Sharp gain bound for scrambled $(0,m,d)$-nets

Put $r=\mathrm{\mid }u\mathrm{\mid }$ and $k=\mathrm{\mid }\kappa \mathrm{\mid }$. At total digit depth $\mathrm{\ell }\le m$, every cell contains $b^{m-\mathrm{\ell }}$ points by partitioning it into depth-$m$ cells, so its ordered-pair count is $n^2{b}^{-\mathrm{\ell }}$. At depth $\mathrm{\ell }\ge m$, each cell lies in a depth-$m$ cell containing one point, so the total ordered-pair count is $n$.

Substitute these counts into [F2]. If $k+r\le m$, every term uses the first count and the alternating sum is zero. If $k\ge m$, every term uses the second count and the binomial theorem gives ${\mathrm{\Gamma }}_{u,\kappa }=1$. In the remaining case $m-r<k<m$, let $c=m-k$, so $1\le c<r$. Grouping subsets by cardinality and using the partial binomial identity gives ${\mathrm{\Gamma }}_{u,\kappa }=g_{r,c}:=1+\frac{(-1{)}^r}{(b-1{)}^{r-1}}{\sum }_{j=0}^{c-1}\left(\genfrac{}{}{0px}{}{r-1}{j}\right)(-b{)}^j\mathrm{.}$

Suppose $r=c+2L$ for an integer $L\ge 1$. For $T_j=\left(\genfrac{}{}{0px}{}{r-1}{j}\right)b^j$ and $1\le j\le c-1$, the ratio $T_j\mathrm{/}{T}_{j-1}=b(r-j)\mathrm{/}j\ge 1$, since $b\ge r$ and $j\le r-2$. Pairing adjacent terms in the alternating sum shows its sign is $(-1{)}^{c-1}$ and its absolute value is at most $T_{c-1}$. Since $(-1{)}^r=(-1{)}^c$, the correction to $1$ in $g_{r,c}$ is nonpositive. Hence $g_{r,c}\le 1$.

Suppose $r=c+2L-1$ for an integer $L\ge 1$. Define $S_{r,c-1}={\sum }_{j=0}^{c-1}\left(\genfrac{}{}{0px}{}{r-1}{j}\right)(-b{)}^j$ and $T_j=\left(\genfrac{}{}{0px}{}{r-1}{j}\right)(-b{)}^j$, with $T_{-1}=0$. Pascal's identity, applied termwise, gives $(b-1{)}^2{S}_{r,c-1}-S_{r+2,c-1}=b(b-2)T_{c-1}+b^2{T}_{c-2}$. Using the formula in step 2.1 and the parity $r=c+2L-1$ therefore gives $g_{r,c}-g_{r+2,c}=\frac{b^c}{(b-1{)}^{r+1}}((b-2)\left(\genfrac{}{}{0px}{}{r-1}{c-1}\right)-\left(\genfrac{}{}{0px}{}{r-1}{c-2}\right))\mathrm{.}$ If $c=1$, the second binomial is zero. If $c\ge 2$ and $r+2\le d\le b$, then $\left(\genfrac{}{}{0px}{}{r-1}{c-2}\right)\mathrm{/}\left(\genfrac{}{}{0px}{}{r-1}{c-1}\right)=(c-1)\mathrm{/}(2L)$. The condition $r+2=c+2L+1\le b$ gives $(c-1)\mathrm{/}(2L)\le (b-2L-2)\mathrm{/}(2L)\le b-2$. Thus $g_{r,c}\ge g_{r+2,c}$ whenever the larger dimension $r+2$ is allowed.

Repeatedly applying step 3.2 shows every allowed odd-offset gain is at most $g_{c+1,c}$. By the binomial theorem, ${\sum }_{j=0}^{c-1}\left(\genfrac{}{}{0px}{}{c}{j}\right)(-b{)}^j=(1-b{)}^c-(-b{)}^c$, so step 2.1 gives $g_{c+1,c}=(b\mathrm{/}(b-1){)}^c$. The even-offset case is at most $1$ by step 3.1, and the other two cases are $0$ or $1$. Since $c=m-k\le m$ and $c<r\le d$, every gain is at most $(b\mathrm{/}(b-1){)}^{\min (m,d-1)}$.