Randomized quasi-Monte Carlo and replicate error — Examples

The first input digits of $0$ and $1\mathrm{/}2$ differ, so the first permutation sends them to opposite halves. The net count from [F1] is preserved, leaving one point in each half.

The two input-prefix paths remain distinct after the first digit; hence all their subsequent node permutations are independent. Each tail is a sequence of independent uniform binary digits, so conditional on the first permutation the two scrambled points are independent uniform variables in their assigned half intervals.

A uniform variable on an interval of width $1\mathrm{/}2$ has variance $(1\mathrm{/}2{)}^2\mathrm{/}12=1\mathrm{/}48$. Under either value of the first permutation, step 2.1 gives the same law for $Z$: half the sum of independent uniforms on the lower and upper halves. Thus this is also its unconditional law. The single-rule estimate $Z=(X_1+X_2)\mathrm{/}2$ is unbiased for $I=1\mathrm{/}2$ by [F3], and [F6]–[F7] give $\mathrm{Var}(Z)=(1\mathrm{/}48+1\mathrm{/}48)\mathrm{/}4=1\mathrm{/}96$ by [F5].

One realized estimate $Z$ cannot form the centered replicate variance formula because $R(R-1)=0$ at $R=1$. With $R\ge 2$ independent full-set scrambles, [F4] gives an unbiased estimate of $\mathrm{Var}({\bar{Z}}_R)=1\mathrm{/}(96R)$. The target integral is fixed, so this is computation-randomization uncertainty only.

The point in $[0,1\mathrm{/}2)$ is uniform there by [F2], while the point in $[1\mathrm{/}2,1)$ never enters $[0,1\mathrm{/}100)$. Thus one rule estimate is $Z=\frac{1}{2}B$, where $B$ is Bernoulli with success probability $(1\mathrm{/}100)\mathrm{/}(1\mathrm{/}2)=1\mathrm{/}50$. Consequently [F8] gives $\mathbb{E}Z=1\mathrm{/}100=I$, and $Z$ has a skewed two-point law.

For two independent estimates, if both $B$ values agree then ${\widehat{V}}_2=0$ by [F3], so the interval is the singleton $\{0\}$ or $\{1\mathrm{/}2\}$ and misses $I=1\mathrm{/}100$. This event has probability $(49\mathrm{/}50{)}^2+(1\mathrm{/}50{)}^2=2402\mathrm{/}2500$.

If the two $B$ values differ, then ${\bar{Z}}_2=1\mathrm{/}4$ and ${\widehat{V}}_2=1\mathrm{/}16$, so the interval half-width is $c\mathrm{/}4$ with $c$ as in the Example. By [F4]–[F6], $T_1$ can be represented as $G\mathrm{/}\sqrt{Q_1}$ with $G^2$ and $Q_1$ independent and identically distributed; their continuous law gives $\mathbb{P}(G^2=Q_1)=0$. Exchangeability then gives $\mathbb{P}(G^2<Q_1)=\mathbb{P}(Q_1<G^2)=1\mathrm{/}2$, so $\mathbb{P}(\mathrm{\mid }{T}_1\mathrm{\mid }<1)=1\mathrm{/}2$. Symmetry and continuity of the CDF [F7] imply $F_{t_1}(1)=3\mathrm{/}4$, hence its $0.975$ quantile satisfies $c>1$. Therefore the interval half-width $c\mathrm{/}4>1\mathrm{/}4>\mathrm{\mid }1\mathrm{/}4-1\mathrm{/}100\mathrm{\mid }=6\mathrm{/}25$, and it contains $I$. This event has probability $2(49\mathrm{/}50)(1\mathrm{/}50)=98\mathrm{/}2500$.

The resulting two-sided interval coverage is $98\mathrm{/}2500=0.0392$, far below $0.95$. The replicate variance formula remains unbiased; this finite-$R$ example shows that unbiased variance estimation alone does not guarantee accurate Student interval coverage.

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.

The function $f_1(x)=x$ has continuous derivative $1$ on $[0,1]$. With $n=2^m$, the hypotheses of [F1] hold, so its randomized-rule variance is $O(n^{-3})$.

For $N\ge 1$, the nonnegative Borel truncations $g_N=f_2{\mathbf{1}}_{[1\mathrm{/}N,1]}$ increase pointwise to $f_2$ and satisfy ${\int }_0^1{g}_N(x)dx=4(1-N^{-1\mathrm{/}4})$. Likewise, $h_N=f_2^2{\mathbf{1}}_{[1\mathrm{/}N,1]}$ increase pointwise to $f_2^2$ and satisfy ${\int }_0^1{h}_N(x)dx=2(\sqrt{N}-1)$. By [F5], ${\int }_0^1{f}_2=4$ while ${\int }_0^1{f}_2^2=+\mathrm{\infty }$; $f_2$ is not continuously differentiable at $0$.

By [F2], ${\widehat{I}}_n(f_2)=n^{-1}{\sum }_i{f}_2(X_i)$; [F4] gives every point uniform law. Since $f_2\ge 0$, [F6] and step 1.2 give $\mathbb{E}[f_2(X_1{)}^2]=+\mathrm{\infty }$. The squared estimate obeys ${\widehat{I}}_n(f_2{)}^2\ge n^{-2}{f}_2(X_1{)}^2$, so its second moment is infinite. By [F7] it has no finite variance, and the smooth variance theorem does not apply to this integrable singularity.