$X_i = \mathcal{N}(0, \sigma_i^2)$, with $\sigma_1^2 \geq \sigma_2^2 \geq \dots \geq 0$ and $\sum_i \sigma^2_i = 1$

Question

Let $\{X_k\}$ be independent random variables such that $X_i = \mathcal{N}(0, \sigma_i^2)$, with $\{\sigma_i^2\}$ such that $\sigma_1^2 \geq \sigma_2^2 \geq \dots \geq 0$ and $\sum_i \sigma^2_i = 1$.

What is the limit of $\{S_n\}$ in $L^2(\Omega. \mathcal{F}, \mathbb{P})$? Before, show that $S_n$ converges in distribution.

My attempt:

I guess that $S_n$ converges to $\mathcal{N}(0,1)$, because $\sum_{i=1}^{+\infty} \sigma_i^2 = 1$.

I demonstrated that $\{S_n\}$ is Cauchy in $L^2$:

Given $\epsilon > 0$, as $\sum_{i=1}^{+\infty} \sigma_i^2 = 1$ and $\sigma_1^2 \geq \sigma_2^2 \geq \dots$, there is $N \in \mathbb{N}$ such that $\sum_{i=N}^{+\infty} \sigma_i^2 < \epsilon$.

\medskip

Then, for $m \geq n \geq N$ we have that:

\begin{align*} E[|S_m - S_n|^2] &= E[|\sum_{i=n+1}^{m} X_i|^2] \\ &= Var[\sum_{i=n+1}^m X_i] \\ &= \sum_{i=n+1}^m Var[X_i] \\ &= \sum_{i= n+1}^m \sigma_i^2 \\ &\leq \sum_{i = n+1}^{+\infty} \sigma_i^2\\ &< \epsilon \end{align*}

Answer 1

Lemma 1 (Levy's Equivalence Theorem)： Let $X_1,X_2,\cdots,X_n,\cdots$ be independent random variables and $S_n=\sum_{i=1}^{n}{X_i}$. Then $$S_n\overset{a.s.}{\rightarrow}S\Leftrightarrow S_n\overset{\mathbb{P}}{\rightarrow}S\Leftrightarrow S_n\overset{d}{\rightarrow}S \ .$$

Lemma 2 ： Let $\left( \Omega,\mathcal{F},\mu \right)$ be a finite measure space and $1\leq p<\infty$. Suppose $f$ and $\left\{ f_n \right\}$ are functions in $L^p\left( \Omega \right)$ and $f_n \rightarrow f$ almost everywhere or in measure. Then $$f_n\overset{L^p}{\rightarrow}f \Leftrightarrow \left\| f_n \right\|_p \rightarrow \left\| f \right\|_p \ .$$

Now we can easily show that $S_n$ converges to a $\mathcal{N}(0,1)$ random variable $S$ in distribution by using characteristic function. Then by lemma 1, we have that $S_n\overset{a.s.}{\rightarrow}S$.

By lemma 2, if we want to prove that $S_n\overset{L^2}{\rightarrow}S$, we only need to prove that $$\lim_{n\rightarrow\infty}{\mathbb{E}\left( S_n^2 \right)}=\mathbb{E}\left( S^2 \right)=1\ .$$ Obviously, it is trivial.