1
$\begingroup$

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.

Improve this question
$\endgroup$
5
  • $\begingroup$ It could indeed help if you provide proofs of your statements (or references to them, if available). $\endgroup$ Commented Sep 8, 2024 at 0:20
  • $\begingroup$ 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, $$E[\|\hat Y - h\|_V^2]=E[(\hat Y - h)^t E[X X^t] (\hat Y - h)]=E[(\hat Y - h)^t X X^t (\hat Y - h)]=E[(X^t (\hat Y - h))^2]=E[(X^t \hat Y - E[X^t \hat Y | X])^2]$$ $\endgroup$ Commented Sep 8, 2024 at 8:03
  • $\begingroup$ 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.. $\endgroup$ Commented Sep 8, 2024 at 8:05
  • 1
    $\begingroup$ I have something that may help. As the first proof start from the fact that $$\mathbb E[X^\top Y]^2\le d\mathbb E[(X^\top \hat Y)^2],$$ it may be useful to generalize it to the absolute value, getting $$\mathbb E[|X^\top Y|]^2\le d\mathbb E[(X^\top \hat Y)^2].$$ We can prove this by defining an auxiliary random vector, $Y'$ that is equal to $Y$ when $X^\top Y>0$ and $-Y$ otherwise. Applying the original inequality to $Y'$, we get the result. $\endgroup$ Commented Sep 23, 2024 at 22:59
  • 1
    $\begingroup$ Indeed, $|X^\top Y|=X^\top Y'$ almost surely. Moreover, as for any vector $v$ we have $(v^\top \hat Y')^2=(v^\top Y')^2=(v^\top Y)^2=(v^\top \hat Y)^2$, we have also $$\mathbb E[(X^\top \hat Y')^2]=\mathbb E[(X^\top \hat Y)^2]$$ $\endgroup$ Commented Sep 23, 2024 at 23:11

0

Reset to default

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.