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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 $q_n = \dfrac{1}{n}\sum_{i=1}^n x_i^2 \overset{p}{\longrightarrow}1$.

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    $\begingroup$ I think you should replace $H$ by $H_n$. Otherwise its irritatig. $\endgroup$ – Dieter Kadelka Sep 22 '20 at 11:55
  • $\begingroup$ Indeed. Updated. $\endgroup$ – dohmatob Sep 22 '20 at 12:17
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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 $$ \mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= \sum_{i=1}^n \lambda_i(H_n)/n. $$

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  • $\begingroup$ Ah, I should've thought of rotational invariance and SVD. Thanks! $\endgroup$ – dohmatob Sep 22 '20 at 12:48
  • $\begingroup$ So what kind of convergence (and to what) does that condition imply ? (My holy grail would be convergence to a constant...) $\endgroup$ – dohmatob Sep 22 '20 at 13:03
  • $\begingroup$ Ok, i guess if $\sum_{i=1}^n \lambda_i(H_n)/n^2 \to 0$, then we have $q_n \overset{L^2}{\longrightarrow} \lim_n \mathbb E[q_n] = \lim_n \sum_{i=1}^n\lambda_i(H_n)/n =: c$, and therefore (by Markov's inequality) $q_n \overset{p}{\longrightarrow} c$ $\endgroup$ – dohmatob Sep 22 '20 at 16:45

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