P r o j e c t s

N o t e s

Maximum of Variance

Suppose that a vector x in mathbb{R^n} consists of the elements which are in [0, c] for a positive real number c. Then
\begin{align}
\sum_{i=1}^n x_i^2 \leq \sum_{i=1}^n c x_i = c \sum_{i=1}^n x_i = c n \bar{x}
\end{align}
where bar{x} is the mean of x elements. Let Var(x) be the variance of x elements, then
\begin{align} Var(x) &= \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 = \frac{1}{n} \sum_{i=1}^n (x_i^2 - 2\bar{x} x_i + \bar{x}^2) \\ &= \frac{1}{n} \left\{ \sum_{i=1}^n x_i^2 - 2\bar{x} \sum_{i=1}^n x_i + n \bar{x}^2 \right\} = \frac{1}{n} \left\{ \sum_{i=1}^n x_i^2 - 2n \bar{x}^2 + n \bar{x}^2 \right\} \\ &= \frac{1}{n}\left\{ \sum_{i=1}^n x_i^2 - n \bar{x}^2 \right\} \leq \frac{1}{n} (c n \bar{x}  - n \bar{x}^2) = c \bar{x} - \bar{x}^2
\end{align}
If we define f(bar{x})=cbar{x}-bar{x}^2, then its extreme value is as follows:

f(\bar{x}) = c \bar{x} - \bar{x}^2 \\
f^{\prime}(\bar{x}) = c - 2\bar{x} = 0 \Rightarrow \bar{x} = \frac{c}{2} \\
f\left( \frac{c}{2} \right) = \frac{c^2}{2} - \frac{c^2}{4} = \frac{c^2}{4}

It yields the following boundary.
\begin{align}
Var(x) \leq c\bar{x} - \bar{x}^2 \leq \frac{c^2}{4}
\end{align}

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