Let $w \sim N(0, \sigma^2 I_d)$ and $w \in \mathbb{R}^d$. Denote $w_i$ as one value from the vector $w$. How can we prove that $\Bbb E[\|w\|^{j-1} \cdot w_i\cdot w_k] = 0$ for $i \neq k$ and $j \ge 2, j\in \mathbb{N}$?

Can it be said that $\ell^2$ norm to the power of $j-1$ (that is, $\|w\|^{j-1}$) is independent from the individual values of vector $w$? If so, then automatically $$\Bbb E[\|w\|^{j-1} \cdot w_i\cdot w_k] = \Bbb E[\|w\|^{j-1}] \cdot \Bbb E[w_i\cdot w_k] = 0.$$

I just need a clue whether this can be done or not, and provide me with the possible idea how I should tackle this? This is the university textbook course question and it has been part of the much complex question that I cannot disclose due to it being too long, but the part I just wrote seemed to me as the smallest understandable part of the question. By the way, I have a Computer Science background, and the course I am attending is related to High Dimensional Statistics book by Wainwright.

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    $\begingroup$ This question would benefit from added context. For example, simply telling us where you got the problem from or explaining what you tried would be a big step forward! (For further feedback/help with asking questions, you can ask here.) $\endgroup$
    – user1729
    Aug 11, 2021 at 13:18
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    $\begingroup$ @GreatDuke I'm sorry once again, but we will end this once you edit in the fact that your background is from computer science. I'd not noticed that you did not add this, so I request you to add it. Also, if you can locate your question in some other nice CS resource (or a similar type of question), we will then add that to the post as well if necessary. $\endgroup$ Aug 12, 2021 at 19:49
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    $\begingroup$ A hint in the spirit of the argument you had proposed (which doesn't quite work, but almost) - Denote $u_i = w_i/\|w\|$. Then the expectation is the same as $\mathbb{E}[\|w\|^{j+1} u_i u_k]$. Now show that $\|w\|$ and $u = w/\|w\|$ are independent, and further that $u$ is a uniform random variable on the sphere. Then argue that $\mathbb{E}[u_i u_k] = 0$. $\endgroup$ Aug 12, 2021 at 22:15


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