Finding the smallest possible value of $|S_n|$ for Sequence and Series in real analysis [closed]

Question

Let $𝑋_1,𝑋_2, … , 𝑋_𝑛$ be a sequence of independent random variables with $𝑃(𝑋_𝑛 = 4^𝑛) = 𝑃(𝑋_𝑛 = −4^𝑛) = \frac12$. Let $𝑆_𝑛 = 𝑋_1 + 𝑋_2 + ⋯ + 𝑋_𝑛$. If $A_n=\sup\, \{𝑟 ∈ \Bbb R: 𝑃(|𝑆_𝑛| ≥ 𝑟) = 1 \}$ for $𝑛 ∈ \Bbb N$, show that $A_𝑛 = \dfrac23(4^𝑛 + 2)$?

for both extreme cases,

$S_𝑛=(4/3)(4^𝑛-1)$ and $S_𝑛=(-4/3)(4^𝑛-1)$

How to proceed with the proof after this?

Answer 1

Let $$a_n:=\dfrac23(4^n + 2).$$ By the triangle inequality, with probability $1$ we have $|S_{n-1}|\le\sum_{k=1}^{n-1}|X_k|=\sum_{k=1}^{n-1}4^k=4^n-a_n$; that is, $P(|S_{n-1}|\le4^n-a_n)=1$. The equality $X_n=S_n-S_{n-1}$ and the triangle inequality imply that on the event $\{|S_n|<a_n,|S_{n-1}|\le4^n-a_n\}$ we have $$|X_n|\le|S_n|+|S_{n-1}|<a_n+(4^n-a_n)=4^n$$ and hence $|X_n|<4^n$. So, on the event $\{|X_n|=4^n,|S_{n-1}|\le4^n-a_n\}$ we have $|S_n|\ge a_n$. So, $$P(|S_n|\ge a_n)\ge P(|X_n|=4^n,|S_{n-1}|\le4^n-a_n)=1,$$ because $P(|X_n|=4^n)=1$ and $P(|S_{n-1}|\le4^n-a_n)=1$. Hence, $$P(|S_n|\ge a_n)=1. \tag{1}\label{1}$$ On the other hand, $$P(|S_n|=a_n)\ge P(X_n=4^n,S_{n-1}=-(4^n-a_n)) \\ =P(X_n=4^n,X_{n-1}=-4^{n-1},\dots,X_1=-4^1)=2^{-n}>0$$ and hence for all $r>a_n$ we have $$P(|S_n|\ge r)\le P(|S_n|>a_n)\le1-2^{-n}<1. \tag{2}\label{2}$$ So, $$A_n=a_n.\quad\Box$$

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Details on the conclusion $A_n=a_n$, in response to a comment by the OP: We have $$A_n=\sup E,\quad\text{where}\quad E:=\{r\in\Bbb R\colon P(|S_n|\ge r)=1\}. $$ By \eqref{2}, for any $r>a_n$ we have $r\notin E$; that is, $E\subseteq(-\infty,a_n]$. Also, by \eqref{1}, $a_n\in E$. We conclude that $\sup E=\max E=a_n$; that is, $A_n=a_n$.