Is the statistical standard deviation of elements of x=(x1,x2,...,xn) a convex function?

Question

For $x\in R^n$, is the statistical standard deviation of its elements, formulated as below, a convex function?

$(\frac{1}{n} \sum_{i=1}^n x_i^2 - (\frac{1}{n}\sum_{i=1}^n x_i)^2)^{\frac{1}{2}}$

Answer 1

Let $\sigma : \Bbb R^n\to \Bbb R$ be the standard deviation. Then, $\sigma$ is convex if and only if for each $x, y\in \Bbb R^n$ we have that $$\sigma(tx+(1-t)y)\leq t\sigma(x)+(1-t)\sigma(y) $$

You just need to verify if this relation holds.

Answer 2

Using matrix-vector notation: $$ \begin{aligned} f(x) &= \left(\frac{1}{n}x^Tx - \left(\frac{1}{n}x^T 1\right)^2 \right)^{1/2} \\ &= \left(\frac{1}{n}\left(x- \frac{1}{n}1^Tx \ 1\right)^T \left(x- \frac{1}{n}1^Tx \ 1\right) \right)^{1/2} \\ &= \frac{1}{\sqrt{n}} \left\|\left(I- \frac{1}{n}11^T\right)x \right\|_2 \\ &= \frac{1}{\sqrt{n}} \left\| (I-P) x \right\|_2 \end{aligned} $$ Where $P=\frac{1}{n}11^T$ is the idempotent projector to the ones vector. So $f$ is a composition of a linear map with the $\ell_2$ norm, so it's convex by the triangle inequality. In words, $f$ is proportional to the length of the component of $x$ which is orthogonal to the ones vector.