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Let $(X_j)_{j\in \mathbb Z}$ be an strictly stationary sequence of random variables with: $$E[X_j]=0, \quad E[X_j^2] < \infty$$ I want to show that for each positive $\varepsilon$: $$ \sum_{n=1}^\infty \mathbb P\left(\max_{1\leq j\leq 2^n}\lvert X_j\rvert>2^{n/2}\varepsilon\right)<\infty. $$ Note that: $$ \mathbb P\left(\max_{1\leq j\leq 2^n}\lvert X_j\rvert>2^{n/2}\varepsilon\right)= \mathbb P\left(\bigcup_{j=1}^{2^n} \left[ |X_j| >2^{n/2}\varepsilon\right]\right)\leq \sum_{j=1}^{2^n} \mathbb P\left( |X_j| >2^{n/2}\varepsilon\right) $$ By stationarity, we have $$\sum_{j=1}^{2^n} \mathbb P\left( |X_j| >2^{n/2}\varepsilon\right)= 2^n \mathbb P\left( |X_1|>2^{n/2}\varepsilon\right)= 2^n \mathbb P\left( |X_1|^2>2^{n}\varepsilon^2\right)$$ My initial idea is to use a comparison test (I accept another solution). However, I am unable to finish.

Could you help me finish?

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    $\begingroup$ If $E[X_j]=0$, then most people would assume that $E[X_j]^2=0$ also, and so the second inequality would be unnecessary.. Or did you mean to write $E[X_j^2] < \infty$ (which makes a lot more sense) instead of $E[X_j]^2 < \infty $? $\endgroup$ Commented Jun 17, 2023 at 19:10
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    $\begingroup$ Can you add the source of this result? $\endgroup$
    – Zhanxiong
    Commented Jun 18, 2023 at 3:39
  • $\begingroup$ Please, see the comment of the answer of this question. math.stackexchange.com/questions/4719818/… I spent a few hours thinking about the suggestion the respondent gave me, but I couldn't. I'm afraid it's something very trivial that I'm missing. Since I don't like to take too much time from one respondent, I decided to post this question here. $\endgroup$ Commented Jun 18, 2023 at 3:49

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To finish the proof, use the equality $E[X_1^2] = \int_0^\infty P[X_1^2 > t]dt$, whence \begin{align*} & E[X_1^2] = \int_0^\infty P[X_1^2 > t]dt \\ \geq & \sum_{n = 1}^\infty \int_{2^{n - 1}\varepsilon^2}^{2^n\varepsilon^2}P[X_1^2 > t]dt \\ \geq & \sum_{n = 1}^\infty 2^{n - 1}\varepsilon^2P[X_1^2 > 2^n\varepsilon^2]. \end{align*} Linking the summand in the last expression above to the last probability in your post and using the condition $E[X_1^2] < \infty$ finishes the proof.

This answer applied the similar trick.

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  • $\begingroup$ Dear, I had already tried to do this: $$2^n \mathbb P\left( |X_1|^2>2^{n}\varepsilon^2\right)= \frac{\mathbb E [|X_1|^2] }{\varepsilon^2}$$ However, the series $\sum_{n=1}^\infty \frac{\mathbb E [|X_1|^2] }{\varepsilon^2}$ Does not converge. $\endgroup$ Commented Jun 17, 2023 at 20:57

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