Largest Variance on an Interval

Since $|X|\le 1$, we have $X^2\le 1$ and therefore

The bound is attained. Take $\mathbb{P}(X=1)=\mathbb{P}(X=-1)=\tfrac12$. Then $\mathbb{E}[X]=0$ and $X^2=1$, so $\operatorname{Var}(X)=1$.

The bound is attained only by this distribution. Equality needs $\mathbb{E}[X]=0$ and $\mathbb{E}[X^2]=1$. Since $1-X^2\ge 0$ and has mean $0$, $|X|=1$ almost surely. Then $\mathbb{E}[X]=0$ forces equal mass at $\pm1$.

Generalisation. If $X\in[a,b]$, then

This is Popoviciu’s inequality. Equality holds for mass $\tfrac12$ at each endpoint.