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Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$$$ \lim_{n\to \infty}\mathbb{P}\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ \lim_{n\to \infty}\mathbb{P}\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

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Hermi
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Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

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Hermi
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Can we find the following $k$ so that the following inequality holds for asymptotic normal?

Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ P\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$