Let $X_{0}> 0$ a.s. and $\mathbb P( S > x \lvert \mathcal{F}_{0})= \frac{X_{0}}{x}\land 1$. Why would this imply that $S$ is distributed under $\mathcal F_{0}$ as $\frac{X_{0}}{U}$ where $U$ is the uniform distribution on $[0,1]$,i.e. $\;\mathcal{U}[0,1]$? ($U$ is independent of $\mathcal{F}_{0}$ and $X_{0}$.)
My problem:
If $X_{0} < x \;,\; \; \text{then we have }\mathbb P( S > x \lvert \mathcal{F}_{0})= \frac{X_{0}}{x}$
If $X_{0} \geq x \;,\; \; \text{then we have }\mathbb P( S > x \lvert \mathcal{F}_{0})= 1$
I would have thought that $U$ would have to be $\mathcal{U}([X_{0},\infty))$, i.e. uniformly distributed on $[X_{0},\infty)$. Why is $[0,1]$ used?
Let $\mathbb P_0:=\mathbb P(\cdot\mid\mathcal F_0)$ be the probability under $\mathcal F_0$. The fact that $\mathbb P_0(S>x)=\frac{X_0}x\wedge1$ for all $x>0$ means that the random variable $S$ under $\mathbb P_0$ follows the Pareto distribution with scale parameter $X_0$ and shape parameter $\alpha:=1$. One way to sample the Pareto distribution is to take the inverse of a uniform variable, as explained here. More precisely, if $U$ is a uniformly distributed random variable on $(0,1)$, then the law of $X_0/U$ (treating $X_0$ as a “constant”, i.e., given $\mathcal F_0$) is Pareto with scale and shape parameters $X_0$ and $1$ respectively, so it has indeed the distribution of $S$ under $\mathbb P_0$.