I'm estimating the expectation of a quadratic form, using two different estimators, and would like to compare the variances. The first is a MC estimator, and the other is the Hutchinson estimator. I would like to show that the Hutchinson estimator has a lower variance than the MC estimator.
The MC estimator of the expectation of the quadratic form looks like follows. $$ E( x^T A x) \approx \frac{1}{N}\sum_{i=1}^N x_i^T A x_i, $$ where $A$ is a symmetric matrix, and $x \sim \mathcal{N}(\mu,B)$ is a normal distributed R.V. $\textit{The Matrix Cookbook}$ gives the variance of stochastic quadratic forms, and the variance of the MC estimator, using one sample of $x$, is \begin{align} \text{Var}\,(x^T A x) & = \text{tr}\,(AB (A+A^T)B) + \mu^T(A+A^T) B (A+A^T)\mu \\\ & = 2 \left(\text{tr}\,((AB)^2) + 2||RA\mu||_2^2\right), \end{align} where $R$ is a Cholesky factor of the symmetric $B$.
Instead of estimating the expectation using the MC method above, one could use the identity $$ E(x^TAx) = \text{tr}\, (AB) + \mu^T A \mu, $$ and use the Hutchinson estimator on the trace instead. The variance of the Hutchinson estimator for the trace of AB is $$ \text{Var}(e^TABe) = 2\sum_ {i\neq j} (AB)_{ij}^2 = 2\left(||AB||_F^2 - \sum_{i=1}^n(AB)_{ii}^2\right) $$ Here, the probing vectors $e$ are Rademacher vectors. To compare these expressions, I rewrite the Frobenius norm as a trace: $$ ||AB||_F^2 = \text{tr}\, (A^2B^2) $$ My question is then: is it true that $$ \text{tr}\, (A^2B^2) -\sum_{i=1}^n(AB)_{ii}^2 - 2||RA\mu||_2^2 \leq \text{tr}\,((AB)^2)? $$