A digital shift and a nested scramble give different error laws

Under a binary digital shift, the two output first digits are opposite and their later digits agree. Outside the null event that the shared tail is eventually all ones, their numeric half-open representatives therefore have average ${\widehat{I}}_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}=1\mathrm{/}4+T$, where $T={\sum }_{k\ge 2}{S}_k{2}^{-k}$. For any fixed starting position, the event that the independent fair digits are all one from there onward has probability at most $2^{-L}$ for every $L$; [F10] makes each such tail event null, so their countable union is null. A prescribed prefix of $K$ tail digits has probability $2^{-K}$ and places $T$ in a half-open dyadic interval of length $2^{-K-1}$, exactly its probability under $\mathrm{Unif}[0,1\mathrm{/}2)$. These intervals generate the Borel sets of $[0,1\mathrm{/}2)$, and [F7] gives $T\sim \mathrm{Unif}[0,1\mathrm{/}2)$. Hence [F5] and [F6] give $\mathbb{E}{\widehat{I}}_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}}=1\mathrm{/}2$ and $\mathrm{Var}({\widehat{I}}_{\mathrm{s}\mathrm{h}\mathrm{i}\mathrm{f}\mathrm{t}})=1\mathrm{/}48$; the null representative choice does not change these moments.

Under nested scrambling, the two distinct first input digits are sent to opposite halves. After that first digit their input-prefix nodes are distinct, so the two subsequent digit streams are independently scrambled. Conditional on which point enters which half, they are independent uniforms on their respective half intervals, and either assignment gives the same law for their average. Hence this is also the unconditional law. Each has variance $(1\mathrm{/}2{)}^2\mathrm{/}12=1\mathrm{/}48$, so [F8]–[F9] give the average variance $(1\mathrm{/}48+1\mathrm{/}48)\mathrm{/}4=1\mathrm{/}96$.

By [F3] and [F4], both rules have uniform marginals and preserve the same finite net almost surely. Their estimates nevertheless have variances $1\mathrm{/}48$ and $1\mathrm{/}96$, respectively, for this $f$. Thus marginal uniformity and almost-sure exact net preservation alone do not determine the randomization error law.