ThmDex – An index of mathematical definitions, results, and conjectures.
Entropy of a standard bernoulli number
Formulation 0
Let $X \in \text{Bernoulli}(1/2)$ be a D3999: Standard Bernoulli random boolean number.
Let $a \in (0, \infty) \setminus \{ 1 \}$ be a D5407: Positive real number.
Then \begin{equation} H_a(X) = \log_a 2 \end{equation}
Proofs
Proof 0
Let $X \in \text{Bernoulli}(1/2)$ be a D3999: Standard Bernoulli random boolean number.
Let $a \in (0, \infty) \setminus \{ 1 \}$ be a D5407: Positive real number.
Proceeding directly from the definitions and applying results
(i) R4826: Logarithm of a ratio
(ii) R4828: Logarithm of one is zero

we have \begin{equation} \begin{split} H_a(X) & = - \sum_{x \in \mathcal{X}} \mathbb{P}(X = x) \log_a \mathbb{P}(X = x) \\ & = - \sum_{x \in \{ 0, 1 \}} \mathbb{P}(X = x) \log_a \mathbb{P}(X = x) \\ & = - \mathbb{P}(X = 0) \log_a \mathbb{P}(X = 0) - \mathbb{P}(X = 1) \log_a \mathbb{P}(X = 1) \\ & = - \frac{1}{2} \log_a \frac{1}{2} - \frac{1}{2} \log_a \frac{1}{2} \\ & = - \log_a \frac{1}{2} \\ & = - (\log_a 1 - \log_a 2) \\ & = - \log_a 1 + \log_a 2 \\ & = - 0 + \log_a 2 \\ & = \log_a 2 \end{split} \end{equation}