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After taking a course on SDEs I have started studying time series on my own. However, I am having difficulties in drawing parallelisms between the two subjects.

I have the following definition of $L^p$ convergence in my time series book. For the sake of simplicity take $T_n = \{t\in\mathbb{Z}: \lvert t \rvert \leq n\}$.

Definition: A series $\sum_t X_t$ converges in pth mean if there is a random variable $Y$ such that $Y_n := \sum_{t\in T_n}X_t \rightarrow Y$ in pth mean, i.e. $E[\lvert Y_n - Y\rvert^p] \rightarrow 0$. The limit is then denoted $\sum_t X_t$.

Here comes the part where I am having difficulties.

Lemma: Let $(X_t: t\in\mathbb{Z})$ be an arbitrary countable set of random variables. If $\sum_t E[\lvert X_t\rvert] < \infty$, then the series $\sum_t X_t$ converges absolutely almost surely and in mean. Furthermore, $E[\sum_t X_t] = \sum_tE[X_t]$.

My first question is about the term "converges absolutely almost surely". I am not sure what this means. I know absolute convergence for an ordinary series and almost sure convergence for a sequence of random variables but absolute almost sure convergence is new to me. If I had to guess I would formulate this concept as follows. We say $\sum_t X_t$ converges absolutely almost surely if there exists a random variable $Y$ such that $Y_n := \sum_{t\in T_n}\lvert X_t\rvert \rightarrow Y$ almost surely and $Y < \infty$ almost surely. (I could put these two conditions under one "almost surely" but that won't make a difference.) If that is the case, then the proof of the first claim in the lemma simply follows from the monotone convergence theorem.

$$E\left[\sum_t \lvert X_t\rvert\right] = \sum_t E[\lvert X_t\rvert] < \infty$$ This gives $\sum_t \lvert X_t\rvert < \infty$ almost surely. Can someone correct the mistakes/fill the gaps in my reasoning and/or understanding?

My second question is about showing convergence in mean, i.e. whether $E[\lvert \sum_{t\in T_n} X_t - Z\rvert]\rightarrow 0$ where $Z = \sum_{t\in \mathbb{Z}} X_t$. $$E[\lvert \sum_{t\in T_n} X_t - Z\rvert] = E[\lvert \sum_{t\in Z\setminus T_n} X_t \rvert] \leq E[ \sum_{t\in Z\setminus T_n} \lvert X_t \rvert] = \sum_{t\in Z\setminus T_n} E[\lvert X_t \rvert]$$

The RHS are the right tails of the sum $\sum_{t\in Z} E[\lvert X_t \rvert]$, which is given to be finite. So the tails must converge to $0$. Is this correct?

Calculon
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  • You are correct about absolute a.s. convergence. The answer for the second question (in more general form) was given here. – Budenn Jul 16 '15 at 09:50
  • I don't see anything wrong with the argument you gave for the second question. – Stephen Montgomery-Smith Jul 20 '15 at 03:39
  • However, in the Lemma you cite, shouldn't the equality be an inequality? – Stephen Montgomery-Smith Jul 20 '15 at 03:40
  • @StephenMontgomery-Smith I checked it again. It is an equality and the proof follows from the dominated convergence theorem. – Calculon Jul 20 '15 at 12:54
  • @StephenMontgomery-Smith I have one question, which is probably very silly. For an ordinary series $\sum_i a_i$ when does it make sense to talk about convergence of $\sum_{i=1}^n a_i - \sum_i a_i$? Obviously when the series itself is convergent the difference of these two sums must converge to 0 as well. When the series is divergent, I am not sure if you can still talk about convergence of the difference. It feels like (whatever that means) there might be cases (such as the harmonic series) where the difference still converges to 0. – Calculon Jul 20 '15 at 13:21
  • First, I don't see how it could be equality if any of the $X_i$ are negative with non-zero probability. Maybe there should be some extra absolute values somewhere. – Stephen Montgomery-Smith Jul 20 '15 at 16:46
  • Second, even in the case of the harmonic series, I don't see why the difference converges to 0. – Stephen Montgomery-Smith Jul 20 '15 at 16:46
  • @StephenMontgomery-Smith You are aware that there is no absolute value involved on either side of the equality right? It is just the fact you can interchange the order of expectation and summation under a particular condition. – Calculon Jul 21 '15 at 15:40
  • @StephenMontgomery-Smith Forget about my remark about the harmonic series. It was dumb of me to think such a thing. Sorry. – Calculon Jul 21 '15 at 15:42
  • Sorry. I had misread the square brackets as absolute vales. – Stephen Montgomery-Smith Jul 21 '15 at 16:02

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