Can someone detail why the following limit is true?

Question

We have some $m \times m$ matrix $\Theta_i$ such that $\underset{i \rightarrow \infty}{\lim} \Theta_i = \bar{\Theta}$ where it is understood that all entries are finite. Furthermore, we have $\Sigma$, a positive semi-definite matrix. We are interested in the limit \begin{align*} \underset{h \rightarrow \infty}{\lim} h^{-1} \sum_{i=0}^{h-1} \Theta_i \Sigma \Theta_i'. \end{align*} This comes from an old paper in econometrics and I am trying to follow the proof of a theorem in that paper. The author says that by Cesaro summation and $\Theta_i \rightarrow \bar{\Theta}$, the value of this limit is $\bar{\Theta} \Sigma \bar{\Theta}'$. Can someone detail the steps implicit in this claim?

Answer 1

A sequnce $\{a_n\}_n \subset \mathbb{R}^n$ is Cesarò convergent if the following limit exists: $$\lim_{N \to \infty} \frac{1}{N} \sum_{i = 1}^Na_i = a \in \mathbb{R}^n$$ It can be proved that if a sequence converges in the standard sense then it is also Cesarò convergent to the same limit. In your case $$\lim_{i \to \infty} \Theta_i \Sigma \Theta_i' = \Theta \Sigma \Theta'$$ implies that $$\lim_{h \to \infty} \frac{1}{h} \sum_{i = 0}^{h-1}\Theta_i \Sigma \Theta_i' = \Theta \Sigma \Theta'$$ I don't know exacty what is the sequence $\{\Theta_i'\}_i$ but I'm assuming that it converges to $\Theta'$ in the usual sense.

Answer 2

Given $$\forall \epsilon>0,\exists n,\forall i>n:|\Theta_i-\bar\Theta|<\epsilon,$$ we want to prove that $$\forall \epsilon>0,\exists n,\forall h>n:\left|\frac 1h\sum_{i=0}^{h-1}\Theta_i\Sigma\Theta_i^\intercal-\bar\Theta\Sigma\bar\Theta^\intercal\right|<\epsilon,$$

First, let's notice that: $$ S=\frac 1h\sum_{i=0}^{h-1}\Theta_i\Sigma\Theta_i^\intercal-\bar\Theta\Sigma\bar\Theta^\intercal = \frac 1h\sum_{i=0}^{h-1}\left(\Theta_i\Sigma\Theta_i^\intercal-\Theta_i\Sigma\bar\Theta^\intercal\right) +\frac 1h\sum_{i=0}^{h-1}\left(\Theta_i\Sigma\bar\Theta^\intercal-\bar\Theta\Sigma\bar\Theta^\intercal\right) = \\ \frac 1h\sum_{i=0}^{h-1}\Theta_i\Sigma\left(\Theta_i-\bar\Theta\right)^\intercal +\frac 1h\sum_{i=0}^{h-1}\left(\Theta_i-\bar\Theta\right)\Sigma\bar\Theta^\intercal $$

Then, for $m\times m$ matrices $A,B,C$ we know that $|ABC|\le m^2|A||B||C|$ (L1 norm). Thus: $$ S \le 2m^2(\max |\Theta_i|)|\Sigma|\frac 1h \sum_{i=0}^{h-1}|\Theta_i-\bar\Theta| $$

What's left is to prove the Cesaro convergence $\frac 1h \sum_{i=0}^{h-1}|\Theta_i-\bar\Theta| \to 0$.

If $\theta_i=|\Theta_i-\bar\Theta|$, then for any $\epsilon>0$ we can choose $n$, so every $\theta_i < \epsilon/2$ for $i>n$, then: $$ s=\frac 1h \sum_{i=0}^{h-1}\theta_i = \frac1h\sum_{i=0}^{n-1}\theta_i + \frac1h\sum_{i=n}^{h-1}\theta_i \le \frac nh\max \{\theta_0,\ldots\theta_{n-1}\} + \frac{h-n}{h}\frac\epsilon2 $$

If we select $h> \frac{2n}\epsilon \max \{\theta_0,\ldots\theta_{n-1}\} $, then $s<\epsilon$.