熵的公理化定义
熵的公理化定義的提出
若對稱函數序列Hm(p1,p2,...,pm)滿足下列性質(1)標準化:H21212=1(2)連續性:H2(p,1?p)為p的連續函數(3)組合法則:Hm(p1,p2,...,pm)=Hm?1(p1+p2,...,pm)+(p1+p2)H2(p1p1+p2,p2p1+p2)\begin{array}{l} 若對稱函數序列{H_m}({p_1},{p_2},...,{p_m})滿足下列性質\\ (1)標準化:{H_2}\frac{1}{2}\frac{1}{2}{\rm{ = }}1\\ (2)連續性:{H_2}(p,1 - p)為p的連續函數\\ (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}若對稱函數序列Hm?(p1?,p2?,...,pm?)滿足下列性質(1)標準化:H2?21?21?=1(2)連續性:H2?(p,1?p)為p的連續函數(3)組合法則: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
由上述熵的性質可做如下推導:
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}X~U,g(MN)=fMN(MN1?,MN1?...,MN1?)??=f(M1?,M1?,...,M1?)+i=1∑M?M1?f(N1?,N1?,...,N1?)=g(M)+g(N)?
g(sm)≤g(tn)<g(sm+1)?mg(s)≤ng(t)<(m+1)g(s)mn≤g(t)g(s)<m+1n                  ?∣mn?g(t)g(s)∣≤1nnlog?s≤nlog?t<(m+1)log?s∣mn?nlog?tnlog?s∣≤1n                      ?∣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}) < g({s^{m + 1}}) \Rightarrow mg(s) \le ng(t) < (m + 1)g(s)\\ \frac{m}{n} \le \frac{{g(t)}}{{g(s)}} < \frac{{m + 1}}{n}\;\;\;\;\;\;\;\;\; \Rightarrow |\frac{m}{n} - \frac{{g(t)}}{{g(s)}}| \le \frac{1}{n}\\ n\log s \le n\log t < (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?nlogs≤nlogt<(m+1)logs∣nm??nlogsnlogt?∣≤n1??∣g(s)g(t)??nlogsnlogt?∣≤n2?n→∞,g(s)g(t)?→logslogt??g(x)=αlog(x)p(n)=Mmn???
進一步推出
∴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=1∑M?Mmn??g(mn?)
f(p1,p2,...,pN)=g(M)?∑i=1Mpig(mi)                                              =αlog?(M)?∑i=1Mpiαlog?(mi)                                              =?α∑i=1Mpi[log?(mi)?log?(M)]                                              =?α∑pilog?pi由H(12,12)=1,可取常數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\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=1∑M?pi?g(mi?)=αlog(M)?i=1∑M?pi?αlog(mi?)=?αi=1∑M?pi?[log(mi?)?log(M)]=?α∑pi?logpi?由H(21?,21?)=1,可取常數α=1因此熵的形式可以寫成Hm?(p1?,p2?,...,pm?)=?i=1∑m?pi?logpi?,m=2,3,...?
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有關信息量的描述形式的推導
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我們知道用概率描述信息量的時候,滿足以下條件:
(1) 事件發生的概率越大,信息量越小;
(2) 事件發生的概率越小,信息量越大;
(3) 兩個事件同時發生的信息量等于兩個事件的信息量之和。
用數學語言,描述為
設f(x)為事件A發生時所攜帶的信息量,x為事件A發生的概率,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}x→∞lim?f(x)=+∞,f(1)=0,f(x1?x2?)=f(x1?)+f(x2?),x1?,x2?∈(0,1]f(xy)=f(x)+f(y)?
我們對上述方程進行求解
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                                                                                                                                    =∫x1f(t)+f(1+dtt)?f(t)dtdt                                                                                                                                    =∫x1f(1+dtt)?f(1)dtdt                                                                                                                                    =∫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'(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'(1)dt \Rightarrow 0 - f(x) = f'(1)\int_x^1 {\frac{1}{t}} f'(1)dt\\ \Rightarrow f(x) = - f'(1)\ln x\\ \Rightarrow f(x) = - \frac{{f'(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)?dtΔt→0,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?
其實這個縮放系數可以理解為信息量的單位,不影響信息量的本質計算。因此,我們把信息量的函數表達式寫成取對數求和的形式。
總結
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