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熵的公理化定义

發(fā)布時(shí)間:2025/3/15 编程问答 33 豆豆
生活随笔 收集整理的這篇文章主要介紹了 熵的公理化定义 小編覺得挺不錯(cuò)的,現(xiàn)在分享給大家,幫大家做個(gè)參考.

熵的公理化定義的提出

若對(duì)稱函數(shù)序列Hm(p1,p2,...,pm)滿足下列性質(zhì)(1)標(biāo)準(zhǔn)化:H21212=1(2)連續(xù)性:H2(p,1?p)為p的連續(xù)函數(shù)(3)組合法則:Hm(p1,p2,...,pm)=Hm?1(p1+p2,...,pm)+(p1+p2)H2(p1p1+p2,p2p1+p2)\begin{array}{l} 若對(duì)稱函數(shù)序列{H_m}({p_1},{p_2},...,{p_m})滿足下列性質(zhì)\\ (1)標(biāo)準(zhǔn)化:{H_2}\frac{1}{2}\frac{1}{2}{\rm{ = }}1\\ (2)連續(xù)性:{H_2}(p,1 - p)為p的連續(xù)函數(shù)\\ (3)組合法則:{H_m}({p_1},{p_2},...,{p_m}){\rm{ = }}{H_{m{\rm{ - }}1}}({p_1} + {p_2},...,{p_m}) + ({p_1} + {p_2}){H_2}(\frac{{{p_1}}}{{{p_1} + {p_2}}},\frac{{{p_2}}}{{{p_1} + {p_2}}}) \end{array}對(duì)數(shù)Hm?(p1?,p2?,...,pm?)滿質(zhì)(1)標(biāo)準(zhǔn):H2?21?21?=1(2)續(xù):H2?(p,1?p)p續(xù)數(shù)(3)Hm?(p1?,p2?,...,pm?)=Hm?1?(p1?+p2?,...,pm?)+(p1?+p2?)H2?(p1?+p2?p1??,p1?+p2?p2??)?

熵的公理化定義的推導(dǎo)

由上述熵的性質(zhì)可做如下推導(dǎo):
X~U,g(MN)=f(1MN,1MN...,1MN)?MN=f(1M,1M,...,1M)+∑i=1M1Mf(1N,1N,...,1N)                                                                                                                  =g(M)+g(N)\begin{array}{l} X \sim U,g(MN) = f\underbrace {(\frac{1}{{MN}},\frac{1}{{MN}}...,\frac{1}{{MN}})}_{MN} = f(\frac{1}{M},\frac{1}{M},...,\frac{1}{M}) + \sum\limits_{i = 1}^M {\frac{1}{M}} f(\frac{1}{N},\frac{1}{N},...,\frac{1}{N})\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = g(M) + g(N) \end{array}XU,g(MN)=fMN(MN1?,MN1?...,MN1?)??=f(M1?,M1?,...,M1?)+i=1M?M1?f(N1?,N1?,...,N1?)=g(M)+g(N)?

g(sm)≤g(tn)&lt;g(sm+1)?mg(s)≤ng(t)&lt;(m+1)g(s)mn≤g(t)g(s)&lt;m+1n&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;?∣mn?g(t)g(s)∣≤1nnlog?s≤nlog?t&lt;(m+1)log?s∣mn?nlog?tnlog?s∣≤1n&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;?∣g(t)g(s)?nlog?tnlog?s∣≤2nn→∞,g(t)g(s)→log?tlog?s?g(x)=αlog?(x)p(n)=mnM\begin{array}{l} g({s^m}) \le g({t^n}) &lt; g({s^{m + 1}}) \Rightarrow mg(s) \le ng(t) &lt; (m + 1)g(s)\\ \frac{m}{n} \le \frac{{g(t)}}{{g(s)}} &lt; \frac{{m + 1}}{n}\;\;\;\;\;\;\;\;\; \Rightarrow |\frac{m}{n} - \frac{{g(t)}}{{g(s)}}| \le \frac{1}{n}\\ n\log s \le n\log t &lt; (m + 1)\log s\\ |\frac{m}{n} - \frac{{n\log t}}{{n\log s}}| \le \frac{1}{n}\;\;\;\;\;\;\;\;\;\;\; \Rightarrow |\frac{{g(t)}}{{g(s)}} - \frac{{n\log t}}{{n\log s}}| \le \frac{2}{n}\\ n \to \infty ,\frac{{g(t)}}{{g(s)}} \to \frac{{\log t}}{{\log s}} \Rightarrow g(x) = \alpha \log (x)\\ p(n) = \frac{{{m_n}}}{M} \end{array}g(sm)g(tn)<g(sm+1)?mg(s)ng(t)<(m+1)g(s)nm?g(s)g(t)?<nm+1??nm??g(s)g(t)?n1?nlogsnlogt<(m+1)logsnm??nlogsnlogt?n1??g(s)g(t)??nlogsnlogt?n2?n,g(s)g(t)?logslogt??g(x)=αlog(x)p(n)=Mmn???

