0
$\begingroup$

I recently asked a question related to a proof I'm working on as well as a follow-up to that. The detailed counter-examples I received elucidated a lot, but it quickly transpired (i) that I was leaving out information that seemed irrelevant to me at first, but that turned out to affect the answer, and (ii) that the bounds that would finish my proof did not exist.

So, to close off my trilogy, I though I'd pose the broader conjecture I'm trying to prove, with the hope that someone can help me work towards (dis)proving it. I'll first state the conjecture, and afterwards add my current (failing) approach as an optional read.

The conjecture

For each $1\le n\le N$, let $(X_n, Y_n)$ be two correlated random variables. When $n\ne m$, $(X_n, Y_n)$ and $(X_m, Y_m)$ are i.i.d. All finite moments of $X_n$ and $Y_n$ exist. We also have a function $f(x)\ge0$, which is locally Lipschitz with at most polynomial growth at infinity. Define $g(x) = \exp(-f(x))$ such that $0<g(x)\le1$.

I aim to show that, for any $p\ge2$, $$ \left\|\left(\frac{\sum_{n=1}^NX_ng(X_n)}{\sum_{n=1}^Ng(X_n)} - \frac{\sum_{n=1}^NY_ng(Y_n)}{\sum_{n=1}^Ng(Y_n)}\right) - \left(\frac{\mathbb E[Xg(X)]}{\mathbb E[g(X)]} - \frac{\mathbb E[Yg(Y)]}{\mathbb E[g(Y)]}\right)\right\|_p \le CN^{-1/2}\|X-Y\|_r^\beta \tag{1}\label{1} $$ where

  • $r$ may depend only on $p$,
  • I abbreviated $X=X_1$ and $Y=Y_1$, and
  • the constant $C$ may depend on $p$ and on the distributions of $X$ and $Y$.

I hope to obtain a bound with $\beta=1$. If that is proven not to exist, however, I am interested to find out whether a bound exists for a smaller value of $\beta$. It is sufficient if \eqref{1} holds whenever $\|X-Z\|_r\le d$ and $\|Y-Z\|_r\le d$, for some $Z$ and some $d>0$, and with $C$ allowed to depend on $Z$ and $d$.

The question of whether \eqref{1} holds comes up in a paper I'm currently writing. If I receive a (positive or negative) answer here and I use (part of) it in the paper, I will naturally make sure to acknowledge the author! (If they approve, of course.)

My attempt

I have attempted to tackle the bound using a variant of Lemma 2 of this paper, which I'll state here for clarity. Define $\Delta = \frac{N_2}{D_2} - \frac{N_1}{D_1}$, with $D_1, D_2 > 0$. Then, for all $0\le\alpha\le1$, $$ D_2|\Delta| \le |N_2 - N_1| + \frac{|N_1|}{D_2}|D_2 - D_1| + \left|\frac{N_1}{D_1}\right|\frac{|D_2-D_1|^{1+\alpha}}{D_2^\alpha}. \tag{2}\label{2} $$ I then bound the l.h.s. of \eqref{1} by $$ \left\|\frac{\sum_n(X_n-Y_n)g(X_n)}{\sum_ng(X_n)} - \frac{\mathbb E[(X-Y)g(X)]}{\mathbb E[g(X)]}\right\|_p + \left\| \left(\frac{\sum_nY_ng(X_n)}{\sum_ng(X_n)} - \frac{\sum_nY_ng(Y_n)}{\sum_ng(Y_n)}\right) - \left(\frac{\mathbb E[Yg(X)]}{\mathbb E[g(X)]} - \frac{\mathbb E[Yg(Y)]}{\mathbb E[g(Y)]}\right) \right\|_p. $$ To the first term, one can immediately apply \eqref{2} to get a bound as in \eqref{1}. For the second term, I brought the part with the sums to the same denominator, and did the same with the part with expectations. After then applying \eqref{2}, the only remaining issue is in the third term in \eqref{2}'s r.h.s.: the $|D_2-D_1|^{1+\alpha}$ (with $\alpha=1$) can be bounded as $\sim N^{-1}$, which means that the $\left|\frac{N_1}{D_1}\right|$ part must take care of a $|X-Y|^\beta$ factor. This is where my first question originated. Based on the replies I received, I am no longer convinced this is a fruitful line of investigation.

$\endgroup$
2
  • $\begingroup$ What is $p$ here? Is it any number $\ge2$, as before? $\endgroup$ Commented Oct 15, 2023 at 12:08
  • $\begingroup$ Yes! That's an oversight. I'll add that to the question. $\endgroup$
    – ArBo
    Commented Oct 15, 2023 at 12:21

0

You must log in to answer this question.

Browse other questions tagged .