I was studying Durrett's proof of Strong Law of Large numbers and I got stuck while trying to understand the following Lemma:

Let $X_1,X_2,\dots$ be pairwise independent identically distributed random variables with $E|X_i|<\infty$. Let $EX_i=\mu$ and $S_n=X_1+\dots+X_n$. Let $Y_k = X_k1_{(|X_k|\leq k)}$ and $T_n = Y_1 + \dots + Y_n$. To show that $S_n/n\to\mu$ a.s., it is sufficient to prove that $T_n/n \to \mu$ a.s. as $n\to\infty$.

Durrett first shows that $P(X_k\neq Y_k \text{ i.o.})=0$. This means that for almost every $\omega$, $|S_n(\omega)-T_n(\omega)|\leq R(\omega)<\infty$ for all $n$. Durrett concludes by saying that from here the desired result follows.

My attempt to complete the proof is as follows:

To show that $S_n/n \to \mu$ a.s. it is enough to show that for any $\epsilon>0$, $P(|S_n/n - \mu|>\epsilon \text{ i.o.})=0$. By the Borel Cantelli Lemma, the above statement is true when $\sum_{n=1}^{\infty}P(|S_n/n - \mu|>\epsilon)<\infty$. By using a union bound, $\sum_{n=1}^{\infty}P(|S_n/n - \mu|>\epsilon)\leq \sum_{n=1}^{\infty} P(|T_n/n - \mu|>\epsilon/2)+\sum_{n=1}^{\infty} P(|S_n-T_n|/n>\epsilon/2)$. The expression $P(|T_n/n - \mu|>\epsilon/2)$ is finite by our assumption.

I can't figure out how to show that the other expression is finite. I know that $P(|S_n-T_n|/n>\epsilon/2) = P(|R|/n>\epsilon/2)$ but I don't know what to do from here. How do I show $\sum_{n=1}^{\infty}P(|R|/n>\epsilon/2)<\infty$?


1 Answer 1


Since $X_n = Y_n$ for $n \geq N$, $$\frac{S_n - T_n}{n} = \frac{S_N - T_N}{n} \to 0 \text{ as }n \to \infty.$$


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