Two competing scores

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. The bounds of Rearrangement bounds for one cell and Ehrhard concavity and the loss of one score concern one cell at a time. This lesson looks at the cell with the shortest moment and uses its moment identity to constrain how the other scores are correlated. The result is Corollary 3.1: two inequalities between the three largest normalized lengths and the smallest one.

Relabel the cells so that \[ \begin{gathered} r_1\ge r_2\ge\dots\ge r_m=t>0,\\ L=r_1^2+r_2^2,\qquad c=r_3^2 . \end{gathered} \] Since \(m\ge5\), the indices \(1,2,3\) and \(m\) are distinct, also when some lengths coincide.

1. Residual scores

Let \(g\) be standard Gaussian in \(V\) and put \[ \begin{gathered} e=\frac{z_m}{|z_m|},\qquad T=\langle e,g\rangle,\\ a_j=-\frac{\langle z_m,z_j\rangle}{|z_m|^2}\quad(j<m). \end{gathered} \] By (S2), \(a_j>0\), and \(\sum_{j<m}a_j=-\langle z_m,\sum_{j<m}z_j\rangle/|z_m|^2=1\). Define the residual vectors and residual scores \[ \begin{gathered} w_j=\frac{z_j}{\sqrt C}+a_jt\,e,\\ Y_j=\langle w_j,g\rangle\qquad(j<m). \end{gathered} \]

Lemma 1.1. (a) Each \(w_j\) is orthogonal to \(e\), and \(\sum_{j<m}w_j=0\). (b) The vector \(Y=(Y_j)_{j<m}\) is centered Gaussian and independent of the standard normal variable \(T\), and \[ \frac{X_j}{\sqrt C}=Y_j-a_jtT\qquad(j<m). \] (c) The covariances \(H_{jk}=\mathbb EY_jY_k=\langle w_j,w_k\rangle\) are \[ \begin{gathered} H_{jk}=\frac{\langle z_j,z_k\rangle}C-a_ja_kt^2,\\ h_j:=H_{jj}=r_j^2-a_j^2t^2 . \end{gathered} \] Every row of \(H\) sums to zero, and \(H_{jk}<0\) for \(j\ne k\). (d) For distinct \(j,k<m\), \(D_{jk}=\sqrt{h_jh_k-H_{jk}^2}\) is positive. In particular every \(h_j\) is positive.

Proof. (a) Since \(t=|z_m|/\sqrt C\), \(\langle z_j,z_m\rangle/(\sqrt C|z_m|)=-a_jt\), so \(\langle w_j,e\rangle=-a_jt+a_jt=0\). Also \(\sum_{j<m}w_j=-z_m/\sqrt C+te=0\).

(b) In an orthonormal basis of \(V\) whose first vector is \(e\), \(T\) is the first coordinate of \(g\) and each \(Y_j\) is a linear function of the other coordinates, which are independent of the first (product rule, Gaussian first moments of partitions, Section 1). The identity follows from \(z_j/\sqrt C=w_j-a_jte\).

(c) Expand \(\langle w_j,w_k\rangle\) and use \(\langle e,z_k\rangle/\sqrt C=-a_kt\). The row sums vanish by (a). For \(j\ne k\) both terms of \(H_{jk}\) are negative by (S2).

(d) \(D_{jk}^2\) is the Gram determinant of \(w_j\) and \(w_k\), which vanishes only if they are linearly dependent. A relation \(\beta w_j+\beta'w_k=0\) is a relation among \(z_j,z_k,z_m\), a proper subset of the moments because \(m\ge5\); by (S3) it is trivial. \(\square\)

2. A pair bound

Lemma 2.1. For distinct \(j,k<m\), \[ D_{jk}<\frac3\pi\,t\,(1+a_j)(1+a_k). \]

Proof. Let \(K=K_m\) be the cone of the shortest moment. By (S4), \(\mathbb E[\mathbf 1_Kg]=z_m\), so \[ \begin{gathered} \mathbb E[\mathbf 1_KT]=|z_m|=\sqrt C\,t>Bt,\\ \mathbb E[\mathbf 1_KY_j]=\langle w_j,z_m\rangle=0 . \end{gathered} \] On \(K\) the score \(X_m=|z_m|T\) is the largest, hence nonnegative by (S5), so \(T\ge0\); and \(X_j\le X_m\) gives \(Y_j-a_jtT\le tT\), that is, \(Y_j\le(1+a_j)tT\).

