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?
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 // Expandresulting in an answer of83566449/106234700. – JimB Mar 19 '21 at 21:11