Can we find such $k$ so that the following inequality holds?

Question

I found this question: Chernoff style concentration bound for ratio of variables. I want to ask if we get similar thing for the ratio of the sum and the one Gaussian variable.

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

Can we find such $k$?

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In these 2020 slides by Andrew Nobel, for $Y\sim \chi_k^2$ where $(Y=\sum_{I=1}^k X_i^2)$, for $t\in (0,1)$ $$ P(Y\ge (1+\epsilon)k)\le \exp(-k(t^2-t^3)/4). $$

Answer 1

$\newcommand\ep\epsilon $In the clever answer by Fedor Petrov, it was shown that \begin{equation*} Q:=P\Big(\frac{X_1^2+\dots+X_k^2}{X_1^2}<C\Big)\le C/k, \tag{1}\label{1} \end{equation*} where $C:=1/\ep^2>0$ and the $X_i$'s are any iid random variables.

Let us show that for the standard normal $X_i$'s as in the OP, the upper bound $C/k$ in \eqref{1} can be replaced by a bound decreasing exponentially in $k$.

For $C\le1$, $Q=0$. So, without loss of generality (wlog) $C>1$. Also, wlog $k\ge2$. For \begin{equation*} c:=C-1>0, \tag{2}\label{2} \end{equation*} any $x\in [0,\sqrt{(k-1)/c}\,]$, and $$h:=\frac{k-1-cx^2}{4(k-1)},$$ we have \begin{equation*} Q=P\Big(\sum_{i=2}^k X_i^2<cX_1^2\Big)\le Q_1+Q_2, \tag{3}\label{3} \end{equation*} where \begin{equation*} Q_1:=P(X_1^2\ge x^2)\le e^{-x^2/2}=:R_1 \tag{4}\label{4} \end{equation*} and \begin{equation*} \begin{aligned} Q_2&:=P\Big(\sum_{i=2}^k X_i^2<cx^2\Big) \\ &=P\Big(\sum_{i=2}^k(1-X_i^2)>k-1-cx^2\Big) \\ &\le\exp\{-h(k-1-cx^2)+(k-1)\ln Ee^{h(1-X_1^2)}\} \\ &=\exp\{-h(k-1-cx^2)+(k-1)(h-\tfrac12\,\ln(1+2h))\} \\ &\le\exp\{-h(k-1-cx^2)+2(k-1)h^2\} \\ &=\exp-\frac{(k-1-cx^2)^2}{8(k-1)}=:R_2. \end{aligned} \tag{5}\label{5} \end{equation*} Choosing now $x$ to be the positive root of the equation $R_1=R_2$, from \eqref{3}, \eqref{4}, and \eqref{5} we get \begin{equation} Q\le 2\exp-\frac{k-1}{2(1+\sqrt C)^2}, \tag{6}\label{6} \end{equation} which is the promised bound, decreasing exponentially in $k$.

One may also note that typically $k$ and $C$ will be large, and then the exponent in the bound in \eqref{6} will be about $-k/(2C)$ -- compare this with the bound in \eqref{1}.

Answer 2

For arbitrary $C>0$ and positive i. i. d. $Y_1,\ldots,Y_k$ for $Z:=\sum_{i=1}^k Y_i$ we have $$\sum_{i=1}^k {\mathbf 1}(Z/Y_i\geqslant C)\geqslant k-C$$ (since $Z/Y_i<C$ means that $Y_i>Z/C$, which may hold for at most $C$ different values of $i$),

and taking the expectation we get $$\mathrm {prob} (Z/Y_1\geqslant C)\geqslant 1-C/k. $$

Answer 3

$\newcommand\ep\epsilon\newcommand{\Ga}{\Gamma}$Somehow, writing my previous answer, I forgot the fact that the random variable (r.v.) \begin{equation*} R:=\frac{X_1^2}{X_1^2+\dots+X_k^2} \end{equation*} has the beta distribution with parameters $1/2,(k-1)/2$ -- which follows because $X_1^2$ and $X_2^2+\dots+X_k^2$ are independent r.v.'s, $X_1^2$ with the gamma distribution with parameters 1/2,2 and $X_2^2+\dots+X_k^2$ with the gamma distribution with parameters $(k-1)/2,2$.

This fact makes it easy to bound \begin{equation*} Q:=P\Big(\frac{X_1^2+\dots+X_k^2}{X_1^2}<C\Big), \tag{10}\label{10} \end{equation*} where $C:=1/\ep^2>1$ and the $X_i$'s are iid standard normal r.v.'s.

Indeed, without loss of generality $k\ge2$. We have \begin{equation*} Q=P(R>1/C)=r_k J, \tag{20}\label{20} \end{equation*} where \begin{equation*} r_k:=\frac{\Ga(k/2)}{\Ga(1/2)\Ga((k-1)/2)} \end{equation*} and \begin{equation*} J:=\int_{1/C}^1 x^{-1/2}(1-x)^{(k-3)/2}\,dx. \end{equation*} By the log-convexity of the gamma function, \begin{equation*} r_k\le\frac1{\Ga(1/2)} \sqrt{\frac{\Ga((k+1)/2)}{\Ga((k-1)/2)}}=\sqrt{\frac{k-1}{2\pi}}. \end{equation*} An easy, even if not quite accurate, way to bound $J$ is as follows: \begin{equation*} J\le\int_{1/C}^1 (1/C)^{-1/2}(1-1/C)^{(k-3)/2}\,dx =C^{1/2}(1-1/C)^{(k-3)/2} \end{equation*} for $k\ge3$; the case $k=2$ is very easy.

Thus, for $k\ge3$, \begin{equation*} Q=P\Big(\frac{X_1^2+\dots+X_k^2}{X_1^2}<C\Big) \le C^{1/2}\sqrt{\frac{k-1}{2\pi}}\,(1-1/C)^{(k-3)/2}. \tag{30}\label{30} \end{equation*}

So, as in the previous answer, we have an upper bound on $Q$ decreasing exponentially in $k$. However, in the latter case the base of the power with exponent $k$ is $(1-1/C)^{1/2}<\exp-\frac1{2C}$, which is strictly less than the corresponding base, $\exp-\frac1{2(1+\sqrt C)^2}$, in the previous answer. So, we now get a faster decreasing upper bound on $Q$.