正方形与圆内的均匀采样

(a) 直接取点 $(U_1,U_2)$，它在单位正方形 $[0,1]^2$ 上均匀。

(b) 一种常用方法是极坐标变换：取

$$\Theta=2\pi U_1,\quad R=\sqrt{U_2},$$

则 $(R\cos\Theta, R\sin\Theta)$ 在单位圆盘内均匀。

（也可用拒绝采样：先生成 $(2U_1-1,2U_2-1)$ 在 $[-1,1]^2$ 上均匀，若落在圆外则重抽。）

(c) 可以。在理论上，一个 $U\sim U(0,1)$ 含有无限位随机性，可以把其二进制展开的奇偶位拆成两个独立的 $U(0,1)$（除去一组测度为 0 的二进制表示不唯一点），从而再用 (b) 的方法生成圆内均匀点。

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英文解析

(a) Directly sample a point $(U_1,U_2)$, which is uniformly distributed on the unit square $[0,1]^2$.

(b) A common method is a polar coordinate transformation: take
$$\Theta=2\pi U_1,\quad R=\sqrt{U_2},$$
Then $(R\cos\Theta, R\sin\Theta)$ is uniformly distributed within the unit disk.

(Alternatively, use rejection sampling: generate $(2U_1-1,2U_2-1)$ uniformly on $[-1,1]^2$, and resample if it falls outside the circle.)

(c) Yes. In theory, a $U\sim U(0,1)$ contains infinite bits of randomness; its binary expansion's even and odd bits can be split into two independent $U(0,1)$ (excluding a set of measure zero of non-unique binary representations), allowing the method in (b) to be used to generate a uniform point inside the circle.