3
$\begingroup$

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:

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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?

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.