Let $Y_1, Y_2, ...$ be random variables and $b_n$ be such that $0 < b_n \uparrow \infty$. Suppose that $\frac{Y_n}{b_n} \stackrel{p}{\to} 0$ (where $\stackrel{p}{\to}$ denotes convergence in probability). I am to show

$ \displaystyle \frac{\max_{1 \le j \le n} |m(Y_j - Y_n)|}{b_n} \to 0 \qquad (n \to \infty) $

where $m(\cdot)$ denotes a median of its argument.

I am embarrassingly clueless (in general I seem to have problems converting statements involving medians into useful bounds). All I know about the result is that it is used in a proof of Feller's Weak Law of Large Numbers.



The missing clue you ask for might be that:

For every random variable $Z$ and positive real number $z$, if $P(|Z|\ge z)\le\frac13$ then $|m(Z)|\le z$.

This follows directly from the definition of the median. In the context of your question, using the inclusion $$ [\,|Y_j-Y_n|\ge2ub_n\,]\subseteq [\,|Y_j|\ge ub_n\,]\cup[\,|Y_n|\ge ub_n\,], $$ which implies the inequality $$ P(|Y_j-Y_n|\ge2ub_n)\le P(|Y_j|\ge ub_n)+P(|Y_n|\ge ub_n), $$ a consequence of the missing clue is that it is enough to prove that:

(0) For every positive real number $u$, there exists a finite integer $N$ such that for every $n\ge N$ and every $j\le n$, $P(|Y_j|\ge ub_n)\le\frac16$ and $P(|Y_n|\ge ub_n)\le\frac16$.

We now prove (0). Fix a positive real number $u$.

First, when $n\to+\infty$, $P(|Y_n|\ge ub_n)\to0$, hence there exists a finite $K$ such that:

$\qquad$(i) for every $n\ge K$, $P(|Y_n|\ge ub_n)\le\frac16$.

Second, for each $j\le K$, $P(|Y_j|\ge ub_n)\to0$ when $n\to+\infty$ because $b_n\to\infty$, hence there exists a finite $M_j$ such that:

$\qquad$(ii) for every $n\ge M_j$, $P(|Y_j|\ge ub_n)\le\frac16$.

Choose finally $N\ge K$ such that $N\ge M_j$ for every $j\le K$. Assume that $n\ge N$ and that $j\le n$.

  1. Using (i), one sees that $P(|Y_n|\ge ub_n)\le\frac16$ because $n\ge K$.
  2. If $j\le K$, using (ii), one sees that $P(|Y_j|\ge ub_n)\le\frac16$ because $n\ge M_j$.
  3. If $j\ge K$, using (i) once again, one sees that $P(|Y_j|\ge ub_n)\le P(|Y_j|\ge ub_j)\le\frac16$ because $b_j\le b_n$.

Hence the proof of (0) is complete.


How could you compute the median of a number?

The problem is solved by using inequalities: you show that your thesis is greater or equal than zero, and lesser or equal than something that converge in probability to 0.

Anyhow I cant help you since I don't understand what is the median of a number. You need at least 3 numbers to compute a nontrivial median.

Edit, possible solution:

$ \displaystyle \max_{1 \le j \le n} |Y_j - Y_n| \le \max_{1 \le j \le n} |Y_j| = |Y_*| $

$ \displaystyle 0 \le \frac{ m ( \max_{1 \le j \le n} |Y_j - Y_n|)}{b_n} \le \frac{ m ( Y_* )}{b_n} $

Now if $m(Y_*)$ is a real number, then you're set.

This is just a sketch. If you want to make this proof more consistent take the definition of convergence:


And apply it to your problem, which I think is the task your professor wants you to do.

  • $\begingroup$ $m(\cdot)$ gives a median of a random variable. For example, if $X$ follows a standard normal distribution then $m(X) = 0$. A median of a random variable $X$ is any $k$ which satisfies $P(X \le k) \ge 1/2$ and $P(X \ge k) \ge 1/2$. $\endgroup$ – guy Jul 1 '11 at 4:37
  • $\begingroup$ so max( Y(j) - Y(n) ) can be considered as a new random variable? $\endgroup$ – Mascarpone Jul 1 '11 at 4:50
  • $\begingroup$ Yes. $m(Y_j - Y_n)$ by contrast is a real number, while $Y_j - Y_n$ is a random variable. $\endgroup$ – guy Jul 1 '11 at 4:57
  • $\begingroup$ possible solution added. $\endgroup$ – Mascarpone Jul 1 '11 at 5:15
  • $\begingroup$ It's max of the median, not median of the max. $\endgroup$ – guy Jul 1 '11 at 5:29

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