Fix \(j\ne k\) and put \[ U=\frac{Y_j}{(1+a_j)t},\qquad W=\frac{Y_k}{(1+a_k)t}. \] The pair \((U,W)\) is centered Gaussian with covariance determinant \(D_{jk}^2/\bigl((1+a_j)^2(1+a_k)^2t^4\bigr)\), so its density is bounded by \[ M=\frac{t^2(1+a_j)(1+a_k)}{2\pi D_{jk}} . \] Let \(\mathcal E\) be the event that \(T\ge0\), \(U\le T\) and \(W\le T\); it contains the event \(\{g\in K\}\). Adding the two zero expectations, taking a positive part and enlarging the region of integration to \(\mathcal E\), \[ \begin{aligned} \mathbb E[\mathbf 1_KT]&=\mathbb E[\mathbf 1_K(T+U+W)]\\ &\le\mathbb E\bigl[\mathbf 1_{\mathcal E}(T+U+W)_+\bigr]. \end{aligned} \] Condition on \(T=s\ge0\); the pair \((U,W)\) is independent of \(T\). With \(a=s-u\) and \(b=s-w\), \[ \begin{aligned} &\iint_{u\le s,\,w\le s}(s+u+w)_+\,du\,dw\\ &\quad=\int_0^\infty\!\!\int_0^\infty(3s-a-b)_+\,db\,da\\ &\quad=\int_0^{3s}\frac{(3s-a)^2}2\,da=\frac{(3s)^3}6 . \end{aligned} \] The integrand is supported on the triangle with vertices \((s,s)\), \((s,-2s)\), \((-2s,s)\). Bounding the density by \(M\) and using \(\mathbb ET_+^3=\int_0^\infty s^3\phi(s)\,ds=2/\sqrt{2\pi}\), \[ \begin{aligned} Bt<\mathbb E[\mathbf 1_KT]&\le\frac{27M}6\cdot\frac2{\sqrt{2\pi}}\\ &=\frac{9t^2(1+a_j)(1+a_k)}{2\pi\sqrt{2\pi}\,D_{jk}} . \end{aligned} \] Since \(B=3/(2\sqrt{2\pi})\), this is the claim. \(\square\)

The negative correlations of the residual scores prevent all pairwise determinants from being small at once; summing the pair bound over pairs among the three longest moments gives the constraints.

3. Two constraints

Corollary 3.1. The numbers \(L\), \(c\) and \(t\) satisfy the inequalities (C1), on the first line, and (C2), on the second: \[ \begin{aligned} (c-t^2)(L-c-2t^2)&\le8(3/\pi)^2t^2,\\ \tfrac32(c-t^2)^2&\le10(3/\pi)^2t^2. \end{aligned} \]

Proof. First, \(c\ge t^2\) by the ordering, and since \(0<a_j\le1\), \[ h_j=r_j^2-a_j^2t^2\ge c-t^2\ge0\qquad(j\le3). \]

(C1). Row 3 of \(H\) sums to zero and its off-diagonal entries are negative, so \(h_3=\sum_{k\ne3}|H_{3k}|\ge|H_{31}|+|H_{32}|\), and \(H_{31}^2+H_{32}^2\le(|H_{31}|+|H_{32}|)^2\le h_3^2\). Hence \[ \begin{aligned} &D_{31}^2+D_{32}^2\\ &\quad=h_3(h_1+h_2)-H_{31}^2-H_{32}^2\\ &\quad\ge h_3(h_1+h_2-h_3). \end{aligned} \] Here \(h_3\ge c-t^2\ge0\) and \[ \begin{aligned} &h_1+h_2-h_3\\ &\quad=L-c-(a_1^2+a_2^2-a_3^2)t^2\\ &\quad\ge L-c-2t^2 . \end{aligned} \] If \(L-c-2t^2\ge0\), then \(D_{31}^2+D_{32}^2\ge(c-t^2)(L-c-2t^2)\); otherwise this holds because the right side is at most zero. On the other side, Lemma 2.1 gives \[ \begin{aligned} &D_{31}^2+D_{32}^2\le\Bigl(\frac3\pi\Bigr)^2t^2(1+a_3)^2\\ &\qquad\cdot\bigl((1+a_1)^2+(1+a_2)^2\bigr). \end{aligned} \] With \(\sigma=1-a_3\ge a_1+a_2\) and \(a_1^2+a_2^2\le(a_1+a_2)^2\), \[ \begin{aligned} (1+a_1)^2+(1+a_2)^2&\le2+2\sigma+\sigma^2\\ &=(2-a_3)^2+1 . \end{aligned} \] For \(0\le x\le1\) the identity \[ \begin{aligned} &8-(1+x)^2\bigl((2-x)^2+1\bigr)\\ &\quad=(1-x)^2(3-x^2)\ge0 \end{aligned} \] then bounds \((1+a_3)^2((2-a_3)^2+1)\) by \(8\), which proves (C1).

