1
$\begingroup$

Let $x_1, x_2, x_3, ...$ be i.i.d random variables with probability $1/2$ equal to $+1$ or $-1$. We know that rescaled partial sums converge in distribution to normal r.v. $$\frac{1}{\sqrt{n}}\sum_{i=1}^{n} x_i \overset{d}{\to} \mathcal{N}(0,1).$$ Now assume that we are given i.i.d gaussian random variables $a_1, a_2, a_3, ... \sim \mathcal{N}(0, 1)$ (independent from $(x_1, x_2, x_3, ...)$), is it true that for almost all sequences $(a_1, a_2, ...)$ we have $$\frac{1}{\sqrt{n}}\sum_{i=1}^{n} a_ix_i \overset{d}{\to} \mathcal{N}(0,\sigma^{2}) \;?$$ Obviously, there are sequences $a_1, a_2, a_3, ...$ where the convergence in distribution holds (e.g. $a_i = 1$) and where it doesn't hold (e.g. $a_i = 2^{i}$), but is it true that it hold almost surely? Note that I consider case with $\sigma^{2} = 0$ also as normal distribution (given by delta measure $\delta_{0}$).

I tried Lindeberg-Feller CLT but it seems that its conditions do not hold almost surely. We can also consider $a_i$ following some arbitrary distribution $D$ with finite moments.

$\endgroup$
4
  • $\begingroup$ Is the sequence $(a_j)_{j\geqslant 1}$ assumed to be independent of $(x_j)_{j\geqslant 1}$? $\endgroup$
    – Davide Giraudo
    Aug 10, 2022 at 11:47
  • $\begingroup$ @DavideGiraudo yes, all independent. Note that we first sample a sequence $(a_j)_{j \geq 1}$ and then examine whether the distribution of weighted sums of Bernoulli random variables converges to normal. $\endgroup$
    – Rybin Dmitry
    Aug 10, 2022 at 11:49
  • $\begingroup$ @geetha290krm added independence to statement and also clarified in which sense i want 'almost sure' convergence to normal distribution $\endgroup$
    – Rybin Dmitry
    Aug 10, 2022 at 12:20
  • 1
    $\begingroup$ @RybinDmitry I have edited my answer. Almost sure convergence is true. $\endgroup$
    – geetha290krm
    Aug 10, 2022 at 12:27

1 Answer 1

Reset to default
1
$\begingroup$

Consider $S\equiv \{(b_k) \in \mathbb R^{\infty}: \frac 1 n \sum\limits_{k=1}^{n} b_k^{2}\to 1\}$. By the ordinary CLT we get $\frac 1 {\sqrt n} \sum\limits_{k=1}^{n} b_kx_k \to N(0,1)$in distribution for any $(b_k) \in S$. By Strong Law of Large Numbers $(a_i) \in S$ with probability $1$. Hence, the result is true.

$\endgroup$
4
  • $\begingroup$ Sequence $(a_ix_i)$ may or may not satisfy ordinary CLT. For example, $a_i = 2^{i}$ doesn't because its variance grows too fast. If you apply ordinary CLT you find that only a subset of $(a_i)$ of measure zero satisfies it. $\endgroup$
    – Rybin Dmitry
    Aug 10, 2022 at 11:51
  • $\begingroup$ @RybinDmitry It is given that $(a_i)$ is i.i.d $N(0,1)$. Variance of $\sum a_ix_i$ can be computed by conditioning on $(a_i)$ and you get the variance as $n$. Examples like $a_i=2^{i}$ are not applicable here. $\endgroup$
    – geetha290krm
    Aug 10, 2022 at 11:54
  • $\begingroup$ No, i'm not interested in distribution of $\sum a_ix_i$ with respect to $a_i$, it is obviously normal and there is nothing to ask. You first sample coefficients $a_i$ e.g. $(a_i) = (1, 1, 1, ...)$ (elementary event of my probability space) and then i compute whether this event gives partial sums that converge in distribution to normal one. Then I'm asking whether good elementary events have probability 1. $\endgroup$
    – Rybin Dmitry
    Aug 10, 2022 at 12:00
  • $\begingroup$ You are right in the edited answer. I was missing the fact that $(a_i) \in S$ almost surely. Because i was thinking that the volume of a sphere of radius $r$ in $n$ dimensions makes it hard to prove that almost surely $r \leq 1$ when $n \to \infty$. $\endgroup$
    – Rybin Dmitry
    Aug 10, 2022 at 15:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .