A question related to the convergence of series in probability

Question

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?

Answer 1

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.