# What is the probability of acceptance for this algorithm?

What is the distribution of $$Y$$ from using the Rejection Sampling algorithm?

• Repeat Sample $$X$$ with distribution function $$F_X = (1-(1+x^\alpha)^{-1})1_{x\ge 0}(x)$$
• Until $$X>x_0$$, where $$x_0$$ is a positive constant.
• Set $$Y=X$$

The solution begins as follows, however i do not understand this part: "The probability of accepting a value is 1 if $$X>x_0$$, so the ratio of $$f_Y$$ to $$f_X$$ is proportional to $$1_{x_0\le x < \infty}$$. Therefore

$$f_{accept}(x)=\frac{f(x;\alpha)1_{x_0 \le x < \infty}(u)}{\int_{x_0}^\infty f(u;\alpha)du}$$

So here are my questions:

1. If $$P(Accept|X>x_0)=1$$, then why is the ratio of the target density to the proposal density is proportional to $$1_{x_0 \le x < \infty}$$? Is this just a property of the algorithm?
2. How do you go from the statement above to the equation of the acceptance density?

Consider an iid sequence $$X_1,X_2,\ldots$$ from $$f(\cdot;\alpha)$$. Then $$Y=X_N$$ when $$N=\min\{n; X_i>x_0\}$$ Hence \begin{align}\mathbb P(Yx_0,X_{i-1} which is indeed the truncated version of the distribution $$f(\cdot;\alpha)$$.
If one was to rephrase the algorithm as an accept-reject algorithm, one would take $$f_Y(y)/f_X(y)\propto\mathbb I_{y>x_0}\le 1$$ Hence following an accept-reject approach by (i) generating $$U\sim\mathcal U(0,1)$$ and (ii) accepting $$X\sim f_X(x)$$ as $$Y=X$$ only when$$U\le \mathbb I_{X>x_0}$$means accepting $$X$$ when $$X>x_0$$ and rejecting $$X$$ otherwise. The uniform generation is superfluous in this particular case since the decision does not depend on $$U$$. If there is no typo in the equation $$f_{accept}(x)=\frac{f(x;\alpha)\mathbb I_{x_0 \le x < \infty}(u)}{\int_{x_0}^\infty f(u;\alpha)\,\text du}$$ produced in the question, the indicator $$\mathbb I_{x_0 \le x < \infty}(u)$$ is constant in $$u$$, the constant being either zero or one. (Note that the use of $$u$$ as the dummy variable for the integration is particularly confusing!)