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I have Chebyshev’s inequality $$ P\left(\big|X-E[X]\big|\geq\epsilon\right)\leq\frac{V[X]}{\epsilon^2}, $$ where $E[X]$ is the expected value and $V[X]$ the variance.

Given I know that the median is $M$, $E[X]=\mu$ and $E[X^2]<\infty$, I have to show that: $$ |M-\mu|\leq\sqrt{2V[X]} $$


I thought about setting $\epsilon=\frac{1}{2}$, as I then will obtain the right side: $$ \sqrt{P\left(\big|X-\mu\big|\geq\frac{1}{2}\right)}\leq\sqrt{2V[X]}, $$ however, this leads to a blind end for me.


Then I thought about wanting to use the information about the median to set: $$ \frac{V[X]}{\epsilon^2}=\frac{1}{2}\quad\quad\Rightarrow\quad\quad \epsilon=\sqrt{2V[X]}, $$ and I will then get: $$ P\left(\big|X-\mu\big|\geq\sqrt{2V[X]}\right)\leq\frac{1}{2}, $$ where I thought I could use the median, when I have $\frac{1}{2}$ on the right side. However, I'm still quite stuck.

Any help regarding this inequality will be greatly appreciated.

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Note that for any real number $y$, $|X-y|\geq |M-y|$ with probability at least $\frac{1}{2}$. Now you can apply the thing you showed to $y=\mathbb{E}X.$

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  • $\begingroup$ Btw, I'm assuming you're doing Stok2 at Uni Copenhagen: Best of luck. There's a different solution to the exercise as far as I recall, which also uses the Chebyshev Inequality. $\endgroup$ – WoolierThanThou Sep 1 at 10:23
  • $\begingroup$ You are correct, I'm attending Stok2. Thank you for the help $\endgroup$ – Frederik Sep 1 at 10:35
  • $\begingroup$ Do you refer to the first or the second idea I had above? $\endgroup$ – Frederik Sep 1 at 10:49

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