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Assume $U_k$ are correlated standard normal random variables.

Let $R_k := a_k U_k^2$, with $a_k > 0$ and $\sum_{k=1}^{\infty} a_k < \infty$.

How can we prove that $S_p:= \frac{1}{p}\sum_{k=1}^{p}R_k$ converge to 0 in probability?

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  • $\begingroup$ Is this a homework problem or textbook question? If so, please tag as self-study. $\endgroup$ – Arya McCarthy Apr 11 at 0:00
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The first thing to try is always Chebyshev's inequality -- especially with well-behaved distributions where moments are likely to be enough to decide the question.

$$\mathrm{var}[S_p]=\frac{1}{p^2}\sum_{i,j=1}^p \mathrm{cov}[a_iU_i^2,a_jU_j^2]$$

Now, correlations are bounded by 1,so: $$\left|\mathrm{cov}[a_iU_i^2,a_jU_j^2]\right|\leq \sqrt{\mathrm{var}[a_iU_i^2]\mathrm{var}[a_jU_j^2]}=|a_ia_j|C$$ for some finite $C$ that we could work out if we cared.

So, $$\mathrm{var}[S_p]\leq C\sum_{i,j=1}^p|a_ia_j|$$ and this bound is sharp; it is achieved when the $U_i$ are perfectly correlated. We need only ask if $$\lim_{p\to\infty}\frac{1}{p^2}\sum_{i,j=1}^p|a_ia_j|= 0$$

There can be only finitely many $a_i>1$; we can delete an initial segment of the series if necessary and assume there are none, so $a_ia_j<a_i$. Then $$\frac{1}{p^2}\sum_{i,j=1}^p|a_ia_j|< \frac{1}{p^2}\sum_{i=1}^pp|a_i|=\frac{1}{p}\sum_{i=1}^p|a_i|$$

Since $\sum_{i=1}^p|a_i|$ is bounded, $$\frac{1}{p}\sum_{i=1}^p|a_i|$$ converges to zero.

So, $S_p$ converges to zero in mean square and thus in probability.

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