進(jìn)一步推出

∴g(M)=f(1M,1M...,1M)?mn=f(p1,p2,...,pN)+∑i=1MmnMg(mn)\therefore g(M) = f\underbrace {(\frac{1}{M},\frac{1}{M}...,\frac{1}{M})}_{{m_n}} = f({p_1},{p_2},...,{p_N}) + \sum\limits_{i = 1}^M {\frac{{{m_n}}}{M}} g({m_n})g(M)=fmn?(M1?,M1?...,M1?)??=f(p1?,p2?,...,pN?)+i=1M?Mmn??g(mn?)
f(p1,p2,...,pN)=g(M)?∑i=1Mpig(mi)&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=αlog?(M)?∑i=1Mpiαlog?(mi)&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=?α∑i=1Mpi[log?(mi)?log?(M)]&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=?α∑pilog?pi由H(12,12)=1,可取常數(shù)α=1因此熵的形式可以寫成Hm(p1,p2,...,pm)=?∑i=1mpilog?pi,m=2,3,...\begin{array}{l} f({p_1},{p_2},...,{p_N}) = g(M) - \sum\limits_{i = 1}^M {{p_i}} g({m_i})\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \alpha \log (M) - \sum\limits_{i = 1}^M {{p_i}} \alpha \log ({m_i})\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm{ = }} - \alpha \sum\limits_{i = 1}^M {{p_i}} [\log ({m_i}) - \log (M)]\\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = - \alpha \sum {{p_i}\log {p_i}} \\ 由H(\frac{1}{2},\frac{1}{2}) = 1,可取常數(shù)\alpha {\rm{ = }}1\\ 因此熵的形式可以寫成{H_m}({p_1},{p_2},...,{p_m}) = - \sum\limits_{i = 1}^m {{p_i}\log {p_i},m = 2,3,...} \end{array}f(p1?,p2?,...,pN?)=g(M)?i=1M?pi?g(mi?)=αlog(M)?i=1M?pi?αlog(mi?)=?αi=1M?pi?[log(mi?)?log(M)]=?αpi?logpi?H(21?,21?)=1數(shù)α=1Hm?(p1?,p2?,...,pm?)=?i=1m?pi?logpi?,m=2,3,...?
**

有關(guān)信息量的描述形式的推導(dǎo)

**:
我們知道用概率描述信息量的時(shí)候,滿足以下條件:
(1) 事件發(fā)生的概率越大,信息量越小;
(2) 事件發(fā)生的概率越小,信息量越大;
(3) 兩個(gè)事件同時(shí)發(fā)生的信息量等于兩個(gè)事件的信息量之和。
用數(shù)學(xué)語言,描述為
設(shè)f(x)為事件A發(fā)生時(shí)所攜帶的信息量,x為事件A發(fā)生的概率,則
lim?x→∞f(x)=+∞,f(1)=0,f(x1x2)=f(x1)+f(x2),x1,x2∈(0,1]f(xy)=f(x)+f(y)\begin{array}{l} \mathop {\lim }\limits_{x \to \infty } f(x) = + \infty ,f(1) = 0,f({x_1}{x_2}) = f\left( {{x_1}} \right) + f\left( {{x_2}} \right),{x_1},{x_2} \in (0,1]\\ f(xy) = f(x) + f(y) \end{array}xlim?f(x)=+,f(1)=0,f(x1?x2?)=f(x1?)+f(x2?),x1?,x2?(0,1]f(xy)=f(x)+f(y)?

我們對(duì)上述方程進(jìn)行求解

x=y=1,f(1)=f(1)+f(1)?f(1)=0x = y = 1,f(1) = f(1) + f(1) \Rightarrow f(1) = 0x=y=1,f(1)=f(1)+f(1)?f(1)=0

由牛頓萊布尼茲公式可得

f(1)?f(x)=∫x1f′(t)dt=∫x1f(t+dt)?f(t)dtdt=∫x1f(tt+dtt)?f(t)dtdt&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=∫x1f(t)+f(1+dtt)?f(t)dtdt&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=∫x1f(1+dtt)?f(1)dtdt&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;&ThickSpace;=∫x11tf(1+dtt)?f(1)dttdtΔt→0,f(1)?f(x)=∫x11tf′(1)dt?0?f(x)=f′(1)∫x11tf′(1)dt?f(x)=?f′(1)ln?x?f(x)=?f′(1)log?aelog?ax\begin{array}{l} f(1) - f(x) = \int_x^1 {f&#x27;(t)dt = } \int_x^1 {\frac{{f(t + dt) - f(t)}}{{dt}}dt = } \int_x^1 {\frac{{f(t\frac{{t + dt}}{t}) - f(t)}}{{dt}}dt} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \int_x^1 {\frac{{f(t) + f(1 + \frac{{dt}}{t}) - f(t)}}{{dt}}dt} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \int_x^1 {\frac{{f(1 + \frac{{dt}}{t}) - f(1)}}{{dt}}dt} \\ \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \int_x^1 {\frac{1}{t}\frac{{f(1 + \frac{{dt}}{t}) - f(1)}}{{\frac{{dt}}{t}}}dt} \\ \Delta t \to 0,f(1) - f(x) = \int_x^1 {\frac{1}{t}} f&#x27;(1)dt \Rightarrow 0 - f(x) = f&#x27;(1)\int_x^1 {\frac{1}{t}} f&#x27;(1)dt\\ \Rightarrow f(x) = - f&#x27;(1)\ln x\\ \Rightarrow f(x) = - \frac{{f&#x27;(1)}}{{{{\log }_a}e}}{\log _a}x \end{array}f(1)?f(x)=x1?f(t)dt=x1?dtf(t+dt)?f(t)?dt=x1?dtf(ttt+dt?)?f(t)?dt=x1?dtf(t)+f(1+tdt?)?f(t)?dt=x1?dtf(1+tdt?)?f(1)?dt=x1?t1?tdt?f(1+tdt?)?f(1)?dtΔt0,f(1)?f(x)=x1?t1?f(1)dt?0?f(x)=f(1)x1?t1?f(1)dt?f(x)=?f(1)lnx?f(x)=?loga?ef(1)?loga?x?
其實(shí)這個(gè)縮放系數(shù)可以理解為信息量的單位,不影響信息量的本質(zhì)計(jì)算。因此,我們把信息量的函數(shù)表達(dá)式寫成取對(duì)數(shù)求和的形式。

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