, while numerical integration is not as desirable, it does seem to work well and quickly. In addition, one can obtain exact results when the beta parameters are integers and also for many sets of rational numbers. – JimB May 17 '20 at 23:28
  • @JimB What does it simplify to when the beta parameters are integers? And perhaps the integral could still be simplified somewhat? – user76284 Mar 19 '21 at 20:14
  • Using Mathematica one gets rational numbers for the mean for any set of positive integer parameters. For example: a = {1, 3, 5, 6}; b = {1, 3, 2, 9}; 1 - Integrate[Product[BetaRegularized[y, a[[i]], b[[i]]], {i, Length[a]}], {y, 0, 1}] // FunctionExpand // Expand resulting in an answer of 83566449/106234700. – JimB Mar 19 '21 at 21:11

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