When does a gaussian quadratic form converge (in probability) to a constant?

Question

Let $(h_{ij})_{i,j \in \mathbb N}$ be a sequence of real numbers (deterministic) and let $x_1,\ldots,x_n,\ldots$ be a sequence of iid $N(0,1)$ randm variables. For each positive integer $n$, consider the quadratic form $q_n:=\dfrac{1}{n}\sum_{i=1}^n\sum_{j=1}^nh_{i,j}x_ix_j$.

Question. Under what conditions on the sequence $(h_{ij})$ does there exist $c \ge 0$ sucht aht $q_n \to c$ in probability ? Is there some other kind of convergence that might hold here ?

Note. In the special case $h_{ij} = \delta_{ij}$, we have $q_n = \dfrac{1}{n}\sum_{i=1}^n x_i^2 \overset{p}{\longrightarrow}1$.

Answer 1

First, notice that w.l.o.g. you can assume that the matrix $H_n$ is diagonal (from rotational invariance of the isotropic Gaussian).

You thus are interested in $$ \frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2. $$

So the condition $$\sum_{i=1}^n \lambda_i^2(H_n)/n^2 \to 0$$

is the key.

Then, one has that $$ Var\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= 2\sum_{i=1}^n \lambda_i^2(H_n)/n^2 \to 0. $$ and $$ \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= \sum_{i=1}^n \lambda_i(H_n)/n. $$