Function that Satisfies Axioms of Uncertainty

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $n \in \N$ be a natural number.

Let $p_1, p_2, \dotsc, p_n$ be real numbers such that:

$\forall i \in \set {1, 2, \dotsc, n}: p_i \ge 0$
$\displaystyle \sum_{i \mathop = 1}^n p_i = 1$

Let $\map H {p_1, p_2, \ldots, p_n}$ be a real-valued function which satisfies the axioms of uncertainty.


Then:

$\map H {p_1, p_2, \ldots, p_n} = \displaystyle -\lambda \sum_{i \mathop = 1}^n p_i \log_b p_i$

where:

$\lambda \in \R_{>0}$
$b \in \R_{>1}$


Thus the uncertainty function satisfies these axioms.


Proof

Let $g: \Z \to \R$ be the mapping defined as:

$(1): \quad \map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n}$


Let $k \in \Z_{>0}$.

We have:

\(\displaystyle \map g {n^k}\) \(=\) \(\displaystyle \map g {n \times n^{k - 1} }\)
\(\displaystyle \) \(=\) \(\displaystyle \map g n + \map g {n^{k - 1} }\) Axiom 7
\(\displaystyle \) \(=\) \(\displaystyle \map g n + \map g n + \map g {n^{k - 2} }\) Axiom 7 again
\(\displaystyle \) \(=\) \(\displaystyle \underbrace {\map g n + \map g n + \dotsb + \map g n}_{\text {$k - 1$ times} } + \map g {n^{k - \paren {k - 1} } }\) after $k - 1$ applications of Axiom 7
\(\displaystyle \) \(=\) \(\displaystyle \underbrace {\map g n + \map g n + \dotsb + \map g n}_{\text {$k - 1$ times} } + \map g n\)
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle k \map g n\)


Let $r, s \in \R$ and $n \in \Z_{>0}$.

Let $m$ satisfy:

$(3): \quad r^m \le s^n \le r^{m + 1}$

From $(2)$ and Axiom 5:

$\map g {r^m} \le \map g {s^n} \le \map g {r^{m + 1} }$

and so:

$m \map g r \le n \map g s \le \paren {m + 1} \map g r$

Taking logarithms of $(3)$:

$m \ln r \le n \ln s \le \paren {m + 1} \ln r$

which holds because Logarithm is Strictly Increasing.

Hence:

$\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } \le \dfrac 1 n$

We have that $n$ is an arbitrary positive integer.

Hence the right hand side of the above can be made arbitrarily small.

Hence:

$\size {\dfrac {\map g s} {\map g r} - \dfrac {\map \ln s} {\map \ln r} } = 0$

and so:

$(4): \quad \dfrac {\map g s} {\map \ln s} = \dfrac {\map g r} {\map \ln r} = A$

for some constant $A$.

That is:

$(5): \quad \map g s = A \map \ln s$

for $s \in \Z$.


Let $p$ be a rational number such that $0 < p < 1$.

That is:

$p = \dfrac t n$

for some integers $t$ and $n$.

Thus we can set:

$q = 1 = p = \dfrac {n - t} n$

Axiom 8 gives:

$\map g n = \map H {\dfrac 1 n, \dfrac 1 n, \dotsc, \dfrac 1 n} = \map H {\dfrac 1 n, \dfrac {n - t} n} + \dfrac t n \map g t + \dfrac {n - t} n \map g {n - t}$

Using $(5)$ and gathering up terms:

$\map H {\dfrac 1 n, \dfrac {n - t} n} = A \paren {\dfrac t n} \ln \dfrac t n + A \paren {\dfrac {n - t} n} \ln \dfrac {n - t} n$

which gives:

$(6): \quad \map H {p, 1 - p} = A p \ln p + A \paren {1 - p} \, \map \ln {1 - p}$

for rational $p$.

Axiom 6 gives that $H$ is continuous.

Thus $(6)$ extends to all real $p$ such that $0 < p < 1$.


It remains to be demonstrated that:

$(7): \quad \map H {p_1, p_2, \ldots, p_N} = A \displaystyle \sum_{i \mathop = 1}^N p_i \ln p_i$

where:

$0 < p_i < 1$ for all $p_i$
$p_1 + p_2 + \dotsb + p_N$

for all $N \in \Z_{\ge 2}$.

We have demonstrated that $(7)$ is true for $N = 2$.

Assume the induction hypothesis that $(7)$ holds for $N = k$.

Consider:

$\map H {p_1, p_2, \ldots, p_k, p_{k + 1} }$

Let:

$p = p_1 + p_2 + \dotsb + p_k$
$q = p_{k + 1}$

and apply Axiom 8:

\(\displaystyle \map H {p_1, p_2, \ldots, p_k, p_{k + 1} }\) \(=\) \(\displaystyle \map H {p, q} + p \map H {\dfrac {p_1} p, \dfrac {p_2} p, \dotsc, \dfrac {p_k} p} + q \map H 1\)
\(\displaystyle \) \(=\) \(\displaystyle A p \ln p + A q \ln q + p A \sum_{i \mathop = 1}^k \dfrac {p_i} p \ln \dfrac {p_i} p\) from the induction hypothesis
\(\displaystyle \) \(=\) \(\displaystyle A p \ln p + A p_{k + 1} \ln p_{k + 1} + A \sum_{i \mathop = 1}^k p_i \paren {\ln p_i - \ln p}\) Difference of Logarithms
\(\displaystyle \) \(=\) \(\displaystyle A p \ln p + A p_{k + 1} \ln p_{k + 1} + A \sum_{i \mathop = 1}^k p_i \ln p_i - A p \ln p\) as $A \displaystyle \sum_{i \mathop = 1}^k p_i = p$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map H {p_1, p_2, \ldots, p_k, p_{k + 1} }\) \(=\) \(\displaystyle A \sum_{i \mathop = 1}^{k + 1} p_i \ln p_i\)

Hence the result from the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources