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.


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 $y=\mathbb{E}X.$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.