# Kolmogoroff condition for truncated random variables

Question summary. Does the Kolmogoroff condition $$\sum_{n=1}^\infty\frac{\mathbb V Y_n}{n^2} < \infty$$ hold for truncated random variables $$Y_n := X_n \cdot 1_{\{X_n \le n\}}$$ (see below for a more rigid definition)?

General Definitions. Let $$(\Omega, \mathcal A, \mathbb P)$$ be a probability measure space, let $$\mathbb E$$ denote the expected value and $$\mathbb V$$ the variance. Let $$1_\text{set}$$ be the characteristic function of $$\text{set}$$.

Definition. Let $$(X_n)_{n\in\mathbb N}$$ be a sequence of non-negative, pairwise independent real random variables with finite expected values such that $$\eta :=\displaystyle \lim_{n \to \infty} \mathbb E X_n$$ exists. (Then clearly $$\sup_n \mathbb E X_n < \infty$$.) Let $$Y_n:= X_n \cdot 1_{\{X_n \le n\}}$$.

I want to generalize the following result:

Theorem 1. If the $$X_n$$ are identically distributed, then $$\displaystyle\sum_{n=1}^\infty\frac{\mathbb V Y_n}{n^2} < \infty$$.

To a Theorem where the $$X_n$$ need not be identically distributed:

Statement 2. (is this true?) $$\ \displaystyle\sum_{n=1}^\infty\frac{\mathbb V Y_n}{n^2} < \infty$$ without further restrictions on the $$X_n$$.

Proof of Theorem 1. Consider any random variable $$M$$ with the same distribution as all the $$X_n$$. Then $$$$\begin{split} \sum_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb V Y_j & \le \sum_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1_{\{M \le j \}}) = \lim_{N\to\infty} \sum_{j=1}^N \left(\frac{1}{j^2} \sum_{k=0}^{j-1} \mathbb E(M^2 \cdot g_k)\right)\\ & = \lim_{N\to\infty} \sum_{k=0}^{N-1} \left(\mathbb E (M^2 \cdot g_k) \sum_{j=k+1}^N \frac{1}{j^2}\right)\\ & \le c + \lim_{N\to\infty} \sum_{k=1}^{N-1} \left(\mathbb E(M^2 \cdot g_k) \sum_{j=k+1}^N \frac{1}{(j-1) j}\right)\\ & \le c + \lim_{N\to\infty} \sum_{k=1}^{N-1} \mathbb E(M^2 \cdot g_k) \cdot \frac{1}{k} = c + \sum_{k=1}^{\infty} \frac{1}{k} \cdot \int_{\{k < M \le k + 1 \}} M^2 \, \mathrm d \mathbb P\\ & \le c + \sum_{k=1}^\infty \frac{k+1}{k} \cdot \mathbb E{(M \cdot g_k)} \le c + 2 \cdot \mathbb E M < \infty \qquad \square \end{split}$$$$

I initially thought that I could use the same proof for Statement 2 by considering $$M := \sup_n X_n$$ (by Beppo Levi we would have $$\mathbb E M < \infty$$.) The exact proof works except for the very first inequality (since $$\frac{1}{j^2} \cdot \mathbb V Y_j \le \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1_{\{X_j\le j \}})$$ is still true but $$\frac{1}{j^2} \cdot \mathbb V Y_j \le \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1_{\{M\le j \}})$$ is wrong in general.)

Proving the following Lemma would be enough:

Lemma. If $$\sum_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb E(M^2 \cdot 1_{\{M\le j \}})<\infty$$ then $$\sum_{j=1}^\infty \frac{1}{j^2} \cdot \mathbb V Y_j < \infty$$

Some ideas for the proof of this Lemma. We have

$$$$\begin{split} \mathbb V Y_j -\mathbb E(M^2 1_{\{M\le j \}}) &= \mathbb E(X_j^2 1_{\{X_j\le j\}})-\mathbb E(X_j 1_{\{X_j\le j\}})^2 -\mathbb E(M^2 1_{\{M\le j \}}) \\ &\overset{\text{(*)}}\le \int_{\{X_j\le j\}\setminus\{M\le j\}} X_j^2 \,\mathrm d \mathbb P \le j^2 \cdot \mathbb P(\{X_j\le j\}\setminus\{M\le j\})\\ &\le j^2 \cdot\mathbb P\{M>j\} \overset{\text{Markow}}\le const \cdot j \end{split}$$$$ (*): I feel like I am loosing something in this inequality

However, this gives an additional $$1/j$$ term in the sum and is thus not enough to prove the lemma.

Welcome to MathOverflow! However, your conjecture is false. Indeed, let $$P(X_n=n)=1/n=1-P(X_n=0)$$. Then for all $$n$$ we have $$EX_n=1$$, $$Y_n=X_n$$, $$Var\, Y_n=Var\,X_n=n-1$$. So, $$\sum_n Var\,Y_n/n^2=\infty$$.
Additional note: Your statement that $$EM<\infty$$ does not follow from the Beppo Levi theorem, and it is actually false in general. Indeed, in the above example, by the second Borel–Cantelli lemma, the events $$\{X_n=n\}$$ occur almost surely (a.s.) infinitely often, and hence $$M=\infty$$ a.s.