[proofplan] We prove the estimate by splitting the deviation from the median into its upper and lower tails. On the upper tail $\{X-a\ge t\}$, intersecting with the event $\{X'\le a\}$ forces $X-X'\ge t$, and independence turns this intersection into a product of probabilities. The median condition gives a factor at least $1/2$, yielding a one-sided bound; the lower tail is identical with the roles of $X$ and $X'$ reversed. Adding the two one-sided estimates gives the desired symmetrized inequality. [/proofplan] [step:Bound the upper tail using the event that the copy lies below the median] Let $(\Omega,\mathcal F,\mathbb P)$ denote the probability space on which $X$ and $X'$ are defined, and let $\omega\in\Omega$ denote a sample point. Fix $t>0$. Define $A_+(t)$ to be the event \begin{align*} A_+(t):=\{\omega\in\Omega:X(\omega)-a\ge t\}. \end{align*} Define $B_-$ to be the event \begin{align*} B_-:=\{\omega\in\Omega:X'(\omega)\le a\}. \end{align*} Define $E_+(t)$ to be the event \begin{align*} E_+(t):=\{\omega\in\Omega:X(\omega)-X'(\omega)\ge t\}. \end{align*} If $\omega\in A_+(t)\cap B_-$, then $X(\omega)\ge a+t$ and $X'(\omega)\le a$, hence $X(\omega)-X'(\omega)\ge t$. Thus \begin{align*} A_+(t)\cap B_- \subseteq E_+(t). \end{align*} Since $X$ and $X'$ are independent, the events $A_+(t)$ and $B_-$ are independent. Since $X'$ has the same distribution as $X$, \begin{align*} \mathbb P(B_-)=\mathbb P(X'\le a)=\mathbb P(X\le a)\ge \frac{1}{2}. \end{align*} Therefore monotonicity of probability gives \begin{align*} \mathbb P(E_+(t))\ge \mathbb P(A_+(t)\cap B_-). \end{align*} Independence gives \begin{align*} \mathbb P(A_+(t)\cap B_-)=\mathbb P(A_+(t))\,\mathbb P(B_-). \end{align*} Using $\mathbb P(B_-)\ge 1/2$, we obtain \begin{align*} \mathbb P(E_+(t))\ge \frac{1}{2}\mathbb P(A_+(t)). \end{align*} Equivalently, \begin{align*} \mathbb P(X-a\ge t)\le 2\,\mathbb P(X-X'\ge t). \end{align*} [guided] Let $(\Omega,\mathcal F,\mathbb P)$ denote the probability space on which $X$ and $X'$ are defined, and let $\omega\in\Omega$ denote a sample point. Fix $t>0$. The goal in the upper tail is to compare the event that $X$ is at least $t$ above the median with an event involving the symmetrized difference $X-X'$. Define $A_+(t)$ to be the event \begin{align*} A_+(t):=\{\omega\in\Omega:X(\omega)-a\ge t\}. \end{align*} Define $B_-$ to be the event \begin{align*} B_-:=\{\omega\in\Omega:X'(\omega)\le a\}. \end{align*} Define $E_+(t)$ to be the event \begin{align*} E_+(t):=\{\omega\in\Omega:X(\omega)-X'(\omega)\ge t\}. \end{align*} The reason for introducing $B_-$ is that if $X$ is above $a+t$ while the independent copy $X'$ is at most $a$, then the gap between the two variables is at least $t$. Indeed, if $\omega\in A_+(t)\cap B_-$, then \begin{align*} X(\omega)\ge a+t \qquad\text{and}\qquad X'(\omega)\le a, \end{align*} so subtracting the [second inequality](/theorems/2136) from the first gives \begin{align*} X(\omega)-X'(\omega)\ge t. \end{align*} Hence \begin{align*} A_+(t)\cap B_- \subseteq E_+(t). \end{align*} Taking probabilities and using monotonicity