The rate at which the expectation of the square of the empirical median of i.i.d. $[-1,1]$-valued uniform random variables goes to zero

Question

Suppose that $X_1,X_2,\dots$ is an i.i.d. sequence of $[-1,1]$-valued uniform random variables. Let $\bar{X}_t$ be the empircal mean of the sample $X_1,\dots,X_t$. Then $\mathbb{E}\big[|\bar{X}_t|^2\big] = \operatorname{Var}[\bar{X}_t] = \frac{1}{3t}$. Hence for the empirical mean we have that $\mathbb{E}\big[|\bar{X}_t|^2\big]$ goes to zero as fast as $t^{-1}$, for $t \to \infty$.

I'm wondering at which rate the expected value of the square of the empirical median of these random variables goes to zero as the sample grows.

Specifically, for each $t \in \mathbb{N}$ (if it helps, assume that in what follows $t$ is odd), let $M_t$ be the empirical median of the sample $X_1,\dots,X_t$.

How fast $\mathbb{E}\big[|M_t|^2\big]$ goes to zero as $t \to \infty$?

We can interpret formally the question as finding $\alpha>0$ for which, if $0<\beta < \alpha $ then $\mathbb{E}\big[|M_t|^2\big] \cdot t^{\beta} \to 0, t\to\infty$, while if $\gamma > \alpha$ then $\mathbb{E}\big[|M_t|^2\big] \cdot t^{\gamma} \to \infty, t\to\infty$.

Given that for the empirical median we don't have any simple formula as we do for the empirical mean (where we can leverage the variance to obtain an easy computation), this problem seems way less trivial. Is there a smart way to deduce the rate of convergence?

Answer 1

For $t$ odd, $B_t:=\frac{1+M_t}2$ is a Beta$(\frac{t+1}2,\frac{t+1}2)$ variable (see Order statistics sampled from a uniform distribution), so $$\Bbb E\bigl[|M_t|^2\bigr]=\Bbb E\bigl[{(2B_t-1)}^2\bigr]=4\,\Bbb E[B_t^2]-4\,\Bbb E[B_t]+1 =4\,\Bbb E[B_t^2]-1.$$ Using $$\Bbb E[B_t^2] =\frac14+\frac1{4t}-\frac1{2t^2}+o\!\left(\frac1{t^2}\right)\!,$$ we conclude that $$\Bbb E\bigl[|M_t|^2\bigr]=\frac1t-\frac2{t^2}+o\!\left(\frac1{t^2}\right)\!.$$

Answer 2

Using order statistics $X_{(i)}$: $$M_t=\begin{cases}\displaystyle X_{\left(\frac{t+1}{2}\right)}&t\text{ odd}\\\frac{1}{2}\left[X_{\left(\frac{t}{2}\right)}+X_{\left(\frac{t}{2}+1\right)}\right]&t\text{ even}\end{cases}$$

@nejimban solved the question the case for $t$ odd. Just for fun and for the sake of completeness, we can also attempt to corroborate this rate of decay for $t$ even, i.e. $t=2n$.

Let $Y_i=\frac{1+X_i}{2}\sim U[0,1]$. It is clear then that order is preserved, i.e. $Y_{(i)}=\frac{1+X_{(i)}}{2}$ for all $1\leq i\leq t$, and thus $X_{(i)}=2Y_{(i)}-1$. Hence, $$M_t=\frac{1}{2}\left[X_{\left(n\right)}+X_{\left(n+1\right)}\right]=\frac{1}{2}\left[2Y_{\left(n\right)}-1+2Y_{\left(n+1\right)}-1\right]=Y_{\left(n\right)}+Y_{\left(n+1\right)}-1$$

