In Are values in a probability density function related to standard deviation? the OP wanted to know if for a normal distribution $$ \frac{P(x-\frac\sigma{2} < X < x+\frac\sigma{2} )}{\sigma} $$ was the same as $f(x)$, and of course this is not the case.
But the question got me to speculating: While this does not work for the normal distribution, is there any distribution for which this ill-concieved notion turns out to be true?
That is,
Prove there is no function on the $f(x): \Bbb{R} \mapsto \Bbb{R} $ such that $$\int_{-\infty}^\infty f(x)dx = 1 \\ \int_{-\infty}^\infty xf(x)dx = \mu \mbox{ with } -\infty < \mu < \infty \\ \int_{-\infty}^\infty (x-\mu)^2 f(x)dx = \sigma^2 \mbox{ with } 0 < \sigma < \infty $$ and for all $-\infty < x < \infty$, $$ f(x) = \int_{x-\sigma/2}^{x+\sigma/2}f(u)\,du $$ or show that such a function does exist.
