This is from Durrett. Let $$X_i$$ be $$iid$$ with $$P\{X_i = (-1)^k k\} = \frac{C}{k^2\log k}, k\geq 2$$ and $$C = [\sum_{k=2}^{\infty} \frac{1}{k^2 \log k}]^{-1}$$. Show that $$E|X_1|=\infty$$ but there is a finite constant $$\mu$$ s.t. $$\frac{S_n}{n}\xrightarrow{p} \mu$$.

My attempt:
Step 1) Show $$nP(|X_1|>n)\rightarrow 0$$

Step 2) This means $$\frac{S_n-E\tilde{S_n}}{n}\xrightarrow{p} 0$$ where $$\tilde{S_n} = \sum_{k=1}^{n}\tilde{X_k}$$ and $$\tilde{X_k} = X_k .1_{(|X_k| which means I have to show $$lim_{n\rightarrow \infty} \frac{E\tilde{S_n}}{n}$$ is a constant.

Step 1)

$$P(|X_{1}|>n) = \sum_{k=n}^{\infty} \frac{C}{k^2 \log k}\leq \int_{n}^{\infty}\frac{C}{x^2 \log x}dx\leq \frac{C}{\log n}\int_{n}^{\infty}\frac{1}{x^2}dx = \frac{C}{n\log n}$$

$$\therefore nP(|X_{1}|>n) = \frac{C}{\log n}\xrightarrow{n\rightarrow \infty} 0$$

Step 2) I am stuck here. Since $$X_i$$ are $$iid$$: $$E\tilde{S_n} = nE\tilde{X_1} = n E[X_1 .1_{(|X_1|

So for $$lim_{n\rightarrow \infty} \frac{E\tilde{S_n}}{n}$$ to be a constant, have to show:

$$lim_{n\rightarrow \infty} \sum_{k=2}^{n} (-1)^k.\frac{1}{k\log k}<\infty$$

• The statement $\frac{S_n}{n} \to \frac{E\tilde{S}_n}{n}$ makes no sense. What I think you mean to say is that $\frac{S_n-E\tilde{S}_n}{n} \to 0$. Then you need to refine that to show that $\frac{E\tilde{S}_n}{n}$ changes slowly enough with $n$ that $\frac{S_n-n\mu}{n} \to 0$ as well, for some $\mu$. – Ian Mar 6 at 18:25
• Yes, I had stated that at the beginning of the sentence $\frac{S_n - E\tilde{S_n}}{n}\xrightarrow{p} 0$ which is the same as $\frac{S_n}{n}\xrightarrow{p} \frac{E\tilde{S_n}}{n}$ – manifolded Mar 6 at 18:33
• The point is that $\frac{E\tilde{S}_n}{n}$ isn't exactly constant. But if its limit exists, then you can expect that it is the same as the limit in probability of $\frac{S_n}{n}$. So once you know what this number is, you can try to prove what you actually want to prove. – Ian Mar 6 at 19:29
• True, $\frac{E\tilde{S_n}}{n} = C\sum_{k=-n}^{n}\frac{(-1)^k}{k\log k}$, so should I take the limit of this sum? – manifolded Mar 6 at 19:47
• I'm not sure that expression is actually correct (even after removing $-1,0$ and $1$ obviously); for instance, does $X$ really have any probability to be equal to $+3$? (Note that if that expression were correct, the sum would actually be zero.) – Ian Mar 6 at 19:48

Theorem 2.2.7. Weak Law of Large Numbers. Let $$X_i$$ be i.i.d. with $$xP(|X_i|>x)\to 0\qquad as\quad x\to\infty.$$ Let $$\mu_n=E[X_1{\bf 1}_{|X_1|. Then $$S_n/n-\mu_n\to 0$$ in probability.
You have already shown $$xP(|X_i|>x)\to 0$$, so you now can conclude $$S_n/n-\mu_n\to 0$$. What is $$\mu_n$$? $$\mu_n=\sum_{k=2}^nk(-1)^k\cdot\frac{C}{k^2\log k}.$$ Note $$\lim_n \mu_n$$ exists by the alternating series test. Can you conclude?