# Maximum of the Variance Function for Given Set of Bounded Numbers

Let $\mathbf{x}$ be a vector of $n$ numbers in the range $[0, p]$, where $p$ is a positive real number.

What's is the maximum of the variance function of this $n$ numbers?

The variance of the vector $\mathbf{x}$ is given by:

$$\operatorname{var} (\mathbf{x}) = \frac{1}{n} \sum_{i=1}^{n} {\left( {x}_{i} - \overline{\mathbf{x}} \right )}^2$$

Where the mean $\overline{\mathbf{x}}$ is given by:

$$\overline{\mathbf{x}} = \frac{1}{n} \sum_{i=1}^n x_i .$$

Thank You.

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What is the variance function of a set? The maximum with respect to what? Please try to rephrase your question using more standard terminology. – joriki Nov 17 '11 at 13:44
If $X$ takes on values in an interval of length $c$, then its variance is bounded above by $c^2/4$ with equality when half the probability mass is at one end of the interval and half at the other (assuming the interval is closed at both ends; else we have strict inequality) – Dilip Sarwate Nov 17 '11 at 13:47
@DilipSarwate, Does it work just like Entropy? What if I have only 3 samples in the range [0 1], which variance could they achieve at most? It seems you neglected the factor of how many samples there are. – Drazick Nov 17 '11 at 14:24
@joriki, I've updated the question. Thank You. – Drazick Nov 17 '11 at 14:25
"What if I have only 3 samples in the range [0 1], which variance could they achieve at most?" The maximum is still bounded above by $c^2/4$ which is $1/4$ in this instance. The bound cannot be attained with equality in this instance but that does not invalidate the bound. For odd $n$ and $c = 1$, putting one point at the center (mean) and the rest at the end points gives $\frac{1}{4}\times \frac{n-1}{n}$ which, when $n$ is large, is close enough to the upper bound of $\frac{1}{4}$for gummint purposes. – Dilip Sarwate Nov 17 '11 at 15:50

Since $x_i \leq c$, $\displaystyle \sum_i x_i^2 = \sum_i x_i\cdot x_i \leq \sum_i c\cdot x_i = cn\bar{x}.$ Note also that $0 \leq \bar{x} \leq c$. Then, \begin{align*} n\cdot \text{var}(\mathbf{x}) &= \sum_i (x_i - \bar{x})^2= \sum_i x_i^2 - 2x_i\bar{x} + \bar{x}^2\\ &= \sum_i x_i^2 - 2\bar{x}\sum_i x_i + n\bar{x}^2= \sum_i x_i^2 - n\bar{x}^2\\ &\leq cn\bar{x} - n\bar{x}^2 = n\bar{x}(c-\bar{x}) \end{align*} and thus $$\text{var}(\mathbf{x}) \leq \bar{x}(c-\bar{x}) \leq \frac{c^2}{4}.$$
The result $\text{var}(X) \leq \frac{c^2}{4}$ also applies to random variables taking on values in $[0,c]$, and, as my first comment on the question says, putting half the mass at $0$ and the other half at $c$ gives the maximal variance of $c^2/4$. For the vector $\mathbf x$, if $n$ is even, the maximal variance $c^2/4$ occurs when $n/2$ of the $x_i$ have value $0$ and the rest have value $c$. Someone else posted an answer -- it has since been deleted -- which said the same thing and added that if $n$ is odd, the variance is maximized when $(n+1)/2$ of the $x_i$ have value $0$ and $(n-1)/2$ have value $c$, or vice versa. This gives a variance of $(c^2/4)\cdot(n^2-1)/n^2$ which is slightly smaller than $c^2/4$. Putting the "extra" point at $c/2$ instead of at an endpoint gives a slightly smaller variance of $(c^2/4)\cdot(n-1)/n$, but both choices have variance approaching $c^2/4$ asymptotically as $n \to \infty$.