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{equation} \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} \end{equation}

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{equation} \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} \end{equation} (*): 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.


1 Answer 1


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.


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