= Solution
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> satisfying
$$
\mathbb E X_i=\mathbb E Y_i=0,
\quad
\mathbb E X_i^2=\mathbb E Y_i^2=1,
\quad
\mathbb E X_i^3=\mathbb E Y_i^3=0,
\quad
\mathbb E X_i^4,\mathbb E Y_i^4\leq9.
$$
If $f$ is multilinear of degree at most $k$ and $\lVert\psi^{(4)}\rVert_\infty\leq M$, then
$$
|\mathbb E\psi(f(X))-\mathbb E\psi(f(Y))|
\leq\frac M{12}9^k\sum_{i=1}^n\operatorname{Inf}_i(f)^2.
$$
For the proof, use the <Lindeberg replacement method>. Replace $X_i$ by $Y_i$ one coordinate at a time and write $f=u_i+z_iv_i$, where $u_i=E_i f$ and $v_i=D_i f$ depend only on the other coordinates. A third-order <Taylor theorem>[Taylor expansion] of $\psi(u_i+z_iv_i)$ 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 $M|z_iv_i|^4/24$, so the $i$th replacement costs at most
$$
\frac M{24}(\mathbb E X_i^4+\mathbb E Y_i^4)\mathbb E v_i^4.
$$
Part (i), applied in the hybrid product space, bounds each of the two fourth-moment terms by $9^k\operatorname{Inf}_i(f)^2$. Thus the cost is at most $(M/12)9^k\operatorname{Inf}_i(f)^2$. The <triangle inequality> and summation over $i$ prove the result.
Solved by gpt-5.6-sol high.
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