Prove that $\text{P}(\max\limits_{1\leq k\leq n}|X_k-m_k|\geq t)\leq 2\text{P}(\max\limits_{1\leq k\leq n}|X_k-X_k^{'}|\geq t)$ Let $\{X_k:1\leq k\leq n\}$ and $\{X_k^{'}:1\leq k\leq n\}$ be two independent sequences of random variables (i.e the two sequences are independent,but the variables in the same sequence are not necessary independent).${X_k}$ and ${X_k^{'}}$ have the same joint distribution.Let $m_k$ be the median of $X_k$ (which is also of $X_k^{'}$).Prove that for any $t>0$,$\text{P}(\max\limits_{1\leq k\leq n}|X_k-m_k|\geq t)\leq 2\text{P}(\max\limits_{1\leq k\leq n}|X_k-X_k^{'}|\geq t)$
$m$ is the median of $X$ iff $P(X\geq m)\geq 1/2$ and $P(X\leq m)\geq 1/2$
I tried the simplified case with $n=1$,that is $P(|X-m|\geq t)\leq 2P(|X-X'|\geq t)$ and split $|X-X'|$ into $|X-m|+|X'-m|$ but without further ideas.
Update:Thanks for the answer below but I still have problems to adapt this proof to multiple cases.Let $A_k$ denotes ${X_k-X_k'}$ reaches the maximum at $k$.Then $P(\max (X_k-X_k')\geq t)=\sum_i P(A_i\cap \{X_i-X_i'\geq t\})\geq\sum_i P(A_i\cap \{X_i-m_i\geq t\}\cap\{X_i'\leq m_i\})$
But I got stuck here because of the existence of $A_i$ so we can't split the terms inside $P$
 A: Here is a proof when you only have one random variable in each family. We have
$$P(|X-m| \geqslant t) = P(X \geq m+t) + P(X \leqslant m-t).$$
We want to compare that to
$$P(|X-X'| \geqslant t) = P(X \geqslant X'+t) + P(X \leqslant X'-t)$$
so let's compare $P(X \geqslant m+t) $ and $P(X \geqslant X'+t)$. The idea is the following : half the time, $X' \leq m$ ; therefore, half the time where we have $X \geq m + t$, we automatically have $X \geq X' + t$. Thus $(X \geq X' + t)$ is at least half as probable as $(X \geq m + t)$.
To prove this rigourously, let's apply the law of total probabilities with respect to the events $(X'\leqslant m)$ and $(X' >m)$ :
$$P(X \geqslant X'+t) = P((X' \leqslant m) \cap(X \geqslant X'+t))+P((X' > m) \cap(X \geqslant X'+t))$$
Note that if $X' \leq m$ and $X \geqslant m+t$, then automatically we have $X \geqslant X'+t$. Thus the event $(X' \leqslant m) \cap(X \geqslant m+t)$ contains the event $(X' \leqslant m) \cap(X \geqslant X'+t)$, and we get
$$\begin{array}{rll}
P(X \geqslant X'+t) &\geq&  P((X' \leqslant m) \cap(X \geqslant m+t))+\underbrace{P((X' > m) \cap(X \geqslant X'+t))}_{\geqslant 0} \\
&\geqslant & P((X' \leqslant m) \cap(X \geqslant m+t)) \\
&\geqslant & P(X' \leqslant m).P(X \geqslant m+t) \text{ as $X$ and $X'$ are independent} \\
&\geqslant & \dfrac{1}{2}P(X \geqslant m+t) \text{ as $m$ is the median of $X'$.} \end{array}$$
Similarly, $P(X \geq X'+t) \geq \dfrac{1}{2} P(X \geq m+t)$. We now get
$$\begin{array}{lll}
P(|X-X'| \geqslant t) &=& P(X \geqslant X'+t) + P(X \leqslant X'-t) \\
&\geqslant&  \dfrac{1}{2} P(X \geqslant m+t)+ \dfrac{1}{2} P(X \geq m-t) \\
&\geqslant&  \dfrac{1}{2} \left(P(X \geqslant m+t)+ P(X \geqslant m-t)\right) \\
&\geqslant&  \dfrac{1}{2} P(|X-m|\geqslant t)
\end{array}$$
which is exactly the desired
$$P(|X-m|\geqslant t) \leq 2 P(|X-X'| \geqslant t).$$
You can now try to adapt this proof in the case of multiple random variables $(X_k)$ and $(X'_k)$.
