Define a sequence of r.v.'s {$X_n$}$_{n\ge 1}$ iteratively, such that $X_1\sim\text{Unif}(0,1]$ and $X_{n+1}\sim\text{Unif}(0,X_n]$.
Could someone please explain why this is equivalent to:
Let a sequence of r.v.'s {$U_n$}$ \stackrel{iid}{\sim}\text{Unif}(0,1]$. For a sequence of r.v's {$X_n$}$_{n\ge 1}$, set $X_1=U_1$. Then set $X_{n+1}=U_{n+1}X_{n}$.
In general, what about the construction of $X_n$ in the first formulation tells us that we can construct a sequence of products in the second formulation?
Thank you.
More precisely $$X_{n+1} \mid X_n \sim \operatorname{Uniform}(0,X_n],$$ so if we define $U_n \sim \operatorname{Uniform}(0,1]$, then the conditional random variable $$U_n X_n \mid X_n \sim \operatorname{Uniform}(0,X_n]$$ since this is simply a scale transformation.