Inverse transform sampling for truncated distributions

Assume that you have a random number generator which gives you $U_1, U_2, \dots \stackrel{iid}{\sim} U[0,1]$ (i.e. uniform distribution on the interval $[0,1]$ ) and a cumulative distribution function $F$ . Assume for simplicity that $F$ is strictly increasing, so that $F^{-1}$ is well-defined as a functional inverse. Inverse transform sampling allows us to use this set-up to draw samples from $F$ : If $U \sim U[0,1]$ then $F^{-1}(U) \sim F$ . Thus, we can take $F^{-1}(U_1),F^{-1}(U_2), \dots$ to be our sample.

(If $F$ is not strictly increasing, we can take $F^{-1}(u) = \inf \{ x | F(x) \geq u\}$ .)

Did you know that we can modify this slightly to draw samples $X \sim F$ , conditional on $X \geq t$ , where $t \in \mathbb{R}$ ? Instead of taking $F^{-1}(U)$ as the sample, take $F^{-1}[F(t) + U (1 - F(t))]$ instead.

Here is the proof: Let $G$ be the CDF of $X \sim F$ given $X \geq t$ . Then $G(x) = 0$ if $x < t$ , and for $x \geq t$ , $G(x) = \dfrac{F(x) - F(t)}{1 - F(t)}$ .

Now, for any $x$ ,

$\begin{aligned} \mathbb{P}(F^{-1}[F(t) + U (1 - F(t))] \leq x) &= \mathbb{P}(F(t) + U (1 - F(t)) \leq F(x)) \\ &= \mathbb{P}\left( U \leq \frac{F(x) - F(t)}{1 - F(t)} \right) \\ &= \mathbb{P}\left( U \leq G(x) \right) \\ &= \mathbb{P}(G^{-1}(U) \leq x), \end{aligned}$

which means that $(F^{-1}[F(t) + U (1 - F(t))]$ has the same distribution as $G^{-1}(U)$ , i.e. $G$ (by an application of inverse transform sampling).

Statistical Odds & Ends

Inverse transform sampling for truncated distributions

4 thoughts on “Inverse transform sampling for truncated distributions”

Leave a Reply Cancel reply

Share this:

Like this:

Related

4 thoughts on “Inverse transform sampling for truncated distributions”

Leave a Reply Cancel reply