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


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?

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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

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 $\lim_{n\to \infty}\sum_{k=1}^nX_k(\omega)=-\infty$.

| cite | improve this answer | |
  • $\begingroup$ 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

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.