Inverse transform sampling

If $U$ has a uniform distribution on $(0,1)$ and $Q$ is the quantile function of a cumulative distribution function $F$, then $Q(U)$ has the desired probability distribution. Indeed, right continuity givesso $\mathbb{P}(Q(U)\le x)=\mathbb{P}(U\le F(x))=F(x)$. This short argument works for both jumps and smooth cumulative distribution functions. MIT's continuous-probability notes introduce the inverse-transformation construction. ^[ref].

For the three-value example, send $U\le 0.2$ to $0$, $0.2<U\le 0.7$ to $1$, and $U>0.7$ to $3$. These intervals have exactly the desired widths $0.2$, $0.5$, $0.3$. For the density example, output $\sqrt{U}$: for $0\le y\le 1$, $\mathbb{P}(\sqrt{U}\le y)=\mathbb{P}(U\le y^2)=y^2$. Independent uniform draws give independent random variables after this deterministic transformation. This direction of construction works even when $F$ has jumps; applying the probability integral transform in the other direction requires its own continuity hypotheses.

For positive mixture weights $w_j$ summing to one, set $C_j={\sum }_{i\le j}{w}_i$ and choose $J=j$ when $C_{j-1}\le U<C_j$, where $U$ has a uniform distribution. Then $V=(U-C_{J-1})/w_J$ has a uniform distribution independently of $J$: for $0\le v\le 1$, $\mathrm{Pr}(J=j,V\le v)=w_{j}v$. This permits exact finite mixture model sampling with a recycled continuous random variable.