CONTENTS

Exercise: Computing star-discrepancy by hand

The 1-dimensional star-discrepancy of a point set P={x1,,xN}[0,1]\mathcal{P} = \{x_1, \ldots, x_N\} \subset [0, 1] is

DN(P)=supb[0,1]{xib}Nb.D_N^*(\mathcal{P}) = \sup_{b \in [0, 1]}\left|\frac{|\{x_i \le b\}|}{N} - b\right|.

Tasks

  1. Sort the points P={0.1,0.4,0.6,0.85}\mathcal{P} = \{0.1, 0.4, 0.6, 0.85\}, N=4N = 4. At each sorted value x(k)x_{(k)}, compute the empirical CDF FN(x(k))=k/NF_N(x_{(k)}) = k/N and FN(x(k))=(k1)/NF_N(x_{(k)}^-) = (k-1)/N. For each kk, compute both FN(x(k))x(k)|F_N(x_{(k)}) - x_{(k)}| and FN(x(k))x(k)|F_N(x_{(k)}^-) - x_{(k)}|.

  2. Why does the supremum in the discrepancy definition reduce to a finite maximum over these 2N2N quantities? (Think about where the worst box edge can lie.)

  3. Compute D4(P)D_4^*(\mathcal{P}) for this set.

  4. Compare with the first 44 terms of the 1-dimensional Halton / van der Corput base-2 sequence: {0.5,0.25,0.75,0.125}\{0.5, 0.25, 0.75, 0.125\}. Compute its star-discrepancy and compare with the random set.

  5. For NN large, uniform IID has DN=O(loglogN/N)D_N^* = O(\sqrt{\log\log N / N}) while the van der Corput sequence has DN=O(logN/N)D_N^* = O(\log N / N). At N=106N = 10^6 roughly how many times smaller is the QMC discrepancy?

Hint

For part 2, the discrepancy function FN(b)b|F_N(b) - b| is piecewise affine in bb, with jumps at the sample points. Extrema occur at the breakpoints.