of probability gives \begin{align*} \mathbb P(E_+(t))\ge \mathbb P(A_+(t)\cap B_-). \end{align*} Now the independence hypothesis is used. Since $A_+(t)$ is determined by $X$ and $B_-$ is determined by $X'$, and since $X$ and $X'$ are independent random variables, the events $A_+(t)$ and $B_-$ are independent. Therefore \begin{align*} \mathbb P(A_+(t)\cap B_-)=\mathbb P(A_+(t))\,\mathbb P(B_-). \end{align*} Because $X'$ is a copy of $X$, it has the same distribution as $X$, so \begin{align*} \mathbb P(B_-)=\mathbb P(X'\le a)=\mathbb P(X\le a). \end{align*} The median hypothesis gives $\mathbb P(X\le a)\ge 1/2$, and hence $\mathbb P(B_-)\ge 1/2$. Combining this lower bound with the previous two displays gives \begin{align*} \mathbb P(E_+(t))\ge \frac{1}{2}\mathbb P(A_+(t)). \end{align*} Multiplying by $2$ gives the upper-tail estimate \begin{align*} \mathbb P(X-a\ge t)\le 2\,\mathbb P(X-X'\ge t). \end{align*} [/guided] [/step] [step:Bound the lower tail using the event that the copy lies above the median] Define $A_-(t)$ to be the event \begin{align*} A_-(t):=\{\omega\in\Omega:a-X(\omega)\ge t\}. \end{align*} Define $B_+$ to be the event \begin{align*} B_+:=\{\omega\in\Omega:X'(\omega)\ge a\}. \end{align*} Define $E_-(t)$ to be the event \begin{align*} E_-(t):=\{\omega\in\Omega:X'(\omega)-X(\omega)\ge t\}. \end{align*} If $\omega\in A_-(t)\cap B_+$, then $X(\omega)\le a-t$ and $X'(\omega)\ge a$, hence $X'(\omega)-X(\omega)\ge t$. Thus \begin{align*} A_-(t)\cap B_+ \subseteq E_-(t). \end{align*} The events $A_-(t)$ and $B_+$ are independent because $X$ and $X'$ are independent. Since $X'$ has the same distribution as $X$, \begin{align*} \mathbb P(B_+)=\mathbb P(X'\ge a)=\mathbb P(X\ge a)\ge \frac{1}{2}. \end{align*} Therefore monotonicity of probability gives \begin{align*} \mathbb P(E_-(t))\ge \mathbb P(A_-(t)\cap B_+). \end{align*} Independence gives \begin{align*} \mathbb P(A_-(t)\cap B_+)=\mathbb P(A_-(t))\,\mathbb P(B_+). \end{align*} Using $\mathbb P(B_+)\ge 1/2$, we obtain \begin{align*} \mathbb P(E_-(t))\ge \frac{1}{2}\mathbb P(A_-(t)). \end{align*} Equivalently, \begin{align*} \mathbb P(a-X\ge t)\le 2\,\mathbb P(X'-X\ge t). \end{align*} [/step] [step:Add the two one-sided estimates to obtain the symmetrized bound] Since $t>0$, the events $A_+(t)=\{X-a\ge t\}$ and $A_-(t)=\{a-X\ge t\}$ are disjoint, and \begin{align*} \{|X-a|\ge t\}=A_+(t)\cup A_-(t). \end{align*} Hence additivity of probability over disjoint events gives \begin{align*} \mathbb P(|X-a|\ge t)=\mathbb P(A_+(t))+\mathbb P(A_-(t)). \end{align*} Using the two one-sided estimates gives \begin{align*} \mathbb P(|X-a|\ge t)\le 2\,\mathbb P(X-X'\ge t)+2\,\mathbb P(X'-X\ge t). \end{align*} The events $\{X-X'\ge t\}$ and $\{X'-X\ge t\}$ are disjoint because $t>0$, and their union is contained in $\{|X-X'|\ge t\}$. Therefore \begin{align*} \mathbb P(X-X'\ge t)+\mathbb P(X'-X\ge t) \le \mathbb P(|X-X'|\ge t). \end{align*} Combining the last two displays gives \begin{align*} \mathbb P(|X-a|\ge t)\le 2\,\mathbb P(|X-X'|\ge t), \end{align*} which is the desired inequality. [/step]