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The next exercise is taken from Alan Karr:

If $X_1, X_2,...$ are independent with:

$$P(X_k=k^2)=\frac{1}{k^2}$$ $$P(X_k=-1)=1-\frac{1}{k^2}$$

Prove that $\sum_{k=1}^{n} X_k \xrightarrow{a.s.} -\infty$

We have that $E(X_k)= k^2 \frac{1}{k^2} -1 \left( 1-\frac{1}{k^2} \right) = \frac{1}{k^2}$

EDIT:

And therefore $E(\sum_{k=1}^n X_k)=\sum_{k=1}^{n} \frac{1}{k^2}$ which is CONVERGENT

The next part is however not very clear. I tried Borel-Cantelli, however I need convergence to minus infinity... Any ideas?

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  • $\begingroup$ Actually $1/k^2$ is summable. The Kolmogorov 3-series theorem tells you this converges to a finite value. Uh oh. Maybe you have the wrong statement? $\endgroup$ – Jeff Dec 31 '13 at 16:05
  • $\begingroup$ Maybe... your comment helps though, since I've never heard that Theorem before. Although I think it doesn't help in solving this exercise $\endgroup$ – Salieri Dec 31 '13 at 16:14
  • $\begingroup$ Since $1/k^2$ is summable, the Borel-Cantelli applies and solves your problem. Try again! :) $\endgroup$ – user940 Dec 31 '13 at 16:18
  • $\begingroup$ Well, I am suggesting that your result is false, so I guess it's good that the 3-series theorem doesn't help to prove it $\endgroup$ – Jeff Dec 31 '13 at 16:18
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    $\begingroup$ Use Borel Cantelli on $\Pr(X_k \ne -1) = \frac1{k^2}$, and then use Borel-Cantelli, quoting that the sum of these probabilities is finite. The other part given in the question is a red herring. Jeff is making an elementary error, and when he realizes it, he will kick himself. $\endgroup$ – Stephen Montgomery-Smith Dec 31 '13 at 16:51
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Define $A_k:=\{X_k=k^2\}$. We have $\sum_k\mathbb P(A_k)\lt \infty$ because the series $\sum_k\frac 1{k^2}$ is convergent. Therefore, using Borel-Cantelli's lemma, we have $\Omega'\subset\Omega$ of probability $1$ such that for each $\omega\in\Omega'$, there is $N=N(\omega)$ for which $\omega\notin A_k$ if $k\geqslant N(\omega)$. Since $X_k$ takes only two values, we have $X_k(\omega)=-1$ for these $k$'s. Therefore, for $n\geqslant N(\omega)$, we have $$\sum_{k=1}^nX_k(\omega)\leqslant -(n-N(\omega))+N(\omega)^3,$$ hence $\lim_{n\to \infty}\sum_{k=1}^nX_k(\omega)=-\infty$.

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  • $\begingroup$ I don't get your answer. What is $\Omega' \subset \Omega$ ? Is this your Tail-Sigma Algebra? The other thing that's confusing me; you talk about an $\Omega$ with probability 1. The Borel Cantelli Lemma states that if the sum is convergent then $P({lim sup}_{n \to \infty} A_n) = 0$ why 1 then? Thanks! $\endgroup$ – Salieri Jan 1 '14 at 16:42
  • $\begingroup$ $\Omega'$ is the complement of $\limsup_k A_k$. It is in particular in the tail $\sigma$-algebra. $\endgroup$ – Davide Giraudo Jan 1 '14 at 16:46
  • $\begingroup$ ok... What is the $j$ on your sum? And then, if we have convergence, doesn't that mean that LimSup = LimInf? $\endgroup$ – Salieri Jan 1 '14 at 17:16
  • $\begingroup$ The $j$ should be a $k$, it's corrected now. Why would the $\limsup$ and $\liminf$ be equal? $\endgroup$ – Davide Giraudo Jan 1 '14 at 17:40

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