Suppose $X_1, X_2, ...$ are independent random variables with $P(X_n=\sqrt n)=1/\sqrt n $ and $P(X_n=0)=1-1/\sqrt n$. Let $S_n=X_1+X_2+\cdots+X_n$ for all $n$. Show that $S_n/n \rightarrow 1 \space a.s.$
I try to apply the Borel-Cantelli lemma, but its not working. I also try to truncate the random variable by define a new random variable $Y_n=X_n\Large 1\normalsize\{ X_n<1\}$. But its still not working.
Can anybody help?

  • $\begingroup$ Can you elaborate on why Borel Cantelli failed? What happens if you try Chebyshev's Inequality? Just to be clear, I'm suggesting you try the form of Borel Cantelli which says that if $\sum_i P(A_i>e_i)<\infty$ for some $e_i\rightarrow 0$, then the probability of $A_i>0$ infinitely often is zero. $\endgroup$ – Alex R. Dec 8 '12 at 4:51
  • $\begingroup$ I try to show $\sum_i P(|S_n-1|>e)<\infty$, but I got this series is diverges. $\endgroup$ – BigMike Dec 8 '12 at 5:03
  • $\begingroup$ Aren't you showing $S_n/n$ converges to 1? So shouldn't that be $S_n/n$ in your sum? $\endgroup$ – Alex R. Dec 8 '12 at 5:06
  • $\begingroup$ I'm sorry it is $S_n/n$ in the sum, it was a typo. $\endgroup$ – BigMike Dec 8 '12 at 5:20
  • $\begingroup$ Can you show your work for divergence? Did you try Chebyshev's Inequality? $\endgroup$ – Alex R. Dec 8 '12 at 5:33

I think it's easier to use the Strong Law of Large Numbers. The following theorem is useful:

Theorem (Kolmogorov's strong law) Let $X_j \in L^2$ independent random variables such that $\sum_{j \geq 1} \frac{\text{var}(X_j)}{j^2}<\infty$. Then $(X_j)$ fulfills the Strong Law of Large Numbers, i.e.

$$\frac{1}{n} \sum_{j=1}^n (X_j-\mathbb{E}X_j)=0 \quad \text{a.s.}$$

(See Sen & Singer (1993, Theorem 2.3.10)).

In this case we have $\text{var}(X_j) = \sqrt{j}-1 \leq \sqrt{j}$, hence $\sum_{j \geq 1} \frac{\text{var} X_j}{j^2}<\infty$. Thus (by Kolmogorov's strong law)

$$ \frac{S_n}{n}- \underbrace{\frac{\mathbb{E}S_n}{n}}_{1} \to 0 \quad \text{a.s.}$$

  • $\begingroup$ Thanks for the advice, but I think I figured out how to prove it. The trick is to use the Kronecker's Lemma. $\endgroup$ – BigMike Dec 9 '12 at 3:54
  • $\begingroup$ To which sequence do you apply Kronecker's Lemma? $\endgroup$ – saz Dec 9 '12 at 8:19
  • $\begingroup$ Define a new random variable $Y_n=X_n-1$ Then this new RV has mean zero, and $\sum_{k=1}^\infty Y_n/n$ converges a.s. to 0. And then, apply the Kroneckers Lemma. $\endgroup$ – BigMike Dec 9 '12 at 8:27
  • $\begingroup$ You could post it as an answer (to your own question) ... it would be interesting to see your way of solving it. $\endgroup$ – saz Dec 9 '12 at 9:14

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