# Laws of large numbers: inequalities and convergence

Let $X_1,X_2,...$ be a sequence of independent random variables with $P(X_n = 3^n) = P(X_n = -3^n) = \frac{1}{2}$. Let $S_n = X_1 + \cdots + X_n$.

Compute $E(X_n)$ for each $n$.

I've figured out this part with some assistance and I came to the fact that $E(X_n) = \frac{1}{2}(-a) + \frac{1}{2}(a) = 0$

• For $n \in N$, compute $R_n \equiv \sup\{r \in R; P(|S_n| \ge r) = 1\}$, i.e. the largest number such that $|S_n|$ is always the least $R_n$.

What would be the first step for this part? I had found this post on here that seems relevant and I'm currently trying to figure out how: Probability that, given a set of uniform random variables, the difference between the two smallest values is greater than a certain value

I'm also thinking about the possibility of applying chebychev's inequality to this but I'm still trying to think about how that would work.

Any help is appreciated to help point me in the right direction.

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Note that $|S_n|\geqslant|X_n|-\sum\limits_{k=1}^{n-1}|X_k|=3^n-\sum\limits_{k=1}^{n-1}3^k=x_n$, say. On the other hand, $[|S_n|=x_n]$ happens if every $X_k$ with $k\leqslant n-1$ has the same sign and if $X_n$ has the opposite sign, and this has positive probability. Hence $R_n=x_n$. Computing the geometric sum yields $x_n=\frac12(3^n+3)$, hence $$R_n=\tfrac12(3^n+3).$$ Likewise, $\bar R_n=\min\{r\in\mathbb R\mid\mathbb P(|S_n|\leqslant r)=1\}$ is $\bar R_n=\tfrac12(3^{n+1}-3)$.
$S_n$ can take the values $\sum_{j=1}^n\varepsilon(j)3^j$, where $\varepsilon(j)\in\{-1,1\}$. So, denoting $A:=\sum_{j=1}^n3^j$, if $R>A$ then $P(|S_n|\geqslant R)=1$, and we deduce that $0\leqslant R_n\leqslant A$. If $R<A$, the event $\bigcap_{j=1}^n\{X_j=3^j\}$ has a non-zero probability, so $P(|S_n|\geqslant R)\neq 1$. We deduce that $$R_n=\sum_{j=1}^n3^j=3\frac{3^n-1}2.$$