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 $\tilde{X_k} = X_k .1_{(|X_k|<n)}$ 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 $iid$: $$E\tilde{S_n} = nE\tilde{X_1} = n E[X_1 .1_{(|X_1|<n)}] = n\sum_{k=2}^{n} (-1)^k k.\frac{C}{k^2\log k} = nC\sum_{k=2}^{n} (-1)^k.\frac{1}{k\log k}$$

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

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

  • $\begingroup$ 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 $\mu$. $\endgroup$ – Ian Mar 6 at 18:25
  • $\begingroup$ 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 $\frac{S_n}{n}\xrightarrow{p} \frac{E\tilde{S_n}}{n}$ $\endgroup$ – manifolded Mar 6 at 18:33
  • $\begingroup$ 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. $\endgroup$ – Ian Mar 6 at 19:29
  • $\begingroup$ 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? $\endgroup$ – manifolded Mar 6 at 19:47
  • $\begingroup$ 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.) $\endgroup$ – Ian Mar 6 at 19:48

You should use this theorem which appears in the chapter before the problem:

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 $\mu_n=E[X_1{\bf 1}_{|X_1|<n}]$. 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?

  • $\begingroup$ Yes, I was applying that theorem. I see, didn't think of alternating series test. Thanks. $\endgroup$ – manifolded Mar 7 at 20:03

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