I've encountered the following problem which seems to be quite standard but for which I can't find any proper references (asking on mathematics SE didn't bring up any answers so I'm reposting my question here):

Suppose that for $n \in \mathbb{N}$ we have a sum of random variables $S(n)= \sum_{r=1}^{\infty} X_r(n)$, such that $X_r(n)$ and $X_s(n)$ are uncorrelated for $r \neq s$ and $S(n)$ is uniformly bounded in $L^2$. Denote by $S_k(n)$ the partial sums $\sum_{r=1}^k X_r(n)$. If we know that

  1. for each $k$, $$S_k(n) \xrightarrow{d} N(\mu_k,\sigma^2_k), \qquad (n \to \infty),$$ where $\xrightarrow{d}$ is convergence is distribution and $N(\mu_k,\sigma^2_k)$ denotes a Gaussian with mean $\mu_k$ and variance $\sigma^2_k$


  1. the sequences $(\mu_k)$ and $(\sigma^2_k)$ both converge to some limits $\mu$ and $\sigma^2$ (so that in particular $N(\mu_k,\sigma^2_k) \xrightarrow{d} N(\mu,\sigma^2)$,

what are sufficient conditions to conclude that $S(n) \xrightarrow{d} N(\mu,\sigma^2)$ as $n\to \infty$? In other words, with some abuse of notation, i.e. writing $\lim$ for "limit in distribution", I'm looking for conditions which imply that $$\lim_{n \to \infty} \lim_{k \to \infty} S_k(n) = \lim_{k \to \infty} \lim_{n \to \infty} S_k(n).$$ What about the more general situation where the partial sums are replaced by some doubly indexed random variables and the limiting Gaussians with some generic (continuous) random variable?

Thanks for your help!

  • $\begingroup$ I must be missing something. $S_k(n)$ converges with probability 1 to $S(n)$ as $k\to\infty$ (since an infinite sum is by definition the limit of the partial sums). So if $S_k(n)$ converges in distribution to $N(\mu_n,\sigma^2_n)$, then $S(n)$ has distribution $N(\mu_n, \sigma^2_n)$ (since convergence with probability 1 implies convergence in distribution). But then assumption 2 gives you that the distribution of $S_n$ converges to $N(\mu, \sigma^2)$. So what is the issue? $\endgroup$ – James Martin Apr 29 '15 at 18:06
  • $\begingroup$ @JamesMartin I'm very sorry, actually I interchanged the roles of $n$ and $k$ in my first writeup of the question. I have corrected the mistake and also added some clarification. $\endgroup$ – r_faszanatas Apr 30 '15 at 13:27

There is a general result, which is Theorem 4.2 of Billingsley's book Convergence of probability measures, 1968:

Theorem. For each integers $m$ and $n$, let $X_n$, $X_n^{(m)}$ and $X^{(m)}$ be real-valued random variables such that

  • for each integer $m$, $X_n^{(m)}\to X^{(m)}$ in distribution, and
  • for each positive $\varepsilon$, $\lim_{m\to\infty}\limsup_{n\to\infty}\mathbb P\{|X_n-X_n^{(m)}| >\varepsilon \}=0$.

Then there exists a real valued random variable $X$ such that $X_n\to X$ in distribution.

Applying this in your setting, you have to check that the convergence $$\lim_{k \to \infty}\limsup_{n\to \infty}\mathbb P\left\{\left|\sum_{i=k +1}^{ +\infty}X_i(n) \right|\gt \varepsilon\right\} =0 $$ takes place for each positive $\varepsilon$. If $X_i(n)$ and $X_j(n)$ are uncorrelated for each $n$, this would hold if $$\lim_{k \to \infty}\limsup_{n\to \infty} \sum_{i\geqslant k}\mathrm{Var}(X_i(n))=0.$$

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