Why is $\operatorname{Var}[X]≤(b−a)^2/4$ if $X$ is a random variable with values between $[a,b]$

Question

I don't know how to start here. I tried to proof it with $$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$$ but this wasn't a good idea in the end.

Answer 1

If X has a uniform distribution $\mathcal{U}([a,b])$ then

\begin{equation} var(X) = \frac{(b-a)^{2}}{12} \leq \frac{(b-a)^{2}}{4} \end{equation}

Answer 2

Ask yourself how you would distribute probability mass on the interval $[a,b]$ so as to make the probability mass as much "spread out" as possible. Intuitively, variance measures the "spread-out-ness" of a random variable (or distribution). It should be clear that if some probability mass is inside the interval (in $(a,b)$) then you can increase the spread by moving some of it against $a$, and some against $b$. So, for the maximum all mass should be at the interval ends $a$, $b$.

So assume that $\DeclareMathOperator{\P}{\mathbb{P}} \P(X=a)=p, \P(X=b)=1-p$, the you can calculate that the variance of $X$ is $$ p(1-p)(b^2-a^2) $$ and by maximizing that in $p$ you will get your result.

Answer 3

Let $Y$=$X-\frac{a+b}{2}$. Since they differ by a constant, $\operatorname{Var}(Y)=\operatorname{Var}(X)$. Since $|Y|$ is always less than or equal to $|\frac{b-a}{2}|$, we have $Y^2$ is always less than $\frac{(b-a)^2}{4}$. Thus $\operatorname{Var}(X)=\operatorname{Var}(Y)=E(Y^2)-(EY)^2\le E(Y^2)\le \frac{(b-a)^2}{4}$. QED