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

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

• I think you should replace $H$ by $H_n$. Otherwise its irritatig. – Dieter Kadelka Sep 22 '20 at 11:55
• Indeed. Updated. – dohmatob Sep 22 '20 at 12:17

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 $$\mathbb{E}\left(\frac{1}{n} \sum_{i=1}^n \lambda_i(H_n) X_i^2\right)= \sum_{i=1}^n \lambda_i(H_n)/n.$$

• Ah, I should've thought of rotational invariance and SVD. Thanks! – dohmatob Sep 22 '20 at 12:48
• So what kind of convergence (and to what) does that condition imply ? (My holy grail would be convergence to a constant...) – dohmatob Sep 22 '20 at 13:03
• 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) $q_n \overset{p}{\longrightarrow} c$ – dohmatob Sep 22 '20 at 16:45