Convergence in probability implies convergence of the median?

Question

Suppose that for every $\varepsilon>0$ the sequence of random variables $X_{n}$ satisfies: \begin{align*} P(|X_{n}|>\varepsilon) \to 0 \end{align*} as $n\to \infty$.

Now let $M_{n}$ denote the median of $X_{n}$.

Is there anything that we can say about convergence of the median? In particular, what are the additional conditions needed to ensure that $M_{n} \to 0$ in the above?

Answer 1

Assume $M_n\not\to0$. By definition, for all $\varepsilon>0$ we can find arbitrarily large $n$ with $\lvert M_n\rvert > \varepsilon$; without loss of generality, we can moreover find arbitrarily large $n$ with $M_n > \varepsilon$.

Since $\mathbb{P}\left(\lvert X_n \rvert > \varepsilon \right)\to0$ as $n\to\infty$, we can find $n$ such that we have both of the following simultaneously: \begin{align} M_n &> \varepsilon, \\ \mathbb{P}\left(\lvert X_n \rvert > \varepsilon \right) &< \frac{1}{2}.\end{align} However, by definition we have that $\mathbb{P}\left(X_n \geq M_n\right)\geq \frac{1}{2}$. It then remains to note that $\lvert X_n \rvert \geq X_n$ , and hence $$ \mathbb{P}\left(\lvert X_n \rvert > \varepsilon \right) \geq \mathbb{P}\left(X_n \geq M_n \right)=\frac{1}{2}, $$ producing a contradiction.

Answer 2

Setting $a_{n}=\inf\left\{ x\in\mathbb{R}\mid P\left(X_{n}\leq x\right)\geq0.5\right\} $ and $b_{n}=\sup\left\{ x\in\mathbb{R}\mid P\left(X_{n}\geq x\right)\geq0.5\right\} $ we find $P\left(X_{n}\leq a_n\right)\geq0.5$ and $P\left(X_{n}\geq b_n\right)\geq0.5$.

Further the assumption $b_{n}<x<a_{n}$ leads $P\left(X_{n}\leq x\right)<0.5\wedge P\left(X_{n}\geq x\right)<0.5$ contradicting that $P\left(X_{n}\leq x\right)+P\left(X_{n}\geq x\right)\geq1$.

So we are allowed to conclude that $a_{n}\leq b_{n}$ and that: $$m_n\in\left[a_{n},b_{n}\right]\iff P\left(X_{n}\leq m_n\right)\geq0.5\wedge P\left(X_{n}\geq m_n\right)\geq0.5$$

The RHS states exactly that $m_n$ is a median of the distribution of $X_{n}$.

$X_{n}\stackrel{p}{\to}0$ implies (is even equivalent with) $X_{n}\stackrel{d}{\to}0$.

Now fix some $x>0$. Then $\lim_{n\to\infty}F_{n}\left(\frac{1}{2}x\right)=1$ and $P\left(X_{n}\geq x\right)\geq1-F_{n}\left(\frac{1}{2}x\right)\to0$ so that eventually $b_{n}<x$.

Likewise we can find that eventually $-x<a_{n}$.

Proved is now that for every $x>0$ there exists a positive integer $m$ with $n>m\implies\left[a_{n,}b_{n}\right]\subseteq\left[-x,x\right]$.

This implies that $\lim_{n\to\infty}m_{n}=0$ for every sequence $\left(m_{n}\right)_{n}$ that satisfies $m_{n}\in\left[a_{n,}b_{n}\right]$ (i.e. a sequence of medians).