Write the <multilinear polynomial> as $f=g+x_nh$, where $g$ is independent of $x_n$, $\deg g\leq k$, and $\deg h\leq k-1$. Put $m_3=\mathbb E X_n^3$. The <Cauchy-Schwarz inequality> gives $|m_3|\leq\sqrt C$, and independence gives
$$
\mathbb E(g+X_nh)^4
\leq\lVert g\rVert_4^4+6\lVert g\rVert_4^2\lVert h\rVert_4^2
+4\sqrt C\,\lVert g\rVert_4\lVert h\rVert_4^3+C\lVert h\rVert_4^4.
$$
Choose $A=A(C)\geq1$ so large that
$$
\frac6A+\frac{2\sqrt C}{A^{3/2}}\leq2,
\qquad
\frac C{A^2}+\frac{2\sqrt C}{A^{3/2}}\leq1.
$$
Induction on $n$, using $2ab^3\leq a^2b^2+b^4$ and the orthogonal identity $\lVert f\rVert_2^2=\lVert g\rVert_2^2+\lVert h\rVert_2^2$, now yields
$$
\mathbb E f(X)^4\leq A^{2k}(\mathbb E f(X)^2)^2.
$$

If $\mathbb E X_i^3=0$ and $C\leq9$, the cubic term vanishes. Taking $A=3$ and applying the induction hypothesis gives exactly
$$
\begin{aligned}
\mathbb E f^4
&\leq9^k\lVert g\rVert_2^4
+6\cdot3^k3^{k-1}\lVert g\rVert_2^2\lVert h\rVert_2^2
+9\cdot9^{k-1}\lVert h\rVert_2^4\\
&=9^k(\lVert g\rVert_2^2+\lVert h\rVert_2^2)^2
=9^k(\mathbb E f^2)^2.
\end{aligned}
$$

Solved by gpt-5.6-sol high.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-168/4/i/solution