Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 168 / 1

Write $q=1-p$, $\mu =q-p$, $\sigma =2\sqrt{pq}$, and ${\varphi }_S(x)={\prod }_{j\in S}(x_j-\mu )/\sigma$. The functions ${\varphi }_S$ form the p-biased product measure orthonormal basis, so the Fourier expansion is $f={\sum }_S{\hat{f}}_p(S){\varphi }_S$. The normalized discrete derivative of a Boolean function satisfiesApplying Parseval identity and then exchanging two finite sums gives

The noise operator on the Boolean hypercube acts diagonally on the same basis: $T_{\rho }{\varphi }_S={\rho }^{|S|}{\varphi }_S$. Hence the noise stability isIts derivative isTaking the right-hand value at $\rho =0$ leaves exactly the linear Fourier weight ${\sum }_i{\hat{f}}_p(\{i\}{)}^2$, while taking the left-hand value at $\rho =1$ gives $\mathbf{I}(f)$.

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It is enough to consider three alternatives $A,B,C$. Encode each voter's three pairwise preferences by $(x_i,y_i,z_i)$, where $x_i=1$ means $A>B$, $y_i=1$ means $B>C$, and $z_i=1$ means $C>A$. A valid ranking excludes $(1,1,1)$ and $(-1,-1,-1)$. Independence of irrelevant alternatives gives three Boolean functions $u,v,w$ for the social comparisons. Unanimity and transitivity force $u=v=w$: fixing arbitrary $x$, taking $y=-x$ and $z$ constantly equal to $u(x)$ shows $v(-x)=-u(x)$, and cyclic symmetry gives the claim.

Choose the voters' valid rankings independently and uniformly. A social Condorcet paradox is absent exactly whenThus transitivity for every profile gives $3\mathbb{E}[u(x)u(y)]=-1$. Conditional on $x_i$, the bit $y_i$ equals $x_i$ with probability $1/3$ and differs with probability $2/3$, so $(x,y)$ has correlation $-1/3$. ThereforeThe Fourier formula, valid for negative correlation, givesby Parseval identity. Among the numbers $(-1/3{)}^m$, the unique minimum is $-1/3$, attained at $m=1$. Equality in this weighted average therefore forces all Fourier mass onto level one. Hence $u$ is a linear Boolean function with zero constant term. Such a function can have only one nonzero coefficient: otherwise varying two coordinates would make it assume more than two values. Thus $u(x)=x_j$ or $u(x)=-x_j$ for some $j$, making voter $j$ a dictator and proving Arrow theorem.

Solved by gpt-5.6-sol high.