Suppose $x$ is an isotropic random variable in $\mathbb{R}^d$ with $E[\|x\|^2]=d$ and $v$ is some vector. It appears that $\sum_i x_i^2 v_i \approx \sum_i v_i$ when $d \approx \infty$.
What is an easy way of showing it?
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asked Apr 13, 2023 at 15:34
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$\begingroup$ How does the "IE" follow? I think that's not true. $\endgroup$– whuber ♦Commented Apr 13, 2023 at 17:00
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$\begingroup$ By "IE", I mean that in $\sum_i x_i^2 v_i$, we can replace random $x_i^2$ with 1's, without changing the result much $\endgroup$– Yaroslav BulatovCommented Apr 13, 2023 at 17:29
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$\begingroup$ But the "well approximated" statement appears to be incorrect. $\endgroup$– whuber ♦Commented Apr 13, 2023 at 17:47
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1$\begingroup$ @whuber how about this -- for any linear function $f:\mathbb{R}^d\to\mathbb{R}$ we have $d\approx \infty \implies f(x_1^2,\ldots,x_d^2)\approx f(1, 1, \ldots, 1)$ $\endgroup$– Yaroslav BulatovCommented Apr 13, 2023 at 18:13
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$\begingroup$ Wondering if there's a version of Weak Law of Large Numbers which could apply when there's some constraint on uniformity of $v_i$'s $\endgroup$– Yaroslav BulatovCommented Apr 13, 2023 at 19:09
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