2
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.