Tuesday, April 24, 2018

Some characterization of the Shannon entropy

Lemma. Let $f : \mathbb{Z}^+ \to \mathbb{R}$ satisfy the following two conditions

  • $f(mn) = f(m)+f(n)$ for all $m, n \in \mathbb{Z}^+$,
  • $\lim_{n\to\infty} (f(n+1) - f(n)) = 0$.
Then there exists $c \in \mathbb{R}$ such that $f(n) = c\log n$ for all $n \in \mathbb{Z}^+$.

Proof. Define $\delta_i = \max\{ |f(n+1) - f(n)| : 2^i \leq n < 2^{i+1} \}$. By the assumption, we know that $\delta_i \to 0$ as $i\to\infty$. Now let $n \geq 2$, and for each $k \geq 1$ we write $$ n^k = \sum_{j=0}^{L_k} a_j 2^j, \qquad a_j \in \{0, 1\} \quad \text{and} \quad a_{L_k} = 1. $$ Then by using the fact that $f(a_{L_k}2^{L_k}) = L_k f(2)$, we can write \begin{align*} \left| \frac{f(n^k)}{L_k} - f(2) \right| &\leq \frac{1}{L_k} \sum_{l=0}^{L_k-1} \left| f\left( \sum_{j=l}^{L_k} a_j 2^k \right) - f\left( \sum_{j=l+1}^{L_k} a_j 2^k \right) \right| \\ &= \frac{1}{L_k} \sum_{l=0}^{L_k-1} \left| f\left( \sum_{j=l}^{L_k} a_j 2^{k-l} \right) - f\left( \sum_{j=l+1}^{L_k} a_j 2^{k-l} \right) \right| \\ &\leq \frac{1}{L_k} \sum_{l=0}^{L_k-1} \delta_{L_k - l} = \frac{1}{L_k} \sum_{i=1}^{L_k} \delta_{i}. \end{align*} Since $L_k = \lfloor k \log_2 n \rfloor$ tends to $\infty$ as $k\to\infty$, by Stolz–Cesàro theorem, this bound converges to $0$. So $$ f(2) = \lim_{k\to\infty} \frac{f(n^k)}{L_k} = \lim_{k\to\infty} \frac{k f(n)}{\lfloor k \log_2 n \rfloor} = \frac{f(n)}{\log_2 n}, $$ and $f(n) = c \log n$ with $c = f(2)/\log 2$. When $n = 1$, we have $f(1) = f(1^2) = 2f(1)$ and hence $f(1) = 0 = c \log 1$, completing the proof. ////

This lemma is the key step for proving the theorem of Faddeev. To state this theorem, let $$\Delta^n = \{ (p_1, \cdots, p_n) \in [0, \infty)^n : p_1 + \cdots + p_n = 1 \}$$ and $\mathtt{Prob} = \bigcup_{n=1}^{\infty} \Delta^n$ the set of all probability vectors.

Theorem. (Faddeev)[1] Let $H : \mathtt{Prob} \to [0, \infty)$ satisfy the following properties

  1. $H$ is a symmetric function on each $\Delta^n$.
  2. $H$ is continuous on $\Delta^2$.
  3. $H(\frac{1}{2}, \frac{1}{2}) = 1$.
  4. $H(t p_1, (1-t)p_1, p_2, \cdots, p_n) = H(p_1, \cdots, p_n) + p_1 H(t, 1-t)$.
Then $H$ is the Shannon entropy: $$ H(p_1, \cdots, p_n) = - \sum_{i=1}^{n} p_i \log_2 p_i. $$

Proof. We extend $H$ onto the cone $\{ \lambda p : p \in \mathtt{Prob}, \lambda > 0 \}$ by setting $$H(\lambda p) = \lambda H(p), \qquad \forall p \in \mathtt{Prob}, \lambda > 0.$$ Then the condition 4 can be rephrased as $$H(p_0, p_1, p_2, \cdots, p_n) = H(p_0+p_1, p_2, \cdots, p_n) + H(p_0, p_1). \tag{1}$$ By applying $\text{(1)}$ repeatedly, we find that \begin{align*} H(p_1, \cdots, p_n) &= \sum_{k=2}^{n} H(p_1 + \cdots + p_{k-1}, p_k). \end{align*} This tells that $H$ is continuous on each $\Delta^n$ as well. By the same argument, we also find that \begin{align*} &H(p_1, \cdots, p_n, q_1, \cdots, q_m) \\ &= H(p_1+\cdots+p_n, q_1, \cdots, q_m) + \sum_{k=2}^{n} H(p_1 + \cdots + p_{k-1}, p_k) \\ &= H(p_1+\cdots+p_n, q_1, \cdots, q_m) + H(p_1, \cdots, p_n), \tag{2} \end{align*} which generalizes the property $\text{(1)}$. Now define $f(n) = H(\frac{1}{n}, \cdots, \frac{1}{n})$. By applying $\text{(2)}$, we obtain $$ f(mn) = f(m) + f(n). $$ Next we show that $f(n+1) - f(n) \to 0$ as $n\to\infty$. In view of Stolz–Cesàro theorem together with $f(2^n)/2^n = n/2^n \to 0$, we know that the only candidate for the limit is zero. Also, by $\text{(2)}$ we easily find that $$(n+1)f(n+1) = H(n,1) + nf(n).$$ So $nf(n)$ is increasing and $$ 0 \leq nf(n) \leq 2^{\lceil \log_2 n\rceil } f(2^{\lceil \log_2 n\rceil }) = 2^{\lceil \log_2 n\rceil } \lceil \log_2 n\rceil. $$ This tells that $ f(n) = \mathcal{O}(\log n)$ and hence $$ f(n+1) - f(n) = H\left(\frac{n}{n+1}, \frac{1}{n}\right) - \frac{1}{n+1}f(n) $$ converges as $n\to\infty$ by the continuity of $H$ on $\Delta^2$. So this limit must be zero, and by the lemma, $f(n) = \log_2 n$. Now if $(\frac{a_1}{N}, \cdots, \frac{a_n}{N}) \in \Delta^n$, then by $\text{(2)}$, \begin{align*} H\left( \frac{a_1}{N}, \cdots, \frac{a_n}{N} \right) &= f(N) - \sum_{k=1}^{N} \frac{a_k}{N} f(a_k) \\ &= - \sum_{k=1}^{N} \frac{a_k}{N} \log_2 \frac{a_k}{N} \end{align*} and hence $H$ equals the Shannon entropy for probability vectors with rational coordinates. Then the general case follows by the continuity and the proof is complete.


  • [1] Rényi, Alfréd. On measures of entropy and information. HUNGARIAN ACADEMY OF SCIENCES Budapest Hungary, 1961.