Range of variance If you have a random variable which takes on values in the range $[a, b]$, is it necessary that the variance be in the range $[a, b]$?
What is the range of variance in general? 
I feel like this is important to understanding it intuitively as a measure of ``how spread out the data is".
 A: For $[a,b]=[0,1]$, this question already has some great answers here:


*

*How to prove a random variable taking values in $[0,1]$ range has variance no larger than $\frac{1}{4}$?

*variance inequality

*Variance of random variable
Now for the general case, start with $X$ supported on $[0,1]$ and shift it to $Y=a+(b-a)X$ supported on $[a,b]$; then $\mathrm{Var}(Y)=\mathrm{Var}(X+(b-a)X)=\mathrm{Var}((b-a)X)=(b-a)^2\mathrm{Var}(X)$. So we have $0\leq \mathrm{Var}(Y)\leq \frac14(b-a)^2$.
A: Suppose the range of $X$ is the interval $[a,b]$. For any constant $c$, let $Y=X+c$. The variance of $Y$ is the same as the variance of $X$, but the range of $Y$ is now $[a+c,b+c]$. 
We can therefore shift our random variable to the interval $[0,b-a$ without changing the variance. Now since the mean $\mu$ is in the interval, and the variance is $E(X-\mu)^2$, it is clear that the variance is bounded. 
A: No, but close.  If a random variable takes on values in the range $[a,b]$ the maximum variance is achieved when it is equal to $a$ half the time and $b$ the other half, so we can the variance is in the range $[0,(\frac{b-a}{2})^2]$.  On the other hand, a random variable not constrained to an interval may not have a finite variance.
