Prove that $H(X) = H_0(X)$ when all characters are equally likely (uniform distribution).

Question

Accepted Answer

When $P(x_i) = \frac{1}{n}$ for all $i$, plugging into the entropy formula simplifies to $\log_2(n) = H_0(X)$.
Proof:
Start with the entropy formula and substitute $P(x_i) = \frac{1}{n}$:
$$H(X) = -\sum_{i=1}^{n} P(x_i) \cdot \log_2(P(x_i))$$
$$= -\sum_{i=1}^{n} \frac{1}{n} \cdot \log_2\left(\frac{1}{n}\right)$$
$$= -n \cdot \frac{1}{n} \cdot \log_2\left(\frac{1}{n}\right)$$
$$= -\log_2\left(\frac{1}{n}\right)$$
$$= \log_2(n) = H_0(X) \quad \blacksquare$$
Intuition: When all outcomes are equally uncertain, you need the maximum number of bits to encode which one occurred. Any deviation from uniformity reduces uncertainty (and therefore entropy) because some outcomes become more predictable.