Convergence of quadratic form $y^T Q y$ where $y$ is a random iid sequence of length $n$ and $Q$ is an $n \times n$ random matrix independent of $y$ For each positive integer, let $Q_n=(q_{i,j})_{i,j \in [n]}$ be a random $n \times n$ psd matrix.  In the limit $n \to \infty$, suppose the eigenvalues of this sequence of matrices are uniformly bounded in the following sense

*

*$\lambda_{\max}(Q_n) = \mathcal O_{\mathbb P}(1)$.

Let $y_1,\ldots,y_n,\ldots$ be a sequence of iid Gaussian or Rademacher random variables (independent of the $Q_n$'s) such that $\mathbb E y_1 = 0$, $\mathbb E[y_1^2] = 1$. Define $r_n := y^\top Q_n y = \sum_{i=1}^n y_i y_j q_{i,j}$, a quadratic form in $(y_1,\ldots,y_n)$ with coefficient matrix $Q_n$.

Question 1. Is it true that $r_n - \mathbb E[r_n] \to 0$ almost-surely (or in any other reasonable sense) ?


Update (positive answer to a slightly different question)
The answer below establishes that the answer to Question 1 is No in general.
It turns out that one can answer a slightly different question in the affirmative, namely we show that

Claim. $n^{-1}y^\top Q_n y - n^{-1}trace(Q_n) \to 0$ a.s,

under the assumption that

*

*$\sup_n \|Q_n\|_{op} = \mathcal O(1)$, this is a deterministic bound, unlike the probabilistic bound in the previous question.

*$\mathbb E[y_1] = 0$, $\mathbb E|y_1|^2 = 1$ and $\mathbb E|y_1|^{4+\eta} < \infty$, for some $\eta>0$. Note that these conditions are satisfied by Gaussian and Rademacher random variables considered in the original setting.

The above claim appears as Lemma 7.6 of this manuscript.
So the argument goes like this ...

(Lemma B.26 of Bai and silverstein 2010). For any $p \ge 2$, it holds that
$$
\mathbb E_y[|y^\top Q_n y - trace(Q_n)|^p] \le 2K_ptrace(Q_n^2)^{p/2}(\nu_4+\nu_{2p}).
$$
where $\nu_k := \mathbb E|y_1|^k$.

Now, since $\|Q_n\|_{op} \le C$ uniformly over $n$, then from the above we get
$$
\mathbb E_y[|n^{-1}y^\top Q_n y - n^{-1}trace(Q_n)|^p] \le 2C^2 K_p\nu_{2p}n^{-p/2}
$$
Thus, by Markov's inequality
$$
\mathbb P(|n^{-1}y^\top Q_n y - n^{-1}trace(Q_n)|^p| > t) \le \frac{2C^2K_p\nu_{2p}}{tn^{-p/2}}.
$$
If $p=2+\eta/2$ (and this is the crucial ingredient), we have $\nu_{2p} < \infty$ by hypothesis and so the RHS is $\mathcal O(n^{-p/2})$ and therefore summable. Thus Borel-Cantelli gives $n^{-1}y^\top Q_n y - n^{-1}trace(Q_n) \to 0$ a.s. $\quad\quad\quad\quad\quad\Box$
 A: $\newcommand{\ep}{\varepsilon}\newcommand{\la}{\lambda}$
We inteprete the abuse of notation "$r_n \to \mathbb E[r_n]$" as "$r_n - \mathbb E[r_n] \to 0$".
One may ask instead whether
\begin{equation*}
    \rho_n:=\frac{r_n}{Er_n}
\end{equation*}
converges to $1$ in probability, say. The answer to this question is: usually, no.
Indeed, let
\begin{equation*}
    Q_n=(2+\ep)I_n,
\end{equation*}
where $\ep$ is a Rademacher random variable independent of the $y_i$'s and $I_n$ is the $n\times n$ identity matrix.
Then $1\le\la_{\min}(Q_n)\le\la_{\max}(Q_n)\le3$,
\begin{equation*}
    Er_n=\sum_{i,j}Eq_{ij}\,Ey_iy_j=\sum_{i}Eq_{ii}=2n,
\end{equation*}
and
\begin{equation*}
    \begin{aligned}
    Er_n^2&=\sum_{i,j,k,l}Eq_{ij}q_{kl}\,Ey_iy_jy_ky_l
    =\sum_{i,k}Eq_{ii}q_{kk}=n^2\,Eq_{11}^2=5n^2. 
\end{aligned}
\tag{1}
\end{equation*}
So, $Var\,r_n=n^2$ and hence
\begin{equation*}
E(\rho_n-1)^2=  Var\,\rho_n=1\not\to0. \tag{2}
\end{equation*}
So, $\rho_n\not\to1$ in $L^2$.
Moreover, similarly to (1) we get $Er_n^4=O(n^4)$ and hence $E\rho_n^4=O(1)$. So, if we had $\rho_n\to1$ in probability, then by the uniform integrability we would get $E(\rho_n-1)^2\to0$, which would contradict (2). Thus, $\rho_n\not\to1$ in probability.
