Rejection Method

Study guide: Exam MAS-I

The Rejection Method (acceptance–rejection method) simulates a draw from a density $f$ that is hard to invert by drawing candidates $Y$ from an easier density $g$ and accepting each candidate with probability $f(Y)/\bigl(c\,g(Y)\bigr)$ , where the constant $c$ bounds the ratio $f/g$ .

$c \geq \max_x \frac{f(x)}{g(x)}$

$\text{accept } X = Y \text{ if } U \leq \frac{f(Y)}{c\,g(Y)}$

The algorithm: (1) generate $Y$ from $g$ ; (2) generate $U$ from a Uniform $(0,1)$ distribution, independent of $Y$ ; (3) if $U \leq f(Y)/\bigl(c\,g(Y)\bigr)$ , accept and set $X = Y$ , otherwise return to step 1. An accepted value has density exactly $f$ .
Efficiency: each pass accepts with probability $1/c$ , so the number of passes $N$ needed for one $X$ is geometric with success probability $1/c$ , and $E[N] = c$ . The smallest valid constant, $c = \max_x f(x)/g(x)$ , is therefore the best one; it is always at least 1, and the closer $g$ is to $f$ , the closer $c$ is to 1.
With $g$ uniform on $(0, 1)$ : $g(x) = 1$ , so $c$ is the maximum of the density, $c = \max_x f(x)$ , found by setting $f'(x) = 0$ . The acceptance test is $U \leq f(Y)/c$ .
When to use it: when $F^{-1}$ has no closed form, so the Inversion Method is unavailable for $f$ itself; the candidates $Y$ are usually generated from $g$ by inversion.
Discrete case: identical, with probability mass functions in place of densities: accept $Y$ if $U \leq p(Y)/\bigl(c\,q(Y)\bigr)$ , where $c \geq \max_k p(k)/q(k)$ .

Rejection Sampling a Beta Loss Ratio The proportion $X$ of a policy's premium paid out in losses has density $f(x) = 20x(1 - x)^3$ for $0 < x < 1$ . It is simulated by the rejection method with $g(x) = 1$ on $(0, 1)$ .

(a) Find the smallest valid $c$ , the probability that a pass is accepted, and the expected number of passes per simulated value.

(b) Two passes produce the pairs $(Y, U) = (0.30,\ 0.80)$ and $(0.70,\ 0.20)$ . Which candidates are accepted?
Answer (a) Maximise $f$ : $f'(x) = 20(1-x)^2(1 - 4x) = 0$ gives $x = \frac{1}{4}$ , so
 $\begin{align*} c &= f\!\left(\tfrac{1}{4}\right) \\ &= 20\left(\tfrac{1}{4}\right)\left(\tfrac{3}{4}\right)^3 \\ &= \tfrac{135}{64} \\ &\approx 2.109 \end{align*}$ 
Each pass is accepted with probability $1/c = 64/135 \approx 0.474$ , and on average $c \approx 2.11$ passes are needed per value.

(b) The acceptance test is $U \leq f(Y)/c = \frac{256}{27}\,Y(1 - Y)^3$ .

$Y = 0.30$ : $\frac{256}{27}(0.30)(0.70)^3 = 0.9756$ , and $0.80 \leq 0.9756$ , so it is accepted: $X = 0.30$ .

$Y = 0.70$ : $\frac{256}{27}(0.70)(0.30)^3 = 0.1792$ , and $0.20 > 0.1792$ , so it is rejected.

Simulating a Gamma Severity from an Exponential Claim sizes (in thousands) have the gamma density $f(x) = x e^{-x}$ , $x > 0$ . Candidates are drawn from the exponential density with mean 2, $g(x) = \frac{1}{2}e^{-x/2}$ , by inversion.

(a) Find the smallest valid $c$ and the expected number of passes per simulated claim.

(b) A pass draws $u_1 = 0.60$ for the candidate and $u_2 = 0.95$ for the test. What is the result?
Answer (a) The ratio is $f(x)/g(x) = 2x e^{-x/2}$ . Its derivative $(2 - x)e^{-x/2}$ is zero at $x = 2$ , so
 $\begin{align*} c &= 2(2)e^{-1} \\ &= 4/e \\ &\approx 1.4715 \end{align*}$ 
On average $1.47$ passes are needed per claim (acceptance probability $e/4 \approx 0.680$ ).

(b) By inversion, $Y = -2\ln(1 - 0.60) = 1.8326$ . The acceptance threshold is
 $\begin{align*} \frac{f(Y)}{c\,g(Y)} &= \frac{Y}{2}\,e^{1 - Y/2} \\ &= 0.9163\,e^{0.0837} \\ &= 0.9963 \end{align*}$ 
Since $0.95 \leq 0.9963$ , the candidate is accepted: the simulated claim is $1.8326$ , or about $1,833.