Given a simple image of size 10 × 10, whose histogram is depicted by symbols…

Concept: For a discrete probability distribution over n mutually exclusive outcomes with probabilities p1, p2, …, pn satisfying Σpi = 1, the first-order (Shannon) entropy H = −Σ pi log pi is a strictly concave function of the pi. By Jensen's inequality (equivalently, via a Lagrange multiplier on the constraint Σpi = 1), H attains its unique maximum precisely when every outcome is equally likely, i.e. pi = 1/n for every i, giving Hmax = log n.

Application: Here there are four symbols (n = 4) with probabilities a, b, c, d, so the working is:

1. Maximize H (any log base gives the same maximizer, since changing base only rescales H by a positive constant) subject to a + b + c + d = 1 by forming the Lagrangian L = −Σ p ln p + λ(Σp − 1).

2. Differentiate with respect to each probability p (p being a, b, c, or d): ∂L/∂p = −(ln p + 1) + λ = 0, so ln p is identical for every symbol — every p must be equal.

3. With four symbols, the common value is p = 1/4, so a = b = c = d = 1/4.

4. The maximum entropy value is H = −4·(1/4)·log2(1/4) = log24 = 2 bits.

Cross-check: Computing the entropy for all four candidate distributions independently confirms the theoretical result:

Therefore, the first-order estimate of image entropy is maximum when a = b = c = d = 1/4.