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Question: Show that if $X_k$, $k\in\Bbb N$ are iid standard normal variables, then there exists a constant $C$ such that for sufficiently large $k\ge2$ we have $|X_k|\le C\sqrt{\log k}$ almost surely.

To me this isn't even true at first glance because the law of $X_k$ doesn't depend on $k$ and for any $r\in\Bbb R$ (taking $r$ as an arbitrary bound, in particular it can be equal to $C\sqrt{\log k}$) $\Bbb P(|X|> r)>0$ so $\Bbb P(|X|\le r)<1$ not matter how big we choose $C$ or $k$.

Am I wrong or is the question wrong?

saz
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H. Walter
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1 Answers1

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Since the random variables are identically distributed, we have by Markov's inequality

\begin{align*} \mathbb{P}(|X_n| \geq 6 \sqrt{\log n})=\mathbb{P}(|X_1| \geq 6 \sqrt{\log n}) &= \mathbb{P}\left( \frac{X_1^2}{4} \geq 9 \log n \right) \\ &= \mathbb{P}(e^{X_1^2/4} \geq e^{9 \log n}) \\ &\leq \frac{1}{e^{\log n^9}} \mathbb{E}\exp(X_1^2/4) \\ &= \frac{1}{n^9} \underbrace{\mathbb{E}\exp(X_1^2/4)}_{<\infty}.\end{align*}

(Note that $\mathbb{E}\exp(X_1^2/4)$ is finite because $X_1$ is standard Gaussian.) In particular,

$$\sum_{n \geq 1} \mathbb{P}(|X_n| \geq 6 \sqrt{\log n}) < \infty.$$

By the Borel-Cantelli lemma, this implies that $|X_n| \leq 6 \sqrt{\log n}$ with probability $1$ for $n=n(\omega)$ sufficiently large.

Remark: With some more effort, one can show a stronger result.

saz
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    I think for the OP, one of the main points of your answer is that $n = n(\omega)$ needs to be sufficiently large. If one wants to choose $n$ independently of $\omega$ (i.e., deterministically), then this is impossible precisely by the argument that the OP gives. – PhoemueX Mar 07 '20 at 21:14
  • True. So at what point do we use the fact that $n$ is a random variable? – H. Walter Mar 08 '20 at 09:43
  • Also, why factor 36 as $4\cdot9$ instead of using $36$ the whole time? It should still work (with coefficients ${1\over n^{36}}$ in the sum) – H. Walter Mar 08 '20 at 09:50
  • @H.Walter 1) By the Borel Cantelli lemma, there exists for almost every $\omega \in \Omega$ a number $N \in \mathbb{N}$ such that $|X_n(\omega)| \leq 6 \sqrt{\log n}$ for all $n \geq N$. The number $N$ depends on $\omega$, i.e. we cannot choose it uniformly for all $\omega \in \Omega$. 2) Since you were just asking for some constant (not necessarily the optimal one) I simply picked one which I liked. You can do the same calculation for other constants, i.e. start with $\mathbb{P}(|X_n|>C \sqrt{\log n})$. Note that one would like to choose $C$ as small as possible (because this gives a – saz Mar 08 '20 at 10:51
  • stronger bound on the growth of $X_n$). As you can see from the question, which I linked, the optimal constant is $C=\sqrt{2}$. – saz Mar 08 '20 at 10:51