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Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. I want to upper bound the function $$ L(t):=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-2\lambda_i t}$ for $t>0$.

Indeed, I try to answer the following question:

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

But this upper bound does not work for my question. So I want to get a tight upper bound as $\alpha=0$.

Can we get more tighter upper bound with high probability? For example, $$ L\le \frac{e^{-2\lambda_1t}}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. Assume that $\sum_{i=1}^n X_i^2=n$. I want to upper bound the function $$ L(t):=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-2\lambda_i t}$ for $t>0$.

Indeed, I try to answer the following question:

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

But this upper bound does not work for my question. So I want to get a tight upper bound as $\alpha=0$.

Can we get more tighter upper bound with high probability? For example, $$ L\le \frac{e^{-2\lambda_1t}}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Another way: I try to consider upper bound the difference between $L_1(t)$ and $L_j(t)$ (here I choose the subscript $i$ dependent on $X_i$):

$$ L_1(t)-L_j(t)=\frac{a_1X_1-a_jX_j}{\sqrt{\sum_i a^2_iX^2_i}}=\frac{e^{-2\lambda_1 t}(1-e^{-2t(\lambda_j-\lambda_1)})X_1+e^{-2\lambda_j t}(X_1-X_j)}{\sqrt{\sum_i a^2_iX^2_i}} $$

Can we upper bound $L_1(t)-L_j(t)$ for every $j$ to get our desired result?

Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. I want to upper bound the function $$ L(t):=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-2\lambda_i t}$ for $t>0$.

Indeed, I try to answer the following question:

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Can we get more tighter upper bound with high probability? For example, $$ L\le \frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. I want to upper bound the function $$ L(t):=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-2\lambda_i t}$ for $t>0$.

Indeed, I try to answer the following question:

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

But this upper bound does not work for my question. So I want to get a tight upper bound as $\alpha=0$.

Can we get more tighter upper bound with high probability? For example, $$ L\le \frac{e^{-2\lambda_1t}}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. I want to upper bound $$ L=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-4\lambda_i t}$ for $t>0$.

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

Following this question Anti-concentration of Gaussian quadratic form. We have the following inequality:

Let $X_1,\dots,X_n$ denote i.i.d. standard Gaussian random variables. For every $\epsilon>0$ and real number $a_1,\dots, a_n>0$, we get $$ \mathbb{P}\left(\sum_ia_iX_i^2\le\epsilon\sum_ia_i\right)\le\sqrt{e\epsilon} $$

In my setting, let $A=\{a_{ij}\}_{1\le i,j\le n}$ be an $n$ by $n$ normalized Gaussian random matrix with $E[a_{ij}]=0$ and $E[a_{ij}^2]=1/n$. Ordering its eigenvalues by $\lambda_1\le \lambda_2\le \cdots \lambda_n$ with corresponding eigenvectors $v_1,\dots, v_n \in \mathbb{R}^n$. Here $\lambda_i$ follows the semi-circle law.

Let $X_1,\dots, X_n$ be $n$ i.i.d. Gaussian random variables with mean $0$ and variance $1$. I want to upper bound the function $$ L(t):=\frac{a_1X_1}{\sqrt{\sum_i a^2_iX^2_i}} $$ where $a_i=e^{-2\lambda_i t}$ for $t>0$.

Indeed, I try to answer the following question:

Question: can we upper bound this one to solve this question: For $\epsilon>0$, define $\tau:=\inf_{t\ge 0}\{L\ge \epsilon\}$ Show that $$\lim_{n\to \infty} P(\tau\ge n^{2/3})=1. $$

I can just apply the inequality in that question. Then for any $\epsilon>0$, $$\quad \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le \epsilon\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{e\epsilon}\quad $$

Moreover, based on the above inequality, taking $\epsilon=1/n^{\alpha}$ for $\alpha>0$, we get the following concentration inequality $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \le n^{-\alpha}\sum_{i=1}^n a_i^2|\lambda_1,\dots,\lambda_n\right)\le \sqrt{en^{-\alpha}}\quad. $$

Hence, taking expectation on the both side $$ \mathbb{P}\left(\sum_{i=1}^n X_i^2a_i^2 \ge n^{-\alpha}\sum_{i=1}^n a_i^2\right)\ge 1-\sqrt{en^{-\alpha}} $$

Now, we can lower bound the denominator of $L$ with high probability $1-\sqrt{en^{-\alpha}}$: $$ L\le n^{\alpha/2}\frac{a_1X_1}{\sqrt{\sum_i a_i^2}}=n^{\alpha/2}\frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$

Can we get more tighter upper bound with high probability? For example, $$ L\le \frac{e^{-2\lambda_1t}X_1}{\sqrt{\sum_i e^{-4\lambda_it}}} $$