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Let us say we have $n$ random values $x_i$ chosen from an interval $[a,b]$. What would be the maximum possible variance $Var(X)=\frac{1}{n}\sum_{i=1}^n(x_i-\bar{x})^2$?

The number $(b-a)/2$ sounds plausible, can we prove this guess?

Easy
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  • It is $(b-a)^2/4$. (Though if $n$ is odd, then the highest attainable value is $\frac{n-1}{n}\cdot\frac{(b-a)^2}{4}$.) Have a read of this page: https://stats.stackexchange.com/questions/350270/what-are-the-minimum-and-maximum-values-of-variance. – Minus One-Twelfth May 07 '19 at 07:59

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This is not a rigorous proof (for that have a look here) but can be useful anyhow.

Also it illustrates how handsome it can be to standardize.

Personally I dislike parameters and always seek for possibilities to get rid of them.

First let's solve this for $a=-1, b=1$ and $n$ is even.

Purely on intuition I state that the variance will take maximal value if the $x_i$ take the extreme values $1,-1$ with equal frequency.

So e.g. if $x_i=-1$ if $i$ is even and $x_i=1$ if $i$ is odd.

Then $\overline x=0$ and $\mathsf{Var}(X)=1$.

If $y_i$ is chosen from $[a,b]$ then we take $y_i=\frac12(b-a)x_i+\frac12(a+b)$ and find easily that $\mathsf{Var}(Y)=\frac14(b-a)^2$.

If $n$ is odd then (again purely on intuition) the same idea except that one of $x_i$ takes value $0$.

drhab
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