Convergence of sum of random variables' expectations

Question

The following is Problem 11 from Chapter 3 of Leo Breiman's book, Probability (I'm doing some self-study). It states:

Let $X_1, X_2, ...$ be independent and $X_k \geq 0$. If for some $\delta$, $0 < \delta < 1$, there exists an $x$ such that:

$$\int_{\{X_k > x\}} X_k \,dP \leq \delta E\left[X_k\right] \, \forall k$$

show that:

$$\sum_{k = 1}^{\infty} X_k < \infty \, \text{a.s.} \implies \sum_{k = 1}^{\infty} E\left[X_k\right] < \infty$$

My attempt so far: I have that, since $E\left[X_k\right] = \int_{\{X_k \leq x\}} X_k \,dP + \int_{\{X_k > x\}} X_k \,dP \leq x P(X_k \leq x) + \delta E\left[X_k\right]$, after some algebra, it holds that $E\left[X_k\right] \leq \frac{x P(X_k \leq x) }{1 - \delta}$ $\forall \delta$ and resp. $x$. We can thus say that $\sum_{k = 1}^{n} E\left[X_k\right] \leq \frac{x}{1 - \delta} \sum_{k = 1}^{n} P(X_k \leq x)$. I also see nothing forbidding allowing $\delta$ and $x$ to depend on $k$ and thus saying $\sum_{k = 1}^{n} E\left[X_k\right] \leq \sum_{k = 1}^{n} \frac{x_k}{1 - \delta_k} P(X_k \leq x_k)$. If I could show that a.s. convergence allows for the right-hand-side to converge somehow, I have a proof. But nothing comes to mind to show that.

Another line of thought: It's known that if $Y_k$ are mean-zero independent random variables, that $\sum_{k=1}^{\infty} Y_k < \infty$ a.s. iff $\sum_{k = 1}^{\infty} Var(Y_k) < \infty$. Consider the sum $\sum_{k = 1}^{n} (X_k - E\left[X_k\right])$, a series of mean-zero random variables. Suppose we show that this is finite almost surely, likely by showing that $\sum_{k = 1}^{\infty} E\left[\left(X_k - E\left[X_k\right]\right)^2\right] < \infty$. Then we would have that $\sum_{k = 1}^{\infty} (X_k - E\left[X_k\right]) = \sum_{k = 1}^{\infty} X_k + \sum_{k = 1}^{\infty} E\left[X_k\right] < \infty$ a.s. Since $\sum_{k = 1}^{\infty} X_k < \infty \, \text{a.s.}$, this is only possible if $\sum_{k = 1}^{\infty} E\left[X_k\right] < \infty$. So the question is can we show that the above conditions allow $\sum_{k = 1}^{\infty} E\left[\left(X_k - E\left[X_k\right]\right)^2\right] < \infty$.

Thoughts?

Answer 1

Although not explicit in the wikipedia article for Kolmogorov three-series theorem, the constant $A$ appearing in the condition (i)-(iii) can be chosen to be arbitrary. Now if we set $A = x$, where $x$ such that your condition is satisfied, then with $Y_n = X_n \mathbf{1}_{\{X_n \leq x\}}$ we know that

1. $\sum_{n=1}^{\infty} Y_n$ converges a.s. by comparison test.

2. $\sum_{n=1}^{\infty} \Bbb{E}[Y_n]$ converges (this is where Kolmogorov three-series theorem is required).

3. $\Bbb{E}[X_n] \leq \frac{1}{1-\delta} \Bbb{E}[Y_n]$.

Thus it follows that $\sum_{n=1}^{\infty} \Bbb{E}[X_n]$ converges by comparison test, given Kolmogorov three-series theorem.

I am thinking about whether we can avoid the use of this sledgehammer method by taking advantage of the fact that $X_n$ are non-negative.