Acceptance Rejection Sampling

There are many distributions for which the inverse-transform method will not give us the required random variable (ie inversing the CDF of some distribution to get a r.v). Let’s call such a distribution $f$, which is the target density. Suppose there exists another distribution $g$ for which we know the ratio of $f/g$ is bounded by M for all values in the support of the two functions and the support of both distributions are compatible ($f(x)>0$ when $g(x)>0$).

Then X can be drawn as follows:

  1. Generate a $U\sim U(0,1)$ r.v.
  2. Generate a Y coming from the g.
  3. If \begin{equation} U \leq (1/M) (f(Y)/g(Y)) \end{equation}, we accept this value of Y as our value for X.

Exercise 2.5 The probability of accepting Y is 1/M.

\begin{equation} P(U \leq \frac{f(Y)}{Mg(Y)}) = \int_0^{\frac{f(Y)}{Mg(Y)}}\int_{-\infty}^{\infty} g(y) du dy \end{equation} \begin{equation}= \int_{-\infty}^{\infty} \int_0^{\frac{f(Y)}{Mg(Y)}} du g(y) dy = \frac{1}{M} \int_{-\infty}^{\infty} f(y) dy = \frac{1}{M} \end{equation}

The Y is a random variable here, hence the double integrals. U is a uniform r.v.


Introduction to Monte Carlo Methods with R

Faith Lee
Faith Lee
PhD student - Statistician

I am a PhD student/statistician based in Quebec City, QC.