<p>We study a game between <i>N</i> job applicants who incur a cost <InlineEquation ID="IEq1"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq1.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="64" /> </InlineMediaObject> <EquationSource Format="TEX">\(c\in [0,1)\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>c</mi> <mo>∈</mo> <mo stretchy="false">[</mo> <mn>0</mn> <mo>,</mo> <mn>1</mn> <mo stretchy="false">)</mo> </mrow> </math></EquationSource> </InlineEquation> (relative to the job value) to reveal their type during interviews and an administrator who seeks to maximize the probability of hiring the best applicant. We define a full learning equilibrium and prove its existence, uniqueness, and optimality. In full learning equilibrium, the administrator accepts the current best applicant <i>n</i> with probability <i>c</i> if <InlineEquation ID="IEq2"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq2.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="52" /> </InlineMediaObject> <EquationSource Format="TEX">\(n&lt;n^*\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>&lt;</mo> <msup> <mi>n</mi> <mo>∗</mo> </msup> </mrow> </math></EquationSource> </InlineEquation> and with probability 1 if <InlineEquation ID="IEq3"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq3.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="52" /> </InlineMediaObject> <EquationSource Format="TEX">\(n\ge n^*\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mi>n</mi> <mo>≥</mo> <msup> <mi>n</mi> <mo>∗</mo> </msup> </mrow> </math></EquationSource> </InlineEquation> for a threshold <InlineEquation ID="IEq4"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq4.gif" Format="GIF" Height="14" Rendition="HTML" Resolution="72" Type="Linedraw" Width="19" /> </InlineMediaObject> <EquationSource Format="TEX">\(n^*\)</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>n</mi> <mo>∗</mo> </msup> </math></EquationSource> </InlineEquation> independent of <i>c</i>. In contrast to the case without cost, where the success probability converges to <InlineEquation ID="IEq5"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq5.gif" Format="GIF" Height="19" Rendition="HTML" Resolution="72" Type="Linedraw" Width="77" /> </InlineMediaObject> <EquationSource Format="TEX">\(1/\textrm{e}\approx 0.37\)</EquationSource> <EquationSource Format="MATHML"><math> <mrow> <mn>1</mn> <mo stretchy="false">/</mo> <mtext>e</mtext> <mo>≈</mo> <mn>0.37</mn> </mrow> </math></EquationSource> </InlineEquation> as <i>N</i> tends to infinity, with cost the success probability decays like <InlineEquation ID="IEq6"> <InlineMediaObject> <ImageObject Color="BlackWhite" FileRef="182_2025_922_Article_IEq6.gif" Format="GIF" Height="16" Rendition="HTML" Resolution="72" Type="Linedraw" Width="33" /> </InlineMediaObject> <EquationSource Format="TEX">\(N^{-c}\)</EquationSource> <EquationSource Format="MATHML"><math> <msup> <mi>N</mi> <mrow> <mo>-</mo> <mi>c</mi> </mrow> </msup> </math></EquationSource> </InlineEquation>.</p>

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Incentivizing hidden types in secretary problem

  • Longjian Li,
  • Alexis Akira Toda

摘要

We study a game between N job applicants who incur a cost \(c\in [0,1)\) c [ 0 , 1 ) (relative to the job value) to reveal their type during interviews and an administrator who seeks to maximize the probability of hiring the best applicant. We define a full learning equilibrium and prove its existence, uniqueness, and optimality. In full learning equilibrium, the administrator accepts the current best applicant n with probability c if \(n<n^*\) n < n and with probability 1 if \(n\ge n^*\) n n for a threshold \(n^*\) n independent of c. In contrast to the case without cost, where the success probability converges to \(1/\textrm{e}\approx 0.37\) 1 / e 0.37 as N tends to infinity, with cost the success probability decays like \(N^{-c}\) N - c .