Probability: Median of a Difference 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.
Thanks.
 A: 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 $j\le n$. 


*
*Using (i), one sees that $P(|Y_n|\ge ub_n)\le\frac16$ because $n\ge K$. 

*
If $j\le K$, using (ii), one sees that $P(|Y_j|\ge ub_n)\le\frac16$ because $n\ge M_j$. 

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