Finite bit random streams and variate generation — Examples

Direct iteration gives $0,1,6,7,4,5,2,3,0$. These are the eight residues modulo $8$, so the cycle is full. The parameters satisfy $5\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}4)$ and $1$ is odd, also verifying [F1]'s hypotheses.

Reducing [F2] modulo $2$ gives $x_{n+1}\equiv x_n+1(\mathrm{m}\mathrm{o}\mathrm{d}2)$, because $5\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}2)$ and $c=1$. Thus the low bit alternates deterministically. By [F3], a uniform seed gives each state mass $1\mathrm{/}8$; since the update is a permutation, each time marginal is uniform. The adjacent pair lies on the eight-point graph of the recurrence rather than the 64-point product space, so the pair is dependent.

For $b=8$, $2^b=256$ and the smallest midpoint is $(0+1\mathrm{/}2)\mathrm{/}256=1\mathrm{/}512=2^{-9}$ by [F1].

For $0<u\le v\le 1$, [F3] gives $0\le -2\log v\le -2\log u$. Their nonnegative square roots satisfy $\sqrt{-2\log v}\le \sqrt{-2\log u}$ by [F4]; thus $r(u)=\sqrt{-2\log u}$ is nonincreasing, and the largest radius occurs at $u=2^{-9}$ and equals $\sqrt{18\log 2}$. The logarithm laws [F3] give $\log (2^{-9})=-9\log 2$. By [F2] and [F5], the ideal radial tail beyond the cutoff is $e^{-18\log 2\mathrm{/}2}=e^{\log (2^{-9})}=2^{-9}=1\mathrm{/}512$, while no finite midpoint input has a larger radius.

The raw endpoint event is exactly $K=0$, which has probability $2^{-b}$; the radius expression [F3] has an undefined logarithm at that input.

By [F1], the midpoint input is at least $2^{-b-1}>0$ and is less than $1$, so its logarithm is finite. By [F2], the ideal stream hits $0$ with probability zero; a software interface still needs an explicit endpoint convention if it accepts closed-interval inputs.

For $0<\epsilon \le 1$, [F1] gives $\widehat{x}=\sqrt{\epsilon }\ge 0$ and ${\widehat{x}}^2=\epsilon >0$, so $\widehat{x}>0$. If $\widehat{x}>1$, then ${\widehat{x}}^2>1$, contradicting $\epsilon \le 1$; hence $\widehat{x}\in [0,1]$ and $F(\widehat{x})={\widehat{x}}^2=\epsilon$. Under the stated endpoint convention, the target is $0$, so the residual is $\epsilon$ while the spatial error is $\sqrt{\epsilon }$.

The ratio $\sqrt{\epsilon }\mathrm{/}\epsilon =1\mathrm{/}\sqrt{\epsilon }$ is unbounded as $\epsilon \downarrow 0$: for any $M>0$, take $0<\epsilon <\min (1,1\mathrm{/}(M+1{)}^2)$, so [F1] gives $\sqrt{\epsilon }<1\mathrm{/}(M+1)$ by comparing nonnegative squares, and the ratio exceeds $M$. Thus no constant $C$ independent of $\epsilon$ gives $\mathrm{\mid }\widehat{x}-Q(0)\mathrm{\mid }\le C\epsilon$ for these points; their nonlinear error is exactly $\sqrt{\epsilon }$. The density $f(x)=2x$ has essential infimum $0$ on every $[0,h]$ with $0<h\le 1$, so the sufficient positive-density condition in [F2] fails near zero. The endpoint computation uses the declared convention $Q(0)=0$, independently of [F2]'s interior-target theorem.