Let $\{x_{1,n}\}_{n\in\mathbb{N}},...,\{x_{k,n}\}_{n\in\mathbb{N}}$ be random sequences of zero mean random variables satisfying $$x_{1,n}\overset{d}{\to} N(0,\sigma^2_1),\cdots, x_{k,n}\overset{d}{\to} N(0,\sigma^2_k)$$ as $n\to\infty$, for some finite number $k>0$.

I know that the sum $x_{1,n}+\cdots+x_{k,n}$ does not necessarily converge in distribution to the sum of the limiting gaussian distributions above.

However, I would like to know if one can show that there exists a random variable $X$ (not necessarily Gaussian) with distribution law $F_X$ such that $$\lim_{n\to\infty}F_n(x)=F_X(x)$$ for all continuity points of $F_X$ and where $F_n$ denotes the distribution law of $x_{1,n}+\cdots+x_{k,n}$. In other words, "if each sequence converges in distribution, then their sum converges to some distribution?"

I always worked with cases where the limiting distribution is known up to its parameters. Does this question make sense?

Thanks in advance!


Here I'm not assuming any kind of independence and the joint distribution $F_n$ is unknown. The special cases where the sequences are independent or where the random variables are jointly gaussian with a given dependence structure are clear to me.

  • $\begingroup$ If you want any convergence of the sum then you may want some rescaling so that some measure of the scale of $F_X(x)$ does not depend on $n$ $\endgroup$
    – Henry
    Jun 16, 2020 at 7:46
  • $\begingroup$ @Henry The fact that I'm summing a fixed number of random variables ($k$), changes your comment? $\endgroup$ Jun 16, 2020 at 8:21
  • $\begingroup$ Celine: I suspect it does. If so, what example were you think of when you say "the sum $x_{1,n}+\cdots+x_{k,n}$ does not necessarily converge in distribution to the sum of the limiting gaussian distributions"? $\endgroup$
    – Henry
    Jun 16, 2020 at 8:55
  • $\begingroup$ @Henry Take $k=2$ and denote $x_{1,n}:=X_n$ and $x_{2,n}:=Y_n$ for all $n$. Assume that $X$ follows a $N(0,1)$ distribution, and $Y=X$ if $|X| <1$ and $-X$ if $|X| \geq 1$. Take $X_n=X$ for all $n$ and $Y_n=Y$ for all $n$. $\endgroup$ Jun 16, 2020 at 15:42

2 Answers 2


The answer is in the negative, as shown with a simple counterexample for $k=2:$ let $Z$ have a Normal distribution with variance $1/4,$ set $x_{1,n}=Z$ for all $n\ge 0,$ and let $x_{2,n}=(-1)^nZ.$

All the assumptions in the question (trivially) hold but the sequence of distribution laws $F_n$ cannot possibly converge, as it alternates between the distribution of the standard Normal variable $2Z$ and an atom at $0.$ Indeed, for any real number $x$ observe that

$$(F_n(x))_{n=0,1,2,\ldots}\ =\ \Phi(x), H(x), \Phi(x), H(x), \Phi(x), \ldots$$

where $\Phi$ is the standard normal CDF and $H(x)$ is the indicator of $x\ge 0,$ equal to $1$ for non-negative $x$ and $0$ otherwise. Since $\Phi(x)\ne 0$ and $\Phi(x)\ne 1,$ this sequence alternates forever between two different values, showing

there exists no real number $x$ for which the sequence $n\to F_n(x)$ has a limit.

BTW, the sequence of random variables $(x_{1,n},\ x_{2,n})$ is jointly Gaussian with an easily determined dependence structure.


Assuming independence of the sequences, one can prove the existence of such a random variable X. The characteristic function of the sum of random variables is $E[\exp(\theta(X_{1,n}+\dots+X_{k,n})]$. Using the independence property, this is equal to the product of expectations $E[\exp(\theta(X_{1,n})]\cdot E[\exp(\theta(X_{2,n})]\dots E[\exp(\theta(X_{k,n})]$. By convergence in distribution, each of these characteristic functions is known to converge and hence the characteristic function of the sum also converges, which in turn implies convergence in distribution for the sum of random variables. This follows by Levy's continuity theorem.

  • 2
    $\begingroup$ Doesn't your initial assumption of independence ruin the question? The interest lies in sequences that are not independent. $\endgroup$
    – whuber
    Jun 16, 2020 at 14:12
  • $\begingroup$ Why do you say it ruins the question? OP's curiosity was around the question of when one doesn't know the limiting distribution, can one still prove the existence of such a limiting distribution. $\endgroup$
    – dshirodkar
    Jun 16, 2020 at 14:54
  • $\begingroup$ On the contrary, the OP states the limiting distribution of the sum does not (necessarily) equal the sum of the limiting distributions. This is not an expression of ignorance: it's a condition that you appear implicitly to reject, which makes your answer only narrowly applicable. $\endgroup$
    – whuber
    Jun 16, 2020 at 15:14
  • 1
    $\begingroup$ @dshirodkar thanks for your answer! Unfortunately, you argued about a specific case. It is not sufficient. $\endgroup$ Jun 16, 2020 at 15:58

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