Fourth-moment bound for a centered quadratic form of an independent vector

First fix a deterministic complex symmetric matrix $A$. Since $A_{ji}=A_{ij}$, ${\xi }^{\mathsf{T}}A\xi -\mathrm{tr}A={\sum }_i{A}_{ii}({\xi }_i^2-1)+2{\sum }_{i<j}{A}_{ij}{\xi }_i{\xi }_j\mathrm{.}$ The diagonal summands have mean zero. Distinct diagonal summands are uncorrelated; a diagonal summand and an off-diagonal summand are uncorrelated because at least one independent centered coordinate occurs to the first power; and two off-diagonal summands are uncorrelated unless their unordered index pairs coincide. Therefore $\mathbb{E}{\mid {\xi }^{\mathsf{T}}A\xi -\mathrm{tr}A\mid }^2=({\mu }_4-1){\sum }_i\mathrm{\mid }{A}_{ii}{\mathrm{\mid }}^2+4{\sum }_{i<j}\mathrm{\mid }{A}_{ij}{\mathrm{\mid }}^2\le C_{{\mu }_4}{\sum }_{i,j}\mathrm{\mid }{A}_{ij}{\mathrm{\mid }}^2\mathrm{.}$ Here ${\mu }_4\ge 1$ follows from $\mathbb{E}{\xi }_i^2=1$ and Cauchy–Schwarz.

By [F2], the joint law of $(\xi ,A)$ is the product of its marginal laws. The integrand is integrable: $\mathrm{\mid }{\xi }^{\mathsf{T}}A\xi -\mathrm{tr}A{\mathrm{\mid }}^2\le 2\mathrm{\parallel }A{\mathrm{\parallel }}_F^2\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^4+2d\mathrm{\parallel }A{\mathrm{\parallel }}_F^2$, and $\mathbb{E}\mathrm{\parallel }\xi {\mathrm{\parallel }}_2^4<\mathrm{\infty }$ because $d$ is finite and $\mathbb{E}{\xi }_i^4<\mathrm{\infty }$. Thus [F3] permits integration of the deterministic inequality in step 1.1 against the law of $A$, giving the asserted bound.