Let $X_1, X_2$ be i.i.d. $\mathcal U(0, 1)$ (continuous) r.v.'s, and let $0 \le R \le 1$ be some number. What is $\mathbb E[\min(X_1, X_2) \mid \min(X_1, X_2) \ge R]$?
My attempt: Let $Y = \min(X_1, X_2)$. Let $F$ be the CDF, and $f$ the pmf of $Y$. We can see that $$ \begin{align*} F(y) &= 1 - (1 - y)^2 \\ f(y) &= \frac{\operatorname d F}{\operatorname d y} = 2(1 - y)\text{.} \end{align*} $$ Now I need to get $f(y \mid Y \ge R)$, but I'm a bit lost. How should I do that?
Bayes' Rule is applicable (or just the definition of conditional probability):
$$f_{Y}(y\mid Y\geqslant R) ~=~ \dfrac{f_Y(y)~\mathbf 1_{y\geqslant R}}{1-F_Y(R)}$$