Support of Variance of Random IID Sample (Bounded)

Question

Let $X_i$, $i\in\{1, 2, ..., n\}$ be independent and identically distributed random variables with bounded support $[\alpha, \beta]$, with $\alpha,\beta \in \mathbb{R}$.

What is the support of the random variable that corresponds to the variance of this sample? That is, if $\mu:=\frac{\sum{x_i}}{n}$ and $\sigma^2:=\sum\frac{(x_i-\mu)^2}{n}$, what are the smallest and largest values that $\sigma^2$ can take?

The smallest value $\sigma^2$ can take is clearly $0$ if all $x_i$ are identical, but what's the largest, and how can you prove it's the largest? To be clear, I'm looking for this value to be written in terms of $\alpha, \beta,$ and $n$.

Answer 1

If half of the $x_i$ equal $\alpha$ and the other half equal $\beta$, then $\sigma^2 = \frac{(\beta - \alpha)^2}{4}$. I claim this is the largest possible sample variance.

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Given a sample $x_1, \ldots, x_n$, let $g(t) = \frac{1}{n}\sum (x_i - t)^2$. You can show that this function is minimized at $t = \mu$. Since $g(\mu) = \sigma^2$ we have $$g(t) \ge \sigma^2, \qquad \text{for all $t$}.$$

Now consider $t = (\alpha+\beta)/2$. \begin{align} \sigma^2 &\le g((\alpha+\beta)/2)\\ &= \frac{1}{n} \sum (x_i - (\alpha+\beta)/2)^2 \\ &= \frac{1}{4n} \sum (x_i - \alpha + x_i - \beta)^2 \\ &\le \frac{1}{4n} \sum (x_i - \alpha - (x_i - \beta))^2 \\ &= \frac{1}{4n} \sum (\beta - \alpha)^2 \\ &= \frac{(\beta - \alpha)^2}{4} \end{align} where the last inequality is due to $x_i - \alpha \ge 0$ and $x_i - \beta \le 0$.

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This answer was basically copied from a proof of an analogous result about the variance (not sample variance) of bounded random variables.

Answer 2

Largest value is obtained when all $X_i$ have the distribution $P(X_i=\alpha)=P(X_i=\beta)=\frac{1}{2}$. The mean for one varialbe is $\frac{\alpha+\beta}{2}$. The variance for one variable is $(\frac{\beta-\alpha}{2})^2$ so the variance for $n$ independent variables is $\frac{(\beta-\alpha)^2}{4n}$. The proof may be messy, but it is obvious.