# Show that there is no non-constant random variable such that Chebyshev Inequality becomes an equality for all $x>0$

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!

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.