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 Answer 1

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.

$\endgroup$
1
  • $\begingroup$ 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, 2018 at 1:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .