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Ehrhard concavity and the loss of one score

Written by Claude Opus 5.5 (Anthropic), October 2026. Self-checked by the writing AI. Public domain (CC0).

We continue in the standing setting (S1)–(S6) of Gaussian first moments of partitions, Section 4. Rearrangement bounds for one cell gave \(P_i>Ar_i^2\), a bound that is weak for short moments. This lesson proves a bound that is linear in \(r_i\) (Proposition 5.2): \[ P_i>0.415\,r_i+0.15\,r_i^2 . \] The idea is to delete one score. By optimality the expected maximum of the remaining scores is noticeably smaller, so the deleted score must win by a margin on a substantial set. On the other hand, the probability that the deleted score wins by a margin \(u\) is the Gaussian measure of a translate of its cone, and Ehrhard's inequality makes this probability decay at least as fast as for a half-space.

1. The loss of one score

Fix \(i\), put \(\alpha=m/(m-1)\), \(M_i=\max_{j\ne i}X_j\), and \(u_+=\max\{u,0\}\).

Lemma 1.1. \(1-\alpha r_i^2\ge0\), and \[ \mathbb E(X_i-M_i)_+\ge C\bigl(1-\sqrt{1-\alpha r_i^2}\bigr). \]

Proof. Put \(w_j=z_j+z_i/(m-1)\) for \(j\ne i\). Since \(\sum_{j\ne i}z_j=-z_i\), \[ \begin{aligned} \sum_{j\ne i}|w_j|^2&=C-|z_i|^2-\frac{2|z_i|^2}{m-1}\\ &\quad+\frac{|z_i|^2}{m-1}=C-\alpha|z_i|^2 , \end{aligned} \] which is nonnegative, so \(\alpha r_i^2\le1\). The scores \(\langle w_j,g\rangle\) differ from the \(X_j\) by the common term \(\langle z_i,g\rangle/(m-1)\), whose mean is zero, so \(\mathbb EM_i=\mathbb E\max_{j\ne i}\langle w_j,g\rangle\), which is at most \(\sqrt C\sqrt{C-\alpha|z_i|^2}\) by (S6). Since \(\max_jX_j=M_i+(X_i-M_i)_+\) and \(\mathbb E\max_jX_j=C\) by (S5), \[ \begin{aligned} \mathbb E(X_i-M_i)_+&=C-\mathbb EM_i\\ &\ge C-\sqrt C\sqrt{C-\alpha|z_i|^2} . \end{aligned} \] \(\square\)

2. Gap probabilities

By (S3) the vectors \(z_i-z_j\), \(j\ne i\), form a basis of \(V\), so there is \(h\in V\) with \(\langle h,z_i-z_j\rangle=1\) for every \(j\ne i\); then \(\langle h,z_i\rangle=1/\alpha\) (Gaussian first moments of partitions, Exercise 5.5). For \(u\in\mathbb R\), the translate \(K_i+uh\) is the set of \(x\in V\) with \(\langle z_i-z_j,x\rangle\ge u\) for all \(j\ne i\). So the gap probability \(p_i(u)\), the probability that \(X_i-X_j\ge u\) for all \(j\ne i\), is \[ p_i(u)=\gamma_{m-1}(K_i+uh). \]

Lemma 2.1. \(p_i(0)=P_i\), \(p_i\) is differentiable, and \(p_i'(0)=-1/\alpha\). Moreover \[ \mathbb E(X_i-M_i)_+=\int_0^\infty p_i(u)\,du . \]

Proof. Write \(\varphi(y)=(2\pi)^{-(m-1)/2}e^{-|y|^2/2}\). Translation invariance of Lebesgue measure gives \(p_i(u)=\int_{K_i}\varphi(y+uh)\,dy\). The derivative of the integrand in \(u\) is \(-\langle y+uh,h\rangle\varphi(y+uh)\); for \(|u|\le u_0\) it is bounded by \(|h|(|y|+u_0|h|)\) times \((2\pi)^{-(m-1)/2}e^{-|y|^2/2+u_0|y||h|}\), which is integrable. Differentiation under the integral sign (Fremlin, Measure Theory, Volume 1, 123D, core course Measure and Integration) gives \[ \begin{aligned} p_i'(0)&=-\int_{K_i}\langle y,h\rangle\varphi(y)\,dy\\ &=-\langle h,z_i\rangle=-\frac1\alpha , \end{aligned} \] using (S4). For the last formula, \(W=(X_i-M_i)_+\) is nonnegative, \(\mathbb EW=\int_0^\infty\Pr\{W\ge u\}\,du\) by Fubini's theorem, and for \(u>0\) the event \(\{W\ge u\}\) is \(\{X_i-X_j\ge u\text{ for all }j\ne i\}\). \(\square\)

3. Ehrhard's inequality

The following theorem is Theorem A of the course Ehrhard's inequality for convex sets, where it is proved in full; Corollary 3.4 there is the concavity statement of Lemma 3.1 below.

(EHR) Ehrhard's inequality. Let \(K,L\subseteq\mathbb R^d\) be closed convex sets with \(\gamma_d(K),\gamma_d(L)\in(0,1)\), and \(0\le\lambda\le1\). Then \[ \begin{aligned} &\Phi^{-1}\bigl(\gamma_d(\lambda K+(1-\lambda)L)\bigr)\\ &\quad\ge\lambda\,\Phi^{-1}(\gamma_d(K))\\ &\qquad+(1-\lambda)\,\Phi^{-1}(\gamma_d(L)). \end{aligned} \]

Ehrhard proved it for convex sets by a Gaussian symmetrization [Ehrhard]. Latała extended it to the case in which only one set is convex, and Borell to arbitrary Borel sets; van Handel [vH] and Neeman and Paouris [NP] gave further proofs, by a stochastic game and by interpolation along the Ornstein–Uhlenbeck semigroup; the course cited above follows the interpolation method with the heat semigroup. For parallel half-spaces \(\{\langle e,x\rangle\le a\}\) and \(\{\langle e,x\rangle\le b\}\) both sides equal \(\lambda a+(1-\lambda)b\), so the inequality is sharp.

Lemma 3.1. The function \(u\mapsto\Phi^{-1}(p_i(u))\) is concave on \(\mathbb R\).

Proof. By (S4) the cone \(K_i\) is closed and convex with nonempty interior, and it is not all of \(V\); its complement is open and nonempty. So every translate \(K_i+uh\) has Gaussian measure in \((0,1)\). For convex \(K\), \(\lambda K+(1-\lambda)K=K\), hence \[ \begin{aligned} &\lambda(K_i+u_1h)+(1-\lambda)(K_i+u_2h)\\ &\quad=K_i+(\lambda u_1+(1-\lambda)u_2)h , \end{aligned} \] and (EHR) gives the concavity inequality. \(\square\)

Lemma 3.2 (tangent bound). Let \(q=\Phi^{-1}(P_i)\). For every \(u\ge0\), \[ p_i(u)\le\Phi\Bigl(q-\frac u{\alpha\phi(q)}\Bigr). \]

Proof. The function \(\psi=\Phi^{-1}\circ p_i\) is concave by Lemma 3.1 and differentiable at \(0\) with \(\psi(0)=q\) and \(\psi'(0)=p_i'(0)/\phi(q)=-1/(\alpha\phi(q))\), by Lemma 2.1 and the derivative of \(\Phi^{-1}\). For a concave function the difference quotient \((\psi(u)-\psi(0))/u\) is nonincreasing in \(u>0\), so it is at most its limit \(\psi'(0)\) as \(u\to0^+\). Hence \(\psi(u)\le q+\psi'(0)u\), and \(\Phi\) is increasing. \(\square\)

Corollary 3.3. With \(q=\Phi^{-1}(P_i)\), \[ \mathbb E(X_i-M_i)_+\le\alpha\,\phi(q)\bigl(q\Phi(q)+\phi(q)\bigr). \]

Proof. By Lemmas 2.1 and 3.2 and the substitution \(s=q-u/(\alpha\phi(q))\), \[ \begin{aligned} &\mathbb E(X_i-M_i)_+\\ &\quad\le\int_0^\infty\Phi\Bigl(q-\frac u{\alpha\phi(q)}\Bigr)du\\ &\quad=\alpha\phi(q)\int_{-\infty}^q\Phi(s)\,ds , \end{aligned} \] and \(\int_{-\infty}^q\Phi=q\Phi(q)+\phi(q)\), because the right side has derivative \(\Phi(q)\) and tends to \(0\) as \(q\to-\infty\). \(\square\)

The loss constraint. Put \[ G(q)=\frac{\phi(q)\bigl(q\Phi(q)+\phi(q)\bigr)}{\Phi(q)^2}. \] Lemma 1.1 and Corollary 3.3 give \[ \begin{aligned} &\alpha\phi(q)\bigl(q\Phi(q)+\phi(q)\bigr)\\ &\quad\ge C\bigl(1-\sqrt{1-\alpha r_i^2}\bigr)\\ &\quad=\frac{C\alpha r_i^2}{1+\sqrt{1-\alpha r_i^2}} . \end{aligned} \] Dividing by \(\alpha\), and using \(C>B^2\) and \(\alpha>1\), we obtain the loss constraint (L): \[ P_i^2\,G\bigl(\Phi^{-1}(P_i)\bigr)>\frac{B^2r_i^2}{1+\sqrt{1-r_i^2}} . \]

4. The function \(G\)

Lemma 4.1. Let \(\lambda(q)=\phi(q)/\Phi(q)\) and \(V(q)=\operatorname{Var}(Z\mid Z\le q)\) for a standard normal \(Z\). Then \(G=\lambda(q+\lambda)=1-V\), so \(0<G<1\); \(G\) is nonincreasing; and \(P\mapsto P^2G(\Phi^{-1}(P))\) is increasing on \((0,\frac12]\).

Proof. Integration by parts gives \(\int_{-\infty}^qs\phi(s)\,ds=-\phi(q)\) and \(\int_{-\infty}^qs^2\phi(s)\,ds=\Phi(q)-q\phi(q)\), so \(\mathbb E(Z\mid Z\le q)=-\lambda\) and \(V=1-q\lambda-\lambda^2\). Expanding \(G\) gives \(G=\lambda q+\lambda^2=1-V\). The variable \(U=q-Z\), conditioned on \(Z\le q\), is nonnegative and nonconstant, with mean \(q+\lambda>0\) and variance \(V>0\); so \(G=\lambda(q+\lambda)>0\) and \(G=1-V<1\).

The survival function of \(U\) is \(S(u)=\Phi(q-u)/\Phi(q)\) for \(u\ge0\). Since \(\lambda'=-\lambda(q+\lambda)\), we have \((\log\Phi)''=\lambda'<0\), so \(\log S\) is concave on \([0,\infty)\) with \(\log S(0)=0\). A concave function has nonincreasing increments over intervals of equal length, so \(\log S(u+v)-\log S(u)\) is at most \(\log S(v)-\log S(0)\), that is, \(S(u+v)\le S(u)S(v)\). Integrating over \(u,v\ge0\) by Fubini's theorem, \[ \begin{aligned} \frac12\mathbb EU^2&=\int_0^\infty\!\!\int_0^\infty S(u+v)\,du\,dv\\ &\le\Bigl(\int_0^\infty S\Bigr)^2=(\mathbb EU)^2 . \end{aligned} \] So \(V=\mathbb EU^2-(\mathbb EU)^2\le(\mathbb EU)^2=(q+\lambda)^2\). Differentiating \(V=1-q\lambda-\lambda^2\) with \(\lambda'=-\lambda(q+\lambda)\), \[ \begin{aligned} V'&=-\lambda+\lambda(q+\lambda)(q+2\lambda)\\ &=\lambda\bigl((q+\lambda)^2-V\bigr)\ge0 , \end{aligned} \] so \(G=1-V\) is nonincreasing.

Finally \(P^2G(\Phi^{-1}(P))=\phi(q)(q\Phi(q)+\phi(q))\) with \(q=\Phi^{-1}(P)\). For \(q\le0\) both factors are positive and increasing in \(q\): \(\phi\) increases on \((-\infty,0]\), and \(q\Phi(q)+\phi(q)\) has derivative \(\Phi(q)>0\) and is positive, being \(\Phi(q)\,\mathbb E(q-Z\mid Z\le q)\). Since \(P\le\frac12\) means \(q\le0\), the claim follows. \(\square\)

At the two ends, \(G(0)=\phi(0)^2/\Phi(0)^2=2/\pi\) and \(G(q)\to1\) as \(q\to-\infty\) (Exercise 6.3).

5. The linear–quadratic bound

Put \(P_0(r)=(0.415+0.15r)\,r\). Then \(P_0(r)\le P_0(\frac23)=\frac{103}{300}<\frac12\) for \(0<r\le\frac23\).

Lemma 5.1. For \(0<r\le\frac23\) the following inequality (T) holds: \[ \begin{aligned} &(0.415+0.15r)^2\,G\bigl(\Phi^{-1}(P_0(r))\bigr)\\ &\qquad<\frac{B^2}{1+\sqrt{1-r^2}} . \end{aligned} \]

Proof. Cover \((0,\frac23]\) by the ten intervals \([a,b]\) of Table 1. On each interval we bound \(G(\Phi^{-1}(P_0(r)))\) above by the number \(U\) of the table: in the first row by \(G<1\) (Lemma 4.1); in the other rows, Table 2 gives a rational \(q_a\) with \(\Phi(q_a)<P_0(a)\) and \(G(q_a)<U\), and for \(r\ge a\) we have \(\Phi^{-1}(P_0(r))\ge\Phi^{-1}(P_0(a))>q_a\), so \(G(\Phi^{-1}(P_0(r)))\le G(q_a)<U\) because \(G\) is nonincreasing. The coefficient \((0.415+0.15r)^2\) and the right side of (T) increase with \(r\), so (T) holds on \([a,b]\) as soon as \[ \begin{aligned} &\Delta(a,b,U)=\frac{B^2}{1+\sqrt{1-a^2}}\\ &\quad-U(0.415+0.15b)^2>0 . \end{aligned} \] The last column of Table 1 gives strict lower bounds, obtained with the rational enclosures of Rearrangement bounds for one cell, Section 3 (an upper bound for \(\pi\) in \(B^2=9/(8\pi)\) and an upper bound for \(\sqrt{1-a^2}\)). \(\square\)

Table 1. The interval cover for (T).

\(a\) \(b\) \(U\) \(10^6\,\Delta(a,b,U)>\)
0 0.05 1 542
0.05 0.14 0.91 6173
0.14 0.24 0.87 2976
0.24 0.33 0.836 1328
0.33 0.41 0.809 523
0.41 0.48 0.788 392
0.48 0.54 0.771 1076
0.54 0.59 0.756 2787
0.59 0.64 0.743 4115
0.64 2/3 0.73 8886

Table 2. Quantiles and values of \(G\), from the enclosures of \(\phi\) and \(\Phi\).

\(a\) \(q_a\) \(10^6\,(P_0(a)-\Phi(q_a))>\) \(G(q_a)<\)
0.05 −2.033 98 0.887728
0.14 −1.548 228 0.854397
0.24 −1.238 381 0.826606
0.33 −1.024 366 0.803691
0.41 −0.860 470 0.783737
0.48 −0.728 452 0.766006
0.54 −0.621 539 0.750456
0.59 −0.535 739 0.737150
0.64 −0.450 684 0.723262

In every row of Table 2 the bound for \(G(q_a)\) is below the \(U\) of the same row of Table 1.

Proposition 5.2. For every \(i\), \(P_i>(0.415+0.15r_i)\,r_i\).

