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.


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

  • $\begingroup$ It's not clear to me why denseness is the same as $\cap E_{a,b}$. I'd argue that $$P\big( \{ S_n/\sqrt{n}: n\in \Bbb N\} \text{ is dense in } \Bbb R \big) = P \bigg( \bigcap_{\substack{a<b \\ a,b \in \Bbb Q}} \bigcup_{n} \frac{S_n}{\sqrt n} \in [a,b] \bigg)$$ so it suffices to prove $P\left(\bigcup_{n} \frac{S_n}{\sqrt n} \in [a,b] \right)=1$. In turn, it suffices to prove $P\left( \frac{S_n}{\sqrt n} \in [a,b] \text{ i.o}\right)=1$ (which you prove) $\endgroup$ – Gabriel Romon Sep 25 at 16:03
  • $\begingroup$ Also, you don't need reverse Fatou. $\endgroup$ – Gabriel Romon Sep 25 at 16:06

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 $ $$B_m := \{S_n/\sqrt{n} \in [r- (1/m),r+(1/m)]~~ i.o.\}$$


Here's a more enlightening proof. Let $\displaystyle Y_n = \frac{S_n}{\sqrt n}$. Essentially we prove that $Y_n$ goes to $\infty$ and $–\infty$ infinitely often, and that $Y_{n+1}-Y_n$ converges to $0$ almost surely. A standard result of real analysis then implies that $Y_n$ is dense in $\mathbb R$ (and this happens with probability $1$).

By the CLT, $Y_n$ converges in distribution to $Z$ where $Z\sim \mathcal N(0,\sigma^2)$.

By the equality of events $\left[\left(\limsup_n Y_n\right) = \infty\right] = \bigcap_{N\geq 1} \bigcup_{k\geq 1} \left(Y_k\geq N \right) $, it suffices to prove that for any $N\geq 1$, $P\left(\bigcup_{k\geq 1} \left(Y_k\geq N \right) \right) = 1$. In turn, it suffices to prove $$\forall N\geq 1, P\left(\limsup_{k\geq 1} \left(Y_k\geq N \right) \right) = 1$$

For each $N$, $\limsup_{k\geq 1} \left(Y_k\geq N \right)$ is a tail event and $$P\left(\limsup_{k\geq 1} \left(Y_k\geq N \right) \right)\geq \limsup_{k\geq 1} P\left(Y_k\geq N \right) = P(Z\geq N)>0$$

where I used this bound and the Portmanteau theorem. Kolmogorov's $0$-$1$ law finishes the proof: $$\limsup Y_n = \infty \text{ a.s}\quad \text{and}\quad \liminf Y_n = -\infty \text{ a.s} $$

Note that $$Y_{n+1}-Y_n = S_n\left(\frac 1{\sqrt{n+1}}-\frac 1{\sqrt{n}}\right) + \frac{X_{n+1}}{\sqrt{n+1}}$$

Since $\displaystyle \frac 1{\sqrt{n+1}}-\frac 1{\sqrt{n}}\sim \frac{1}{n \sqrt n}$, the SLLN implies that $\displaystyle S_n\left(\frac 1{\sqrt{n+1}}-\frac 1{\sqrt{n}}\right)$ converges a.s. to $0$.

$\bullet$ If $X_1$ has a moment of order $2+\delta$ where $\delta>0$, then for any $\epsilon>0$, $$P\left(\frac{|X_{n+1}|}{\sqrt{n+1}} > \epsilon \right) \leq \frac{E(|X_1|^{2+\delta})}{\epsilon^{2+\delta} n^{1+\delta/2}}$$ Hence $\sum_n P\left(\frac{|X_{n+1}|}{\sqrt{n+1}} > \epsilon \right)$ converges and $\displaystyle \frac{X_{n+1}}{\sqrt{n+1}} $ goes to $0$ a.s.

$\bullet$ If $X_1$ only has a moment of order $2$, note that for any $\epsilon >0$, $$\sum_n P\left(\frac{|X_n|}{\sqrt n}>\epsilon\right) = \sum_n P(|X|>\epsilon \sqrt n)= \sum_n (1-F_X(\epsilon \sqrt n))$$ Since the function $x\mapsto 1-F_X(x)$ decreases to $0$, the last series converges if and only if the integral $\int_0^\infty 1-F_X(\epsilon \sqrt t)dt$ converges, which after a change of variable is equivalent to the convergence of $$\frac{2}\epsilon \int_0^\infty uP(|X|>u) du= \frac{4}\epsilon E(X^2)<\infty$$ Hence $\displaystyle \frac{X_{n}}{\sqrt{n}} $ goes to $0$ a.s. (and so does $\displaystyle \frac{X_{n+1}}{\sqrt{n+1}} $).

In both cases, we conclude that $Y_n$ is dense in $\mathbb R$ with probability $1$.


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