Necessary and sufficient conditions for $\sum_{n = 1}^\infty \mathbb{E}\vert X_n \vert < \infty$ with $X_n \sim \mathcal{N}(\mu_n, \sigma_n^2)$

Question

As the title says, I want to know if there is a simple necessity and sufficient condition for $$ \sum_{n = 1}^\infty \mathbb{E}\vert X_n \vert < \infty $$ where $X_n \sim \mathcal{N}(\mu_n, \sigma_n^2)$.[ It comes from the fact that if this is true then: $$ \mathbb{E}\left(\sum_{n = 1}^\infty X_n\right) = \sum_{n = 1}^\infty \mathbb{E}(X_n) $$ ]. I am able to show that if $\sum_{n = 1}^\infty \sqrt{\mu_n^2 + \sigma_n^2} < \infty$, then $\sum_{n = 1}^\infty \mathbb{E}\vert X_n \vert < \infty$. Indeed, since $$ \mathbb{E}\vert X_n \vert \le \sqrt{\mathbb{E}(X_n^2)} = \sqrt{\mu_n^2 + \sigma_n^2} $$ Another approach is I try to estimate $\mathbb{E}\vert X_n \vert$ by the identity: $$ \mathbb{E}\vert X_n \vert = \int_0^\infty \mathbb{P}(\vert X_n \vert > t)dt $$ But I couldn't come up with a good idea. Any references or hints are appreciated. Thank you

Answer 1

Suppose $\sum_n E|X_n| <\infty$. Let $Y_n=\frac 1 {\sigma_n}(X_n-\mu_n)$. Then, $\sum_n E|\mu_n+ \sigma_nY_n|<\infty$. Also, $\sum_n |\mu_n|\le \sum_n E|X_n| <\infty$. Hence, $\sum_n \sigma_n E|Y_n|<\infty$. But $Y_n $ is standard normal variable so $\sum_n \sigma_nE|Y|<\infty$ where $Y \sim N(0,1)$. We have proved that $\sum_n |\mu_n|<\infty$ and $\sum_n \sigma_n<\infty$.

Conversely, these two conditions imply that $\sum E|X_n|<\infty$ because $\sum_n E|X_n|=\sum_nE|\mu_n+\sigma_nY_n|\leq \sum_n |\mu_n|+\sum_m \sigma_n E|Y|$.