3
$\begingroup$

I'm trying to show that there is no non-constant random variable with finite mean $\mu$ such that $$P(|X-\mu|\geq x)=\frac{\sigma^2}{x^2}$$ for all $x>0$. I know that the equality can be achieved for a fixed $x$ with random variables with zero mean and unit variance, but I'm not sure how to tackle this problem.

I noticed that if the equality were to happen, for $x<\sigma$, then $P(|X-\mu|\geq x)=\frac{\sigma^2}{x^2}>1$, which is a contradcition as $P$ is a probability.

Is that the only source of contradiction here, or is there something deeper that must be recognized and be proven?

I would appreciate more feedback on my approach or a push in the right direction for another approach!

$\endgroup$
1
$\begingroup$

Yes. Equality is impossible when $ 0 < x < \sigma$, so no distribution with $\sigma \ne 0$ can satisfy this for all $x < 0$.

It's also worth noting that if we were to assume there is such a distribution with finite $\sigma$, it would have to satisfy $$ \frac{d}{dx}\left[\frac{\sigma^2}{x^2} \right]=-2\frac{\sigma^2}{x^3} = \frac{d}{dx} P(|X-\mu|\ge x) = \frac{d}{dx}\left[\int_{-\infty}^{\mu-x} p_X(x)dx + \int_{\mu+x}^\infty p_X(x)dx\right] =-p_X(\mu+x) - p_X(\mu - x), $$ which implies $$ \frac{p_X(\mu+x) + p_X(\mu - x)}{2} = \frac{\sigma^2}{x^3} $$ And any distribution that falls off as $x^{-3}$ has $\sigma = \infty$. So it definitely isn't going to work.

$\endgroup$
  • $\begingroup$ So it suffices to simply note the fact that I mentioned above, and the proof is complete...? Maybe I'm being a doubting Thomas, but I felt that it was too simple for that to be the answer... $\endgroup$ – Mog Oct 31 '18 at 1:58

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.