We also know that $Y_{(i)}\sim\mathrm{Beta}(i,t-i+1)$, which implies $\mathbb{E}\{Y_{(i)}\}=\frac{i}{t+1}$ and $\mathbb{E}\{Y_{(i)}^2\}=\frac{i(i+1)}{(t+1)(t+2)}$. Thus, \begin{align} \mathbb{E}\{|M_t|^2\} &=\mathbb{E}\left\{(Y_{\left(n\right)}+Y_{\left(n+1\right)}-1)^2\right\}\\ &=\mathbb{E}\left\{Y_{(n)}^2+Y_{(n+1)}^2+2Y_{(n)}Y_{(n+1)}-2Y_{(n)}-2Y_{(n+1)}+1\right\}\\ &=\mathbb{E}\left\{Y_{(n)}^2\right\}+\mathbb{E}\left\{Y_{(n+1)}^2\right\}+2\mathbb{E}\left\{Y_{(n)}Y_{(n+1)}\right\}-2\mathbb{E}\left\{Y_{(n)}\right\}-2\mathbb{E}\left\{Y_{(n+1)}\right\}+1\\ &=\frac{n(n+1)}{(t+1)(t+2)}+\frac{(n+1)(n+2)}{(t+1)(t+2)}+2\mathbb{E}\left\{Y_{(n)}Y_{(n+1)}\right\}-2\frac{n}{t+1}-2\frac{n+1}{t+1}+1\\ &=\frac{(2n+2)(n+1)}{(t+1)(t+2)}+2\mathbb{E}\left\{Y_{(n)}Y_{(n+1)}\right\}-2\frac{2n+1}{t+1}+1\\ &=\frac{n+1}{t+1}+2\mathbb{E}\left\{Y_{(n)}Y_{(n+1)}\right\}-1\\ \end{align}

As the $Y_i$ are uniformly distributed in the unit interval $[0,1]$, the joint distribution of the order statistics $Y_{(i)},Y_{(j)}$ is $$f_{Y_{(i)},Y_{(j)}}(u,v)=t!\frac{u^{i-1}}{(i-1)!}\frac{(v-u)^{j-i-1}}{(j-i-1)!}\frac{(1-v)^{t-j}}{(t-j)!}\mathbf{1}_{0\leq u\leq v\leq 1}$$ so the joint distribution of $Y_{(n)},Y_{(n+1)}$ is \begin{align} f_{Y_{(n)},Y_{(n+1)}}(u,v) &=\frac{t!}{\left(n-1\right)!\left(n-1\right)!}u^{n-1}(1-v)^{n-1}\mathbf{1}_{0\leq u\leq v\leq 1}\\ \end{align} so we have that \begin{align} \mathbb{E}\left\{Y_{(n)}Y_{(n+1)}\right\}&=\int\int uv f_{Y_{(n)},Y_{(n+1)}}(u,v)\,\mathrm{d}u\mathrm{d}v\\ &=\frac{t!}{\left(n-1\right)!\left(n-1\right)!}\int_0^1\int_0^v u^{n}v(1-v)^{n-1}\,\mathrm{d}u\mathrm{d}v\\ &=\frac{t!}{\left(n-1\right)!\left(n-1\right)!}\frac{1}{n+1}\int_0^1v^{n+1}v(1-v)^{n-1}\,\mathrm{d}v\\ &=\frac{t!}{\left(n-1\right)!\left(n-1\right)!}\frac{1}{n+1}\frac{\Gamma(n)\Gamma(n+3)}{\Gamma(2n+3)}\\ &=\frac{t!}{\left(n-1\right)!\left(n-1\right)!}\frac{1}{n+1}\frac{(n-1)!(n+2)!}{(2n+2)!}\\ &=\frac{(n+2)n}{(t+2)(t+1)}\\ \end{align}

and we have \begin{align} \mathbb{E}\{|M_t|^2\} &=\frac{n+1}{t+1}+2\frac{(n+2)n}{(t+2)(t+1)}-1\\ &=\frac{(t+2)(n+1)+2n(n+2)}{(t+1)(t+2)}-1\\ &=\frac{t^2+4t+2}{t^2+3t+2}-1\\ &=\frac{t}{t^2+3t+2}\\ \end{align} which also decays at rate $1/t$.