Normalized partial sums of normal random variables are dense in $\mathbb{R}$

Question

I came across an interesting result appearing as an exercise in some lecture notes I'm reading. Suppose $X_{1},X_{2},...$ are IID $N\left(0,1\right)$ RVs all defined on $\left(\Omega,\mathcal{F},\mathbb{P}\right)$ and let $S_{n}=\sum_{i=1}^{n}X_{i}$ . Then with probability 1 (w.r.t to $\mathbb{P}$ ) the sequence $\frac{S_{n}\left(\omega\right)}{\sqrt{n}}$ is dense in $\mathbb{R}$ . That is, with probability 1 for all $x\in\mathbb{R}$ there is a subsequence $\frac{S_{n_{k}}}{\sqrt{n_{k}}}$ converging pointwise to $x$.

I'm curious about what kind of proof approach would be appropriate here. If some cares to reference or write a proof that would be great.

Answer 1

In fact we don't even have to assume that the $X_i$ are normally distributed. Just assume they have mean zero and variance $1$.

Fix some $a<b \in \Bbb R$. By reverse Fatou's Lemma we see that $$P\big( \frac{S_n}{\sqrt{n}} \in [a,b] \;\;\text{ for infinitely many $n$ } \big) \geq \limsup_{n \to \infty} P\big(\frac{S_n}{\sqrt{n}} \in [a,b] \big) = P\big(Z \in [a,b] \big)>0$$ where $Z$ is normally distributed, and the equality after the limsup follows by the central limit theorem. On the other hand, note that the event $E_{a,b} :=\big\{ \frac{S_n}{\sqrt{n}} \in [a,b] \;\;\text{ for infinitely many $n$ } \big\}$ is an exchangeable event and therefore the Hewitt-Savege 0-1 Law together with the above computation implies that $P(E_{a,b})=1$.

Finally, note that $$P\big( \{ S_n/\sqrt{n}: n\in \Bbb N\} \text{ is dense in } \Bbb R \big) = P \bigg( \bigcap_{\substack{a<b \\ a,b \in \Bbb Q}} E_{a,b} \bigg) = 1$$

Answer 2

Here is my attempt at a more direct proof.

Take any interval $[a,b]\,, b > a$, the event $\{S_n/\sqrt{n} \in [a,b]~~ i.o.\}$ is in the tail probability of $\{\mathcal{F}_n\}$, where $\mathcal{F}_n = \sigma(x_n)$, $x_i$ are independent and therefore by Kolmogorov's 0-1 law, the probability of $\{S_n/\sqrt{n} \in [a,b]~~ i.o.\}$ is either 1 or 0.

Now, by CLT, $S_n/\sqrt{n}$ converges in distribution to the standard normal. That is for any $b>a$. $$ \lim_{n \to \infty}P(S_n/\sqrt{n} \in [a,b]) = P(Z \in [a,b])\,,~ Z \sim N(0,1), $$ therefor for all $n$ large enough the probability $P(S_n/\sqrt{n} \in [a,b])>p>0$,

This implies $P(A_m)>p$ where $$ A_m : = \{ \exists n>m~|~S_n /\sqrt{n} \in [a,b]\}\,. $$ Since $A_m \supset A_{m+1} \supset \cdots $, by continuity of probability measure $P(\cap_m A_m) = \lim_m A_m > p$. Now, the event $\{S_n/\sqrt{n} \in [a,b]~~ i.o.\} \supset \left(\cap_m A_m\right)$ thus $P(\{S_n/\sqrt{n} \in [a,b]~~ i.o.\})= 1$.

From here, given $r \in \mathbb{R}$ the a.s. existence of a convergent sequence follows from continuity of a probability measure on events $B_m \supset B_{m+1} \supset \cdots $ $$B_m := \{S_n/\sqrt{n} \in [r- (1/m),r+(1/m)]~~ i.o.\}$$