Let $(X_n)_{n\gt 2}$ be independent, $P(X_n=n)=\dfrac{1}{n\log n}, P(X_n=0)=1-\dfrac{1}{n\log n}$. I want to show that this sequence obeys the WLLN, but not the SLLN.

I am trying to prove the WLLN part using Theorem 5.2.3 from here. In the book, the author does the same for a very similar sequence a few pages later. But in my case, I can't even verify the first condition since $\sum_{k\gt n}^\infty \frac{1}{k\log k}$ diverges. I think I might be getting something wrong here, so I appreciate any advice to guide me in the right direction.

Edit: Now that I wrote it down, I might have found the correct reasoning: $\sum_{k\gt n}^\infty \frac{1}{k\log k}\le \sum_{k\gt n}^\infty \frac{1}{n\log n}\xrightarrow{\scriptscriptstyle n\to\infty}0$ thus condition (i) holds. Is this correct?

  • $\begingroup$ Your reasoning is wrong, since as you say, $\sum_k \frac{1}{k \log k}$ diverges. $\endgroup$ – Robert Israel Dec 9 '14 at 16:24
  • $\begingroup$ Is this statement fixable though? $\endgroup$ – blst Dec 9 '14 at 16:31

With $S_n = \sum_{j>2} X_j$, we have $a_n = E[S_n] = \sum_{j>2} 1/\log j$, and WLLN would say $(S_n - a_n)/n \to 0$ in probability. Conveniently, we do have condition (i) with $b_N = N$, since $\int_{|x| > n} dF_j(x) = P(|X_j| > n) = 0$ for $j \le n$.

Next you have to verify (ii). That's not too hard.


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