Equivalence of conditions for Convergence of Non-negative Random Series

Question

Question

Let $X_n\geq 0$ be independent for $n\geq 1$. The following are equivalent.

1. $\sum_{n=1}^\infty X_n<\infty$ a.s.
2. $\sum_{n=1}^\infty[P(X_n>1)+E(X_nI(X_n\leq 1)]<\infty$
3. $\sum_{n=1}^\infty E(X_n/(1+X_n))<\infty$

My attempt

I was able to prove $(1)$ implies $(2)$ and $(2)$ implies $(3)$ but not $(3)$ implies $(1)$.

For $(1)$ implies $(2)$, observe that by Borel-Cantelli $\sum_{n=1}^\infty P(X_n>1)<\infty$, else $X_n>1$ infinitely often w.p.1 so that $(1)$ is violated. To see that $\sum_{n=1}^\infty E(X_nI(X_n\leq 1)]<\infty$, apply Kolmogorov three-series theorem (I think).

For $(2)$ implies $(3)$ we can write that $$ \begin{align} \frac{X_n}{1+X_n}& =\frac{X_n}{1+X_n}I(X_n\leq 1)+\frac{X_n}{1+X_n}I(X_n>1)\\ &\leq X_nI(X_n\leq 1)+ I(X_n>1). \end{align} $$ Now take the expectation of both sides and sum on $n$ using $(2)$ to conclude $(3)$.

For the last implication $(3)$ implies $(1)$, I tried to use Kolmogorov three series theorem. To this end put $Y_n=X_nI(X_n\leq 1)$. First note that $\sum_{n=1}^\infty P(X_n>1)<\infty$ since $$ P(X_n>1)\leq P\left(\frac{X_n}{1+X_n}>\frac{1}{2}\right)\leq 2E\frac{X_n}{1+X_n} $$ by Markov's inequality. But I am unable to show that $\sum EY_n<\infty$ and $\sum\text{Var}(Y_n)<\infty$. Any help is appreciated.

Answer 1

As

$$\mathbb{E}(Y_n) \leq \mathbb{E}(X_n 1_{\{X_n \leq 1\}}) + \mathbb{P}(X_n>1),$$

it follows from $(2)$ that

$$\sum_{n \geq 1} \mathbb{E}(Y_n) \leq \sum_{n \in \mathbb{N}} [ \mathbb{E}(X_n 1_{\{X_n \leq 1\}}) + \mathbb{P}(X_n>1)] < \infty.$$

To prove $\sum_n \text{var} \, (Y_n)<\infty$, we note that $0 \leq Y_n \leq 1$ implies

$$Y_n^2 \leq Y_n,$$

and so

$$\text{var} \, (Y_n) = \mathbb{E}(Y_n^2)- (\mathbb{E}(Y_n))^2 \leq \mathbb{E}(Y_n) - (\mathbb{E}(Y_n))^2.$$

Thus,

$$\sum_{n \in \mathbb{N}} \text{var} \, (Y_n) \leq \sum_{n \in \mathbb{N}} \mathbb{E}(Y_n) - \sum_{n \geq 1} (\mathbb{E}(Y_n))^2.$$ We have already seen that the first series on the right-hand side converges; for the second one note that $$\sum_{n \in \mathbb{N}} a_n < \infty \implies \sum_{n \in \mathbb{N}} a_n^2 < \infty$$ for any sequence $(a_n)_{n \in \mathbb{N}} \subseteq (0,\infty)$. Applying the Kolmogorov three series theorem, we find that $(3)$ implies $(1)$.