# Analytic center

Given the set, $LaTeX: \{x \in X : g(x) \ge 0 \}$, which we assume is non-empty and compact, such that $LaTeX: g$ is concave on $LaTeX: X$, with a non-empty strict interior, its analytic center is the (unique) solution to the maximum entropy problem:

$LaTeX: \max \left\{ \sum_i \ln(g_i(x)) : x \in X, g(x) > 0 \right\}.$

Note that the analytic center depends on how the set is defined -- i.e., the nature of $LaTeX: g$, rather than just the set, itself. For example, consider the analytic center of the box, $LaTeX: [0,1]^n$. One form is to have $LaTeX: 2n$ functions as: $LaTeX: \{ x : x \ge 0, 1-x \ge 0\}$. In this case, the analytic center is $LaTeX: x^*_j = y^*$ for all $LaTeX: j$, where $LaTeX: y^*$ is the solution to:

$LaTeX: \max \left\{ \ln(y) + \ln(1 - y) : 0 < y < 1 \right\}.$

Since $LaTeX: y^* = 1/2$, the analytical center is what we usually think of as the center of the box. However, the upper bounds could be defined by $LaTeX: 1 - (x_j)^p \ge 0$ for all $LaTeX: j$, where $LaTeX: p > 1$ (so $LaTeX: 1 - (x_j)^p$ is concave). This changes the functional definition, but not the set -- it's still the unit box. The analytic center is skewed towards the corner because the defining mathematical program is:

$LaTeX: \max \left\{ \ln(y) + \ln(1 - y^p) : 0 < y < 1 \right\}.$

The solution is $LaTeX: y^* = (1/(1+p))^{(1/p)}$, so the analytic center approaches $LaTeX: (1,1,...,1)$ as $LaTeX: p \rightarrow \infty$.