I was reading about the Markov inequality and tried to see if I could prove it for a uniform distribution. So say we have $X\sim U(a,b)$ and we are trying to prove

$$\mathop{\mathbb{P}}(X\ge \lambda)\le \frac{\mathop{\mathbb{E}}(X)}{\lambda}$$

So the right hand side is just


and I thought the left hand side should be

$$\int_{\lambda}^{b}\frac{1}{b-a}dx = \frac{b-\lambda}{b-a}$$

So now I would have to prove that

$$\frac{b-\lambda}{b-a} \le \frac{a+b}{2\lambda}$$

So if it's fair to assume (is it?) that $a< \lambda <b$, then after a bit of rearranging we have

$$b \lambda-\lambda ^2 \le \frac{b^2+a^2}{2}$$

and it is clear to me that $b^2+a^2$ is larger than $b \lambda-\lambda ^2$ but I don't see how to prove that it is more than twice as large.

  • 1
    $\begingroup$ Why don't you get (b-a)(b+a)= b$^2$-a$^2$? $\endgroup$ – Michael R. Chernick Jul 26 '17 at 3:10
  • $\begingroup$ @MichaelChernick oh you're right! But that means that the answer by Glen_b is also wrong as he was starting after my error... $\endgroup$ – Dan Nov 8 '17 at 15:44

As the sum of three squares, $(b-\lambda)^2+a^2+\lambda^2\geq 0$. (Hopefully that's obvious.)

Expand and rearrange so that terms in $\lambda$ are on the other side. Presumably you can see how to proceed from there.

  • 1
    $\begingroup$ Thanks, I understand how this proves it. However, it is not obvious to me how you chose that sum of three squares inequality in the first place. i.e. how did you 'know' to start there? $\endgroup$ – Dan Jul 25 '17 at 23:59
  • 2
    $\begingroup$ To get that, I started with $b^2+a^2\geq 2b\lambda-2\lambda^2$, put everything on the LHS, recognized the possibility to make $(b-\lambda)^2$ and then that the remaining terms would also be squares, which would clearly be $\geq 0$ as required. When you're trying to show something is non-negative, completing the square is an obvious thing to do. Ultimately the $2b\lambda$ term is what drives everything there. $\endgroup$ – Glen_b -Reinstate Monica Jul 26 '17 at 0:03

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.