I am confused about a proof involving in what my professor calls the ultimate WLLN. The statement goes as follows:
Let $X_1,X_2,\ldots$ be iid, and $S_n = X_1 + X_2 + \cdots + X_n$, then the following are equivalent:
- $\lim_{t\to\infty} tP(\left\lvert X_1\right\rvert > t) = 0$.
- $\exists \{\mu_n\}_{n\geq 1}\subset \mathbb{R}$ such that $\frac{S_n}{n}-\mu_n \to 0$ in probability.
I understand the direction $(1\implies 2)$, but i don't understand the symmetrization argument the professor used to prove $(2\implies 1)$. It goes something like this:
Let $X_1',X_2',\ldots$ be another iid sequence distributed like $X_1$, and $S_n'=X_1' + \cdots + X_n'$, then it also holds that $\frac{S_n'}{n} - \mu_n \to 0$ in probability. This implies that $\frac{S_n}{n}- \frac{S_n'}{n} \to 0$ in probability. Now, we have that $S_n - S_n' = \sum_{i=1}^{n}(X_i - X_i')$. Let $Z_i = X_i - X_i'$, then $Z_1,Z_2,\ldots$ are iid with $Z_i =^d -Z_i$.
The professor then goes on to show that statement $(1)$ is true for the Z's and that somehow, this implies the property for the X's. I don't understand how showing property (1) for the Z's implies it for the X's. Also, if you know an alternative proof of this, you are more than welcome to share it.
Edit: Corrected the statement. There was a 1/n missing.
Let $m\equiv \operatorname{med}(X_1)$, i.e., the median of $X_1$. Then, for $t\ge |2m|$, $$ \mathsf{P}(|X_1|>t)\le\mathsf{P}(|X_1-m|>t/2)\le 2\mathsf{P}(|Z_1|>t/2). $$
To see the second inequality, let $X$ be such that $$ \min\{\mathsf{P}(X\le 0),\mathsf{P}(X\ge 0)\}\ge \alpha $$ for some $\alpha>0$, and let $Z=X-X'$, where $X'$ is an independent copy of $X$. Then \begin{align} \mathsf{P}(Z>t)&\ge \mathsf{P}(\{X>t\}\cap \{X'\le 0\}) \\ &=\mathsf{P}(X>t)\mathsf{P}(X'\le 0)\ge\alpha\mathsf{P}(X>t), \end{align} and \begin{align} \mathsf{P}(Z<-t)&\ge \mathsf{P}(\{X<-t\}\cap \{X'\ge 0\}) \\ &=\mathsf{P}(X<-t)\mathsf{P}(X'\ge 0)\ge\alpha\mathsf{P}(X<-t). \end{align} That is, $$ \mathsf{P}(|X|>t)\le \alpha^{-1}\mathsf{P}(|Z|>t). $$