Just when I thought I was starting to get my head around the multitudinous uses of convexity in statistics I was thrown by the following definition:

A function f over the interval (a,b) is convex if, for all choices of {x,y,z} satisfying a < x < y < z < b the determinant \[ \left| \begin{matrix} 1 & 1 & 1 \\ x & y & z \\ f(x) & f(y) & f(z) \end{matrix} \right| \] is non-negative.

After expanding the determinant and some algebraic twiddling I realised that this is just a very compact way of requiring that \[ \frac{z-y}{z-x} f(x) + \frac{y-x}{z-x}f(z) \geq f(y) \] which, after noticing that (z-y) + (y-x) = (z-x), of course is the more traditional way of saying a function is convex.

What’s neat about this determinant representation is that it extends nicely to what are known as k^{th}-order convex functions (ones whose derivatives up to order k are convex). Specifically, f is k-convex whenever \(\{x_i\}_{i=0}^{k+1}\) satisfy \(a < x_0 < \ldots < x_{k+1} < b\) and \[
\left|
\begin{matrix}
1 & \cdots & 1 \\
x_0 & \cdots & x_{k+1} \\
x_0^2 & \cdots & x_{k+1}^2 \\
\vdots & \ddots & \vdots \\
x_0^k & \cdots & x_{k+1}^k \\
f(x_1) & \cdots & f(x_{k+1})
\end{matrix}
\right|
\geq 0.
\]

While it is arguably less transparent than explicitly writing out all the convexity inequalities for each of the derivatives of f it certainly makes up for it with compactness.

Mark Reid February 4, 2008 Canberra, Australia