Asymptotic bound of quotient of absolute and squared deviation from mean

Question

The following fraction shows up when trying to show consistency of the OLS slope estimator in a simple linear regression on a log-log scale where the window of observation changes as the sample size $T$ increases and I want to find a constant $c > 0$ such that $$ \frac{\sum_{k = 1}^{N(T)} \vert x_{k, T} - \bar{x}_{N(T)} \vert}{\sum_{k = 1}^{N(T)} (x_{k, T} - \bar{x}_{N(T)})^2} \leq c $$ if $T$ is large enough, where $$ \begin{align*} x_{k, T} &= \log(T^{\varepsilon} + k - 1), \ \varepsilon \in (0, 1), \\ \bar{x}_{N(T)} &= \frac{1}{N(T)} \sum_{k = 1}^{N(T)} x_{k, T}, \\ N(T) &= (m - 1) T^{\varepsilon},\ m > 1 \end{align*} $$

So far, I have tried some form of figuring out for which $k$ it holds that $\vert x_{k, T} - \bar{x}_{N(T)} \vert \leq 1$ and then using that the absolute value is less or equal to its square which helps to bound some part of the fraction but I do not now how to deal with the remaining part.

Another idea of mine is to rewrite the fraction by multiplying both the numerator and denominator by $1 / N(T)$ such that the numerator is the mean absolute deviation of a (non-random) sample $(x_{1, T}, \ldots, x_{N(T), T})$ from its mean whereas the denominator is the mean squared deviation. I was hoping that there maybe is some knowledge about the asymptotic behavior of both numerator and denominator but so far I did not find anything. Also, things are a bit tricky because as $T$ increases the sample not only becomes larger but changes altogether.

Finally, another idea is to use the rewritten fraction from the previous approach and try to show that both numerator and denominator are asymptotically equivalent to integrals of the form $\int_{T^{\varepsilon}}^{mT^{\varepsilon}}...dx$ which are hopefully somewhat easy to evaluate. However, I am not quite sure if such an asymptotic equivalence is feasible.

Answer 1

$\newcommand{\ep}{\varepsilon}\newcommand{\num}{\operatorname{num}}\newcommand{\den}{\operatorname{den}}\newcommand{\R}{\mathbb R}$Let $$x_k:=x_{k,T}=\ln(S+k-1),\quad S:=T^\ep,\quad N:=N(T).$$ Let us ensure that $N$ is an integer, assuming, more generally, just that \begin{equation} N\sim(m-1)S. \end{equation} We have to show that eventually (that is, for all large enough $S>0$) \begin{equation} \frac{\num}{\den}\le c<\infty, \end{equation} where \begin{equation} \num:=\sum_1^N|x_k-\bar x|,\quad \den:=\sum_1^N(x_k-\bar x)^2,\quad \bar x:=\frac1N\,\sum_1^Nx_k, \end{equation} and $c$ may depend only on $m\in(1,\infty)$.

Note that $\sum_1^N|x_k-a|$ is convex in $a\in\R$. Hence, \begin{align*} \num&\le\max\Big(\sum_1^N(x_k-x_1),\sum_1^N(x_N-x_k)\Big) \\ &\le\max\Big(\sum_1^N\ln\Big(1+\frac{k-1}S\Big),\sum_1^N\ln\Big(1+\frac{N-K}S\Big)\Big) \\ &\le\max\Big(\sum_1^N\frac{k-1}S,\sum_1^N \frac{N-K}S\Big) \\ &=\frac{N(N-1)}{2S}\le\frac{(m-1)^2}{2+o(1)}\,S. \end{align*}

On the other hand, by Jensen's inequality for the concave function $\ln$, \begin{equation} \bar x=\frac1N\,\sum_1^N\ln(S+k-1)\le\ln\Big(\frac1N\,\sum_1^N(S+k-1)\Big) =\ln(S+(N-1)/2). \end{equation} So, \begin{align*} \den&\ge\sum_{3N/4\le k\le N}(x_k-\bar x)^2 \\ &\ge\sum_{3N/4\le k\le N}\ln^2\Big(1+\frac{k-1-(N-1)/2}{S+(N-1)/2}\Big) \\ &\ge\frac N{4+o(1)}\,\ln^2\Big(1+\frac{N/4}{S+(m-1)S/2}\Big) \\ \ &\sim\frac{(m-1)S}4\,\ln^2\Big(1+\frac{m-1}{2m+2}\Big). \end{align*} Thus, \begin{equation} \frac{\num}{\den}\le(2+o(1))(m-1)\Big/\ln^2\Big(1+\frac{m-1}{2m+2}\Big), \end{equation} as desired.