# Can someone check if this proof of the strong law of large numbers is correct?

I think I have a proof of the strong law that is slightly different than some of the others I have seen. I would appreciate another eye to see if it makes sense.

I am assuming the following "Kolmogorov Truncation Lemma" which is pretty standard:

Lemma: Suppose that $X_{1},X_{2},\ldots$ are IID with $E(|X|)<\infty$. Define $Y_{n}=X_{n}1_{\{|{X_{n}|}<n\}}$ then:

i) $E(Y_{n})\to E(X)$

ii) $P\left(Y_{n}=X_{n}\text{ eventually }\right)=1$

iii) $\sum_{n=1}^{\infty}n^{-2}Var(Y_{n}^{2})<\infty$

Here is my proof of the strong law now:

Theorem (SLLN): $X_{1},X_{2},\ldots$ are IID with $E(|X|)<\infty$ then $n^{-1} \sum_{k\leq n}X_k \to E(X)$ a.s. Proof: By handling the positive and negative parts separately, we assume WOLOG that $X\geq0$. Also assume WOLOG that $P\left(X<1\right)\neq0$ (you can just subtract a constant from $X$ to make this so, prove the strong law for the rv with the constant subtracted, then add the constant back in). By ii) from the truncation lemma it suffices to show that $n^{-1}\sum_{k\leq n}Y_{k}\to E(X)$ a.s. where $Y_{k}=X_{k}1_{\left\{ X_{k}<k\right\} }$. Our assumption that $P\left(X<1\right)\neq0$ means that $E(Y_{1})>0$.

We will show actually that $\frac{\sum_{k\leq n}Y_{k}}{E\left(\sum_{k\leq n}Y_{k}\right)}\to1$ and then the result will follow since \begin{eqnarray*} n^{-1}\sum_{k\leq n}Y_{k} & = & \frac{\sum_{k\leq n}Y_{k}}{E\left(\sum_{k\leq n}Y_{k}\right)}\frac{E\left(\sum_{k\leq n}Y_{k}\right)}{n}\\ & = & \frac{\sum_{k\leq n}Y_{k}}{E\left(\sum_{k\leq n}Y_{k}\right)}\cdot\frac{\sum_{k\leq n}E(Y_{k})}{n}\\ & \to & 1\cdot E(X) \end{eqnarray*}

Which holds since $E(Y_{k})\to E(X)$ as $n\to\infty$ by i) from the truncation lemma.

Now consider by Cheb. inequality that for any $\delta>0$:

$$E\left(\left|{\sum_{k\leq n}Y_{k}-E\left(\sum_{k\leq n}Y_{k}\right)}\right|>\delta E\left(\sum_{k\leq n}Y_{k}\right)\right)\leq\frac{1}{\delta^{2}E\left(\sum_{k\leq n}Y_{k}\right)^{2}}Var\left(\sum_{k\leq n}Y_{k}\right)$$

Now since $E(Y_{k})\uparrow E(X)$ is an increasing sequence, we know $E(Y_{k})\geq E(Y_{1})>0$ by our WOLOG earlier. Hence $E\left(\sum_{k\leq n}Y_{k}\right)\geq n E(Y_{1})$ So we will have: \begin{eqnarray*} E\left(\left|{\frac{\sum_{k\leq n}Y_{k}}{E\left(\sum_{k\leq n}Y_{k}\right)}-1}\right|>\delta\right) & \leq & \frac{1}{\delta^{2}E\left(\sum_{k\leq n}Y_{k}\right)^{2}} Var\left(\sum_{k\leq n}Y_{k}\right)\\ & \leq & \frac{Var\left(\sum_{k\leq n}Y_{k}\right)}{\delta^{2}E(Y_{1})^{2}n^{2}} \end{eqnarray*}

This is summable by iii) from the truncation lemma, so by a standard application of Borel Cantelli, we have $\frac{\sum_{k\leq n}Y_{k}}{E\left(\sum_{k\leq n}Y_{k}\right)}\to1$. $\square$

The main trick comes from assuming that $E(Y_1) > 0$ and then using that. Please don't hesitate to ask any questions! Thanks!

• "This is summable by iii) from the truncation lemma"... How so? The series involved in (iii) is different, no? – Did Jul 12 '13 at 18:15
• Wooops! Yes you are definetly right....I had a feeling it was too easy. Thanks for pointing that out – Mihai Nica Jul 12 '13 at 18:26