Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 168 / 3 / ii

If $\alpha >1/2$, monotonicity already gives $\mathbb{E}{f}^{(1/2)}\ge \alpha >1/2$, so assume $\alpha \le 1/2$. Suppose for a contradiction that $\mathbb{E}{f}^{(1/2)}\le 1/2$. By the mean value theorem, some $s\in (p,1/2)$ satisfiesThe Margulis-Russo formula identifies this derivative with the appropriately normalized total influence, so ${\mathbf{I}}_s(f)$ is bounded solely in terms of $\zeta$. The $p$-biased Friedgut junta theorem then supplies, for any small $\eta >0$, a Boolean $J$-junta $h$ with $|J|\le r(\zeta ,\eta )$ and

Because $f$ is monotone, ${\mathbb{P}}_s(f=1)\le 1/2$, hence ${\mathbb{P}}_s(h=0)\ge 1/2-\eta \ge 1/4$ when $\eta \le 1/4$. It follows thatFor some assignment $u$ on $J$ with $h(u)=0$, therefore, ${\mathbb{E}}_s{f}_u\le 4\eta$. Monotonicity and $p<s$ imply ${\mathbb{E}}_p{f}_u\le 4\eta$. Choose $\eta <\alpha /5$, set $\epsilon =\eta$, and take $r\ge |J|$. Thencontradicting $(\epsilon ,p,r)$-quasirandomness. Thus $\mathbb{E}{f}^{(1/2)}>1/2$.

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