Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 168 / 4

Write the multilinear polynomial as $f=g+x_{n}h$, where $g$ is independent of $x_n$, $\mathrm{deg}g\le k$, and $\mathrm{deg}h\le k-1$. Put $m_3=\mathbb{E}{X}_n^3$. The Cauchy-Schwarz inequality gives $|m_3|\le \sqrt{C}$, and independence givesChoose $A=A(C)\ge 1$ so large thatInduction on $n$, using $2a{b}^3\le a^2{b}^2+b^4$ and the orthogonal identity $\Vert f{\Vert }_2^2=\Vert g{\Vert }_2^2+\Vert h{\Vert }_2^2$, now yields

If $\mathbb{E}{X}_i^3=0$ and $C\le 9$, the cubic term vanishes. Taking $A=3$ and applying the induction hypothesis gives exactly

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The relevant invariance principle for a low-degree multilinear polynomial is the following. Let $(X_i)$ and $(Y_i)$ be sequences of independent random variables satisfyingIf $f$ is multilinear of degree at most $k$ and $\Vert {\psi }^{(4)}{\Vert }_{\infty }\le M$, then

For the proof, use the Lindeberg replacement method. Replace $X_i$ by $Y_i$ one coordinate at a time and write $f=u_i+z_i{v}_i$, where $u_i=E_{i}f$ and $v_i=D_{i}f$ depend only on the other coordinates. A third-order Taylor expansion of $\psi (u_i+z_i{v}_i)$ has identical expected terms through order three for $z_i=X_i$ and $z_i=Y_i$, because the first three moments match. Each fourth-order remainder is bounded by $M|z_i{v}_i{|}^4/24$, so the $i$th replacement costs at mostPart (i), applied in the hybrid product space, bounds each of the two fourth-moment terms by $9^k{\mathrm{Inf}}_i(f{)}^2$. Thus the cost is at most $(M/12)9^k{\mathrm{Inf}}_i(f{)}^2$. The triangle inequality and summation over $i$ prove the result.

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Apply the coordinate-replacement proof from part (ii) directly to the quadratic formThe zero diagonal makes $f$ multilinear. For coordinate $i$,and the row and column assumptions implyEvery hybrid vector appearing during replacement has independent centered variance-one coordinates with fourth moments at most $9$. The degree-one case of part (i) therefore givesThe fourth-order Taylor remainder from replacing coordinate $i$ is at most $(M/12)\mathbb{E}(D_{i}f{)}^4\le 12M$. Summing the $n$ replacement errors yields the stronger estimate

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