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Let us consider a $n$-dimensional Gaussian blob, i.e. a set of $N$ random vectors $\{\boldsymbol{X}^{(j)}\}_{j=1}^N$, with $n$ independent components, $X_i^{(j)}$, and such that $X_i^{(j)} \sim \mathcal{N}(\mu_i, \sigma^2)$.
I want to prove that the elements on the tails are less collinear. More specifically, let us fix a direction $\boldsymbol{w}/||\boldsymbol{w}||$ and define the projection $$p^{(j)} = \boldsymbol{X}^{(j)} \cdot \frac{\boldsymbol{w}}{||\boldsymbol{w}||} ,$$ where $\boldsymbol{x} \cdot \boldsymbol{y}$ indicates the scalar product between the two vectors.
It follows (see for example this post) that $$\mathbb{E}(p^{(j)}) = \boldsymbol{\mu} \cdot \frac{\boldsymbol{w}}{||\boldsymbol{w}||}$$, with $\boldsymbol{\mu} = \mathbb{E}(\boldsymbol{X}^{(j)})$.
We can then identify the elements on the tail of the gaussian blob w.r.t. the $\boldsymbol{w}$ direction by considering the subset of vectors $\{\boldsymbol{X}^{(t)}\}$ with $\Omega = \{t \, , t \in [1, N] \, , \, p^{(t)}< \epsilon\}$ where $\epsilon$ is a parameter that fix the distance from the center $\mathbb{E}(p^{(j)})$ (we are interested in the case $|\mathbb{E}(p^{(j)}) - \epsilon| \gg 1$).

Is it possible to show that

$$\left\| \frac{1}{N} \sum_{j=1}^N \frac{\boldsymbol{X}^{(j)}}{\left\| \boldsymbol{X}^{(j)} \right\|} \right\|\geq \left\| \frac{1}{|\Omega|} \sum_{t \in \Omega} \frac{\boldsymbol{X}^{(t)}}{\left\| \boldsymbol{X}^{(t)} \right\|} \right\|$$

where $|\Omega|$ indicate the cardinality of the set of index $\Omega$ (i.e. the subset of samples on the tail).

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  • $\begingroup$ $p^{(t)}$ is a random variable, so perhaps we should have $\Omega = \{t \, , t \in [1, N] \, , \, \mathbb{E}(|p^{(t)}|^2)< \epsilon\}$ or something similar? $\endgroup$
    – NN2
    Commented Sep 9, 2023 at 9:29
  • $\begingroup$ Do you want the inequality you wrote to hold with high probability (or perhaps in expectation?). In any case, I doubt a general inequality of this type exists, as letting $n$ grow larger than $N$ suitably, and choosing $\mu=w$, I believe the opposite inequality holds with high probability. $\endgroup$
    – Pax
    Commented Sep 16, 2023 at 18:31

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