How the Total Chess Rating works

Proof. Because $\nabla g$ is nowhere vanishing, every nonempty level set of $g$ is a continuously differentiable hypersurface. Since each such level set is connected, any two of its points can be joined by a piecewise continuously differentiable path lying in the level set.

Let $\gamma : [0,1] \to \mathbb{R}^n$ be such a path. On each continuously differentiable segment, $g(\gamma(t))$ is constant, so

Since $\nabla f(\gamma(t))$ is parallel to $\nabla g(\gamma(t))$, it follows that

Hence $f(\gamma(t))$ is constant along the path. Therefore $f$ is constant on every nonempty level set of $g$. We may consequently define $h(s)$ to be the common value of $f$ on the level set

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Proof. Write $q(r) = 10^{r/400}$ and $q^{-1}(s) = 400\log_{10} s$. For $n = 1$, normalization gives $T_1(r) = r$, which is the stated formula. Now consider the binary rule $M(x,y) = T_2(x,y)$. By marginal Elo-strength consistency, $\frac{M_x(x,y)}{M_y(x,y)} = \frac{q(x)}{q(y)}$. The function $V_2(x,y) = q(x) + q(y)$ has the same marginal ratio: $\frac{(V_2)_x}{(V_2)_y} = \frac{q(x)}{q(y)}$. Hence the gradients of $M$ and $V_2$ are everywhere parallel. Moreover, $\nabla V_2$ is nowhere vanishing, since both of its components are strictly positive, and every level set of $V_2$ is connected: under the change of variables $u_1 = q(x)$ and $u_2 = q(y)$, it becomes the line segment

The Lemma therefore implies that $M$ is constant on the level sets of $V_2$. Thus

for some function $h$ on $(0,\infty)$. Normalization gives $h(2q(r)) = r$. Since $r \mapsto 2q(r)$ is a bijection from $\mathbb{R}$ onto $(0,\infty)$, the entire range of $V_2$, it follows that

Now fix $n \ge 2$ and write $F(R_1,\dots,R_n) = T_n(R_1,\dots,R_n)$. Aggregation consistency applied to singleton blocks implies permutation invariance. Hence it is enough to compare the first two coordinates. Using aggregation consistency for the partition $\{1,2\},\{3\},\dots,\{n\}$, we get $F(x,y,R_3,\dots,R_n) = F(M(x,y), M(x,y), R_3,\dots,R_n)$. Differentiating with respect to $x$ and $y$ gives

and

Since the common factor is positive, $\frac{\partial_1 F(x,y,R_3,\dots,R_n)}{\partial_2 F(x,y,R_3,\dots,R_n)} = \frac{M_x(x,y)}{M_y(x,y)} = \frac{q(x)}{q(y)}$. By permutation invariance, for every pair $i, j$, $\frac{\partial_i T_n(R)}{\partial_j T_n(R)} = \frac{q(R_i)}{q(R_j)}$.

Let $V_n(R) = \sum_{i=1}^{n} q(R_i)$. Since

the previous display shows that $\nabla T_n$ is everywhere parallel to $\nabla V_n$. Moreover, $\nabla V_n$ is nowhere vanishing. The level sets of $V_n$ are connected because, under the change of variables $u_i = q(R_i)$ for $i = 1,\dots,n$, they become the simplices

The Lemma therefore implies that $T_n$ is constant on every level set of $V_n$. Consequently, there exists a function $h_n$ on $(0,\infty)$, the range of $V_n$, such that

Normalization gives

Since $r \mapsto n\,q(r)$ is a bijection from $\mathbb{R}$ onto $(0,\infty)$, the entire range of $V_n$, it follows that

Therefore,

Equivalently,

It remains only to verify that this rule satisfies the axioms. Same-scale normalization is immediate. The rule is continuously differentiable and strictly increasing because $\partial_j T_n(R) = \frac{q(R_j)}{\sum_{i=1}^{n} q(R_i)} > 0$. For aggregation consistency, if $A_j = T_{|B_j|}(R_{B_j})$, then $q(A_j) = \frac{1}{|B_j|}\sum_{i \in B_j} q(R_i)$. Thus replacing each block $B_j$ by $|B_j|$ copies of $A_j$ preserves the overall average of the $q(R_i)$'s. Hence the final combined rating is unchanged. Finally, for $n = 2$, $\frac{\partial_x T_2(x,y)}{\partial_y T_2(x,y)} = \frac{q(x)}{q(y)} = \frac{E(x,y)}{1-E(x,y)}$. Therefore all four axioms hold, and the formula is unique.

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