Proof. Suppose \(P_i\le P_0(r_i)\). Both numbers lie in \((0,\frac12]\), by (S5) and since \(r_i<\frac23\) (Rearrangement bounds for one cell, Proposition 2.4). By Lemma 4.1, \[ \begin{aligned} &P_i^2G(\Phi^{-1}(P_i))\\ &\quad\le P_0(r_i)^2G\bigl(\Phi^{-1}(P_0(r_i))\bigr)\\ &\quad=r_i^2(0.415+0.15r_i)^2\\ &\qquad\cdot G\bigl(\Phi^{-1}(P_0(r_i))\bigr), \end{aligned} \] which is less than \(B^2r_i^2/(1+\sqrt{1-r_i^2})\) by (T). This contradicts (L). \(\square\)

Summary of the scalar bounds. With \(A=0.929\) for \(m=5\) and \(A=0.884\) for \(m\ge6\), every cell satisfies \[ \begin{gathered} r_i<\frac23,\qquad P_i>Ar_i^2,\\ P_i>0.415\,r_i+0.15\,r_i^2 . \end{gathered} \] These three bounds, the identities \(\sum P_i=\sum r_i^2=1\), and the geometric constraints of Two competing scores are all that Five or more cells and the propeller bound uses.

6. Exercises

6.1 (easy). Let \(K=\{x:\langle e,x\rangle\ge0\}\) with \(|e|=1\), and \(h\) a vector with \(\langle e,h\rangle=c>0\). Show that \(\Phi^{-1}(\gamma_d(K+uh))=-cu\) for all \(u\), so Lemma 3.2 holds with equality for a half-space whose parameters match.

6.2 (medium). Suppose only that \(\log p_i\) is concave. (This follows from Lemma 3.1, because \(\log\Phi\) is concave and increasing.) Show that then \(p_i(u)\le P_ie^{-u/(\alpha P_i)}\), and that the argument of Section 3 yields only \(P_i>Br_i/\sqrt{1+\sqrt{1-r_i^2}}\). Evaluate both this bound and \(P_0(r)\) at \(r=0.42\).

6.3 (medium). Show that \(G(q)\to1\) as \(q\to-\infty\), using \(\lambda(q)=-q+1/(-q)+O(|q|^{-3})\).

6.4 (easy). Show that \(\mathbb EW=\int_0^\infty\Pr\{W\ge u\}\,du\) for a nonnegative random variable \(W\).

7. Solutions

6.1. \(K+uh=\{x:\langle e,x\rangle\ge cu\}\), whose measure is \(\Phi(-cu)\). Here \(\gamma_d(K)=\frac12\), \(q=0\), and \(\int_Kx\,d\gamma_d=\phi(0)e\) gives \(p'(0)=-\langle h,\phi(0)e\rangle=-c\phi(0)\); the tangent bound is \(\Phi(0-c\phi(0)u/\phi(0))=\Phi(-cu)\), the exact value.

6.2. The tangent line of \(\log p_i\) at \(0\) has slope \(p_i'(0)/P_i=-1/(\alpha P_i)\), so \(p_i(u)\le P_ie^{-u/(\alpha P_i)}\) and \(\int_0^\infty p_i\le\alpha P_i^2\). With Lemma 1.1, \(\alpha P_i^2\ge C\alpha r_i^2/(1+\sqrt{1-\alpha r_i^2})\), so \(P_i^2>B^2r_i^2/(1+\sqrt{1-r_i^2})\). At \(r=0.42\) this gives only \(P>0.18197\), whereas \(P_0(0.42)=0.20076\). The difference matters: Five or more cells and the propeller bound uses the excess of \(P_0(0.42)\) over \(Ar^2\) with a margin of less than \(0.003\).

6.3. \(G=\lambda(q+\lambda)\). With \(\lambda=-q+1/(-q)+O(|q|^{-3})\), \(q+\lambda=1/(-q)+O(|q|^{-3})\), and \(\lambda(q+\lambda)=1+O(q^{-2})\). (The expansion of \(\lambda\) follows from the asymptotic series of \(\Phi(q)/\phi(q)\) as \(q\to-\infty\), obtained by integrating by parts twice.)

6.4. \(W=\int_0^\infty\mathbf 1_{\{W\ge u\}}\,du\) pointwise; take expectations and exchange the integrals by Fubini's theorem for nonnegative functions.

References