(C2). Let \(\rho_{jk}=H_{jk}/\sqrt{h_jh_k}\) for \(1\le j<k\le3\); these correlations lie in \([-1,0]\). Since \[ 0\le\operatorname{Var}\Bigl(\sum_{j=1}^3\frac{Y_j}{\sqrt{h_j}}\Bigr)=3+2\sum_{j<k\le3}\rho_{jk}, \] we get \(\sum_{j<k\le3}\rho_{jk}^2\le\sum_{j<k\le3}|\rho_{jk}|\le\frac32\), and therefore \[ \begin{aligned} \sum_{j<k\le3}D_{jk}^2&=\sum_{j<k\le3}h_jh_k(1-\rho_{jk}^2)\\ &\ge(c-t^2)^2\Bigl(3-\frac32\Bigr). \end{aligned} \] On the other side, by Lemma 2.1, \(\sum_{j<k\le3}D_{jk}^2\) is at most \((3/\pi)^2t^2\) times \(\sum_{j<k\le3}(1+a_j)^2(1+a_k)^2\). With the elementary symmetric functions \(s_1,s_2,s_3\) of \(a_1,a_2,a_3\), \[ \begin{aligned} &\sum_{j<k\le3}(1+a_j)^2(1+a_k)^2\\ &\quad=3+4s_1+2s_1^2+2s_1s_2+s_2^2\\ &\qquad-(6+2s_1)s_3 . \end{aligned} \] Since \(0\le s_1\le1\) (the \(a_j\) are positive with total \(1\)), \(0\le s_2\le s_1^2/3\le\frac13\) and \(s_3\ge0\), the sum is at most \(3+4+2+\frac23+\frac19=\frac{88}9<10\). This proves (C2). \(\square\)

4. Exercises

4.1 (easy). Show that a centered Gaussian vector in \(\mathbb R^2\) with covariance matrix \(\Sigma\), \(\det\Sigma>0\), has density at most \(1/(2\pi\sqrt{\det\Sigma})\).

4.2 (easy). Verify the expansion of \(\sum_{j<k\le3}(1+a_j)^2(1+a_k)^2\) in terms of \(s_1,s_2,s_3\) at \(a_1=a_2=a_3=\frac13\).

4.3 (medium). Show that \(s_2\le s_1^2/3\) for real \(a_1,a_2,a_3\).

4.4 (medium). For the three moments of the propeller (\(m=3\)), compute the coefficients \(a_1,a_2\) relative to \(z_3\), and the residual vectors \(w_1,w_2\). Why does the argument of Lemma 2.1 need two residual scores, and hence \(m\ge4\)?

5. Solutions

4.1. The density is \((2\pi)^{-1}(\det\Sigma)^{-1/2}\exp(-\frac12y^{\mathsf T}\Sigma^{-1}y)\), and the exponential is at most one.

4.2. Each product \((1+a_j)^2(1+a_k)^2\) equals \((4/3)^4=256/81\), so the sum is \(3\cdot256/81=256/27\). With \(s_1=1\), \(s_2=\frac13\) and \(s_3=\frac1{27}\), the expansion gives \(3+4+2+\frac23+\frac19-\frac8{27}\), which is \(9+\frac{13}{27}=\frac{256}{27}\).

4.3. \[ \begin{aligned} &s_1^2-3s_2\\ &\quad=a_1^2+a_2^2+a_3^2\\ &\qquad-a_1a_2-a_1a_3-a_2a_3\\ &\quad=\frac12\sum_{j<k}(a_j-a_k)^2\ge0 . \end{aligned} \]

4.4. The propeller moments have equal lengths and pairwise angles \(2\pi/3\), so \(a_1=a_2=\frac12\). Then \(w_1=-w_2\) are orthogonal to \(z_3\): with \(m=3\) there is only one independent residual score, \(Y_2=-Y_1\), and the determinant \(D_{12}\) vanishes. The argument needs two linearly independent residual vectors, which (S3) provides when \(m\ge4\).

References