Consider $X,Y,Z$ independently and identically distributed from uniform distribution $[0,1]$.
I am trying to solve for:
$E(X|X>Y)$.
$E(X|X>Y,X>Z)$.
I am just tyring to understand these simplace cases, so that I can extend to the case of finite $n$ IID uniform random variables, and when I take the limit.
By definition: $\mathsf E(X\mid X\in A) = \dfrac{\mathsf E(X\,\mathbf 1_{X\in A})}{\mathsf P(A)} = \dfrac{\displaystyle \int_A x~f_{\small X}(x)\,\mathrm d x}{\displaystyle \int_A f_{\small X}(x)\,\mathrm d x}$
So $\mathsf E(X\mid X>Y) =\dfrac{\mathsf E(X\,\mathbf 1_{X>Y})}{\mathsf P(X > Y)}= \dfrac{\displaystyle\int_0^1 x\,\mathsf P(Y<x)\,\mathrm d x}{\displaystyle\int_0^1 \mathsf P(Y<x)\,\mathrm d x}=\dfrac{\displaystyle\int_0^1\int_0^x x\,\mathrm d y\,\mathrm d x}{\displaystyle\int_0^1\int_0^x 1\,\mathrm d y\,\mathrm d x} = \dfrac{1/3}{1/2} =\dfrac 23\\\displaystyle\phantom{\mathsf P(X_1 {\,=\,}\max\{X_i\}_{i=2}^n) = \int_\Bbb R f_{\small X}(x) \prod_{i=2}^n \mathsf P(x\geqslant X_i)\,\mathrm d x = \int_0^1 x^{n-1}\,\mathrm d x \\\mathsf E(X_1\,\mathbf 1_{X_1=\min\{X_i\}_{i=2}^n}) = \int_\Bbb R x\,f_{\small X}(x)\prod_{i=2}^n\mathsf P(x\geqslant X_i)\,\mathrm d x = \int_0^1 x^n\,\mathrm d x } $