<p>It is well understood that if one is given a set <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="72" /> </InlineMediaObject> <EquationSource Format="TEX">\(X \subset [0,1]\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>X</mi> <mo>⊂</mo> <mo stretchy="false">[</mo> <mn>0</mn> <mo>,</mo> <mn>1</mn> <mo stretchy="false">]</mo> </mrow> </math></EquationSource> </InlineEquation> of <i>n</i> independent uniformly distributed random variables, then <Equation ID="Equ6"> <MediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_Equ6.gif" Format="GIF" Height="43" Rendition="HTML" Resolution="72" Type="Linedraw" Width="442" /> </MediaObject> <EquationSource Format="TEX">\(\begin{aligned} \sup _{0 \le x \le 1} \left| \frac{\# X \cap [0,x]}{\# X} - x \right| \lesssim \frac{\sqrt{\log {n}}}{ \sqrt{n}} \qquad \text{ with } \text{ high } \text{ probability. } \end{aligned}\)</EquationSource> <EquationSource Format="MATHML"><math display="block"> <mrow> <mtable> <mtr> <mtd columnalign="right"> <mrow> <munder> <mo movablelimits="true">sup</mo> <mrow> <mn>0</mn> <mo>≤</mo> <mi>x</mi> <mo>≤</mo> <mn>1</mn> </mrow> </munder> <mfenced close="|" open="|"> <mfrac> <mrow> <mo>#</mo> <mi>X</mi> <mo>∩</mo> <mo stretchy="false">[</mo> <mn>0</mn> <mo>,</mo> <mi>x</mi> <mo stretchy="false">]</mo> </mrow> <mrow> <mo>#</mo> <mi>X</mi> </mrow> </mfrac> <mo>-</mo> <mi>x</mi> </mfenced> <mo>≲</mo> <mfrac> <msqrt> <mrow> <mo>log</mo> <mi>n</mi> </mrow> </msqrt> <msqrt> <mi>n</mi> </msqrt> </mfrac> <mspace width="2em" /> <mspace width="0.333333em" /> <mtext>with</mtext> <mspace width="0.333333em" /> <mspace width="0.333333em" /> <mtext>high</mtext> <mspace width="0.333333em" /> <mspace width="0.333333em" /> <mtext>probability.</mtext> <mspace width="0.333333em" /> </mrow> </mtd> </mtr> </mtable> </mrow> </math></EquationSource> </Equation>We show that one can improve the error term by removing a few of the points. For any <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq2.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="88" /> </InlineMediaObject> <EquationSource Format="TEX">\(m \le 0.001n\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>≤</mo> <mn>0.001</mn> <mi>n</mi> </mrow> </math></EquationSource> </InlineEquation> there exists a subset <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq3.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="55" /> </InlineMediaObject> <EquationSource Format="TEX">\(Y \subset X\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>Y</mi> <mo>⊂</mo> <mi>X</mi> </mrow> </math></EquationSource> </InlineEquation> obtained by deleting at most <i>m</i> points, so that the error term drops from <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq4.gif" Format="GIF" Height="20" Rendition="HTML" Resolution="72" Type="Linedraw" Width="102" /> </InlineMediaObject> <EquationSource Format="TEX">\(\sim \sqrt{\log {n}}/\sqrt{n}\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>∼</mo> <msqrt> <mrow> <mo>log</mo> <mi>n</mi> </mrow> </msqrt> <mo stretchy="false">/</mo> <msqrt> <mi>n</mi> </msqrt> </mrow> </math></EquationSource> </InlineEquation> to <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="72" /> </InlineMediaObject> <EquationSource Format="TEX">\( \log {(n)}/m\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>log</mo> <mrow> <mo stretchy="false">(</mo> <mi>n</mi> <mo stretchy="false">)</mo> </mrow> <mo stretchy="false">/</mo> <mi>m</mi> </mrow> </math></EquationSource> </InlineEquation> with high probability. When <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq6.gif" Format="GIF" Height="10" Rendition="HTML" Resolution="72" Type="Linedraw" Width="58" /> </InlineMediaObject> <EquationSource Format="TEX">\(m=cn\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>m</mi> <mo>=</mo> <mi>c</mi> <mi>n</mi> </mrow> </math></EquationSource> </InlineEquation> for a small <InlineEquation ID="IEq7"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq7.gif" Format="GIF" Height="15" Rendition="HTML" Resolution="72" Type="Linedraw" Width="99" /> </InlineMediaObject> <EquationSource Format="TEX">\(0 \le c \le 0.001\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>0</mn> <mo>≤</mo> <mi>c</mi> <mo>≤</mo> <mn>0.001</mn> </mrow> </math></EquationSource> </InlineEquation>, this achieves the essentially optimal asymptotic order of discrepancy <InlineEquation ID="IEq8"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="12220_2025_2177_Article_IEq8.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="65" /> </InlineMediaObject> <EquationSource Format="TEX">\(\log (n)/n\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mo>log</mo> <mo stretchy="false">(</mo> <mi>n</mi> <mo stretchy="false">)</mo> <mo stretchy="false">/</mo> <mi>n</mi> </mrow> </math></EquationSource> </InlineEquation>. The proof is constructive and works in an online setting (where one is given the points sequentially, one at a time, and has to decide whether to keep or discard it). A change of variables shows the same result for any random variables on the real line with absolutely continuous density.</p>

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Regularizing Random Points by Deleting a Few

  • Dmitriy Bilyk,
  • Stefan Steinerberger

摘要

It is well understood that if one is given a set \(X \subset [0,1]\) X [ 0 , 1 ] of n independent uniformly distributed random variables, then \(\begin{aligned} \sup _{0 \le x \le 1} \left| \frac{\# X \cap [0,x]}{\# X} - x \right| \lesssim \frac{\sqrt{\log {n}}}{ \sqrt{n}} \qquad \text{ with } \text{ high } \text{ probability. } \end{aligned}\) sup 0 x 1 # X [ 0 , x ] # X - x log n n with high probability. We show that one can improve the error term by removing a few of the points. For any \(m \le 0.001n\) m 0.001 n there exists a subset \(Y \subset X\) Y X obtained by deleting at most m points, so that the error term drops from \(\sim \sqrt{\log {n}}/\sqrt{n}\) log n / n to \( \log {(n)}/m\) log ( n ) / m with high probability. When \(m=cn\) m = c n for a small \(0 \le c \le 0.001\) 0 c 0.001 , this achieves the essentially optimal asymptotic order of discrepancy \(\log (n)/n\) log ( n ) / n . The proof is constructive and works in an online setting (where one is given the points sequentially, one at a time, and has to decide whether to keep or discard it). A change of variables shows the same result for any random variables on the real line with absolutely continuous density.