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Let $\mathbb{X}, \mathbb{Y} \subset \mathbb{R}^d$ be finite sets. Suppose random vectors $X \in \mathbb{X}$ and $Y \in \mathbb{Y}$ are sampled according to a joint distribution $\mathbb{P}_{XY}$. Additionally, define a random vector $\hat{Y}$, which has the same marginal distribution as $Y$, but is independent of $X$.

We know that the following inequality holds: $$ \mathbb{E}\left[X^\top Y - \mathbb{E}\left[X^\top \hat{Y} \mid X\right]\right]^2 \leq d \cdot \mathbb{E}\left[ \left(X^\top \hat{Y} - \mathbb{E}\left[X^\top \hat{Y} \mid X\right]\right)^2 \right], $$ which itself is a direct consequence of this inequality: $$ \mathbb{E}\left[X^\top Y\right]^2 \leq d \cdot \mathbb{E}\left[ \left(X^\top \hat{Y}\right)^2 \right]. $$ I think the following inequality also holds: $$ \mathbb{E}\left[|X^\top Y| - \mathbb{E}\left[|X^\top \hat{Y}| \mid X\right]\right]^2 \leq d \cdot \mathbb{E}\left[ \left(|X^\top \hat{Y}| - \mathbb{E}\left[|X^\top \hat{Y}| \mid X\right]\right)^2 \right], $$ but I can't prove it, so I'm looking for a proof!

P.s. I can provide the proof of the other inequalities if needed.

Alireza Bakhtiari
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  • Yes please provide the proofs. – Wei Sep 07 '24 at 13:23
  • Let $h = E[Y], and V = E[X X^t]$ be invertible for simplicity. The LHS is $$E[X^t Y - E[X^t \hat Y | X ]]=E[X^t (Y - h)]\leq E[|X|_{V^{-1}}|Y - h|_V]\leq\sqrt{E[|X|_{V^{-1}}^2]E[|\hat Y - h|V^2]}$$ Further $$E[|X|{V^{-1}}^2]=E[tr(X^t E[X X^t]^{-1} X)]=tr(E[X X^t] E[X X^t]^{-1})= d.$$ Moreover, $$E[|\hat Y - h|_V^2]=E[(\hat Y - h)^t E[X X^t] (\hat Y - h)]=E[(\hat Y - h)^t X X^t (\hat Y - h)]=E[(X^t (\hat Y - h))^2]=E[(X^t \hat Y - E[X^t \hat Y | X])^2]$$ – Alireza Bakhtiari Sep 08 '24 at 08:02
  • I hope it's clear:) I didn't include the proof in the main question since from my experience long questions have significantly less chance for getting a response.. – Alireza Bakhtiari Sep 08 '24 at 08:06

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