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:

*

*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?

*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(Y<y)&=\mathbb P(X_N<y)\\&=\sum_{i=1}^\infty\mathbb P(N=i,X_i<y)\\&=\sum_{i=1}^\infty\mathbb P(X_i<y,X_i>x_0,X_{i-1}<x_0,\ldots,X_1<x_0)\\&=\sum_{i=1}^\infty\mathbb P(x_0<X_i<y)\mathbb P(X_1<x_0)^{i-1}\\&=\mathbb P(x_0<X_1<y)\sum_{i=1}^\infty\mathbb P(X_1<x_0)^{i-1}\\&=\mathbb P(x_0<X_1<y)\big/\mathbb P(x_0<X_1)
\end{align}
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!)
