Application of Three Series Theorem on Strong Convergence

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Let $\phi$ be a positive, even and continuous function on $(-\infty,\infty)$ such that as $|x|$ increases, $$\frac{\phi(x)}{|x|}\uparrow,\;\frac{\phi(x)}{x^2}\downarrow.$$ Let $\{X_n\}$ be a sequence of independent random variables with d.f.'s $F_n$ and $\mathscr{E}(X_n)=0$ and $0<a_n\uparrow\infty$.  If, additionally, $\phi$ satisfies $$\sum_n\frac{\mathscr{E}\left(\phi(X_n)\right)}{\phi(a_n)}<\infty,$$ then $$\sum_n\frac{X_n}{a_n}\mbox{ converges  a.e.}$$



$\bullet$ Proof.
Define $$Y_n=\begin{cases}X_n, &\mbox{if }X_n\leq a_n,\\ 0,&\mbox{if }X_n>a_n.\end{cases}$$  This definition seems to violate the requirement of Three Series Theorem.  But, in fact, it fits if we check the convergence of the three series with $Y_n/a_n$ since $|Y_n/a_n|\leq 1$.  In another view, it might be clear if we define $Y_n/a_n=X_n/a_n$ if $|X_n/a_n|\leq1$ and $Y_n/a_n=0$ otherwise.  This variable do truncate $X_n/a_n$ by a fixed value $1$ and is reasonable to use Three Series Theorem.

Write $$\sum_n\mathscr{E}\left(\frac{Y_n^2}{a_n^2}\right)=\sum_n\int_{|x|\leq a_n}\frac{x^2}{a_n^2}\,dF_n(x).$$  By the second monotone property of $\phi$, we have $$\frac{\phi(a_n)}{a_n^2}\leq\frac{\phi(x)}{x^2}\implies\frac{x^2}{a_n^2}\leq\frac{\phi(x)}{\phi(a_n)}\mbox{ for }|x|\leq a_n.$$  It follows that $$\begin{array}{rl}\sum_n\sigma^2\left(\frac{Y_n}{a_n}\right) & \leq\sum_n\mathscr{E}\left(\frac{Y_n^2}{a_n^2}\right) \\ & \leq\sum_n\int_{|x|\leq a_n}\frac{\phi(x)}{\phi(a_n)}\,dF_n(x) \\ & \leq\sum_n\int\frac{\phi(x)}{\phi(a_n)}\,dF_n(x) \\ & =\sum_n\mathscr{E}\left(\frac{\phi(x)}{\phi(a_n)}\right)<\infty. \end{array}$$ The third series of Three Series Theorem converges.

$$\begin{array}{rl}\sum_n\frac{|\mathscr{E}(Y_n)|}{a_n}
&=\sum_n\frac{1}{a_n}\left|\int_{|x|\leq a_n}x\,dF_n(x)\right| \\
&=\sum_n\frac{1}{a_n}\left|\int_{|x|> a_n}x\,dF_n(x)\right| \quad(\because \mathscr{E}(X_n)=0)\\
&\leq\sum_n\int_{|x|> a_n}\frac{|x|}{a_n}\,dF_n(x). \end{array}$$ By the first monotone property of $\phi$, we have $$\frac{\phi(a_n)}{a_n}\leq\frac{\phi(x)}{|x|}\implies\frac{|x|}{a_n}\leq\frac{\phi(x)}{\phi(a_n)}\mbox{ for }|x|> a_n.$$ It follows that $$\sum_n\frac{|\mathscr{E}(Y_n)|}{a_n}\leq\sum_n\int_{|x|> a_n}\frac{\phi(x)}{\phi(a_n)}\,dF_n(x)\leq\sum_n\mathscr{E}\left(\frac{\phi(x)}{\phi(a_n)}\right)<\infty.$$ The second series of Three Series Theorem converges.

$$\begin{array}{rl}\sum_n\mathscr{P}\left\{\frac{X_n}{a_n}\neq\frac{Y_n}{a_n}\right\}&=\sum_n\mathscr{P}\{X_n\neq Y_n\} \\ &=\sum_n\int_{|x|>a_n}\,dF_n(x) \\ &\leq \sum_n\int_{|x|>a_n}\frac{\phi(x)}{\phi(a_n)}\,dF_n(x) \quad(\because\phi>0)\\ &\leq\sum_n\mathscr{E}\left(\frac{\phi(x)}{\phi(a_n)}\right)<\infty.\end{array}$$  The first series of Three Series Theorem converges.

Since $\sum_n\mathscr{P}\left\{\frac{X_n}{a_n}\neq\frac{Y_n}{a_n}\right\}$, $\sum_n\mathscr{E}\left(\frac{Y_n^2}{a_n^2}\right)$ and $\sum_n\sigma^2\left(\frac{Y_n}{a_n}\right)$ converge, we have $$\sum_n\frac{X_n}{a_n}\mbox{ converges  a.e.}$$ by Three Series Theorem.

$\Box$