Upper bound on difference of correlated ratios

Question

Suppose $(x_n, y_n)$ are i.i.d. samples (that is, $x_n$ and $y_n$ are not independent, but $(x_n, y_n)$ is i.i.d. with regards to $(x_m, y_m)$ if $n\ne m$) from a joint distribution, with $0 < x_n, y_n \le 1$. The quantity I am interested in is $$ \left\|\frac{Nx_1}{\sum_{n=1}^Nx_n} - \frac{Ny_1}{\sum_{n=1}^Ny_n}\right\|_p, $$ where I used the $p$-norm $\|u\|_p = \mathbb E[|u|^p]^{1/p}$. I'm not entirely sure how best to describe this quantity in words (hence the vague title), but it is something that comes up as part of a proof I'm working on.

The conjecture I'd like to prove is that, for all $p\ge2$, $$ \left\|\frac{Nx_1}{\sum_{n=1}^Nx_n} - \frac{Ny_1}{\sum_{n=1}^Ny_n}\right\|_p \le \frac{2\|x\|_p}{\mathbb E[x]\mathbb E[y]}\|x - y\|_p. $$ I believe this bound will hold, as it does so for $N=1$ -- in which case the left-hand side is zero -- and in the "limit to infinity" when $\sum_{n=1}^Nx_n/N$ is replaced by $\mathbb E[x]$ (and similarly for $y$). It seems logical to me that the left-hand side grows monotonically with $N$, from which my conjecture follows. I've spent quite some time trying to prove the statement, but have so far been unsuccessful. Lemma 2 of this paper seemed promising, but did not result in a proof for me.

Any help would be greatly appreciated! If you can (dis)prove the conjecture, that would be awesome, but pointers/ideas for me to pursue are welcome as well. For my purpose, I could also work with several relaxations of the conjecture:

• The bound may use $\|x-y\|_r$ for any $r$ instead of $p$. This $r$ can only depend on $p$.
• The bound may include a factor $N$ if need be (but no higher powers $N^\alpha$ with $\alpha>1$).
• The factor $\frac{2\|x\|_p}{\mathbb E[x]\mathbb E[y]}$ can be replaced by pretty much anything that depends on the distributions of $x$ and $y$ and that doesn't "blow up". Something like $\|1/x\|_p$ cannot be included, as it might not exist.

Thank you in advance for taking the time to read through my post. I look forward to seeing your opinions on this problem.

Answer 1

Such bounds do not exist.

Indeed, suppose that $N=2$; $X_1,X_2$ are independent random variables each uniformly distributed on the interval $(0,1)$; $Y_1=h+(1-h)X_1$ and $Y_2=h+(1-h)X_2$, where $h\in(0,1)$.

Then \begin{equation} \|X_1-Y_1\|_r=\frac h{(1+r)^{1/r}} \end{equation} for all real $r>0$.

On the other hand, \begin{equation} \frac1h\Big(\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big) =\frac{X_1-X_2}{(X_1+X_2)(X_1+X_2+h(2-(X_1+X_2)))}. \end{equation} So, letting $h\downarrow0$ and $c_p:=2^p/3^{2p}$, by monotone convergence we get \begin{equation} \frac1{h^p}\Big\|\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big\|_p^p \to\int_0^1\int_0^1 dx_1\,dx_2\,\frac{|x_1-x_2|^p}{(x_1+x_2)^{2p}} \\ \ge\int_0^1 dx_1\int_0^{x_1/2} dx_2\,\frac{c_p}{x_1^p}=\infty, \end{equation} since $p\ge2$. So, the inequality \begin{equation} \Big\|\frac{X_1}{X_1+X_2}-\frac{Y_1}{Y_1+Y_2}\Big\|_p\le C\|X_1-Y_1\|_r \end{equation} cannot hold for any real $C$, any real $r>0$, and all $h\in(0,1)$.