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一文看懂信息熵的本质——谈谈自己对信息熵的理解
置顶 StudentWang_ 2020-04-24 10:33:39 3152 收藏 57 原力计划 分类专栏: 信息融合 文章标签: 信息论 信息熵 人工智能 版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。 本文链接: https://blog.csdn/qq_35624030/article/details/105704098 版权 信息融合 专栏收录该内容 2 篇文章 0 订阅 订阅专栏一、序言
初次看到信息熵的公式有很多不理解的地方,只知道信息熵如何进行计算,却不懂得公式背后的原理,我通过查阅了一些资料,加深了对信息熵的理解,现在将这些理解分享给大家。如有疑问欢迎评论,若对你有帮助,麻烦点个赞。未经允许、请勿转载。(本文适合只知道信息熵的公式,但是不明白其中原理的人进行阅读)
二、什么是信息熵
正如我们想要衡量某个物体的质量引入了克这个单位、我们想衡量时间,我们设计一秒钟这么长。香农老人家想要量化一条消息中带有的“信息量”的大小,提出了信息熵。
那么,首先明确一个问题,什么样的消息算作“信息量大”呢?什么样的消息又算作“信息量小呢”?举个例子昨天小明和我说:“今天罗志祥又和周扬青秀恩爱了!”,我就觉得这有啥的,他们天天秀恩爱。也就是说小明的这条消息并不能给我带来很大的信息量。
BUT今天小明和我说:”周扬青怒锤罗志祥!!!罗志祥人设崩塌!!!“,我就会很惊讶,因为这条消息给我的信息量很大。(类似的信息量很大的消息还有:小明告诉我今天太阳会从西边升起)
我们用信息熵来描述一个事件混乱程度的大小(一个事件我们一定知道结果,那么这个事件的混乱程度就是0;一个时间充满随机性,我们猜不到或者很难猜到结果,那么他的混乱度就很大)
引用下面一个在箱子里面摸球的例子,我们来更具体的了解信息熵。
(此例引自:Youtube的一个视频)
图左侧中有一个装有四个球的封闭箱子(这个箱子里面有三个红球、一个蓝球),现在我们从箱子中随机取出一个球,记录它的颜色后再放回箱子中,重复四次操作。如果你依次取出的序列为右上角所示(第一次取到红色、第二次取到红色、第三次取到红色、第四次取到蓝色),则你可以获得奖金;否则你就输了。大家用初中数学来算一算,我赢得奖金的概率是
P
(
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=
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4
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3
4
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4
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4
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27
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P(x)=\frac{3}{4}*\frac{3}{4}*\frac{3}{4}*\frac{1}{4}=\frac{27}{256}
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vertical-align: -0.68em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 1.32em;"><span style="top: -2.31em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">4</span></span></span><span style="top: -3.23em;"><span class="pstrut" style="height: 3em;"></span><span class="frac-line" style="border-bottom-width: 0.04em;"></span></span><span style="top: -3.67em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">3</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.68em;"></span></span></span></span><span class="mclose nulldelimiter"></span></span><span class="mspace" style="margin-right: 0.22em;"></span><span class="mbin">∗</span><span class="mspace" style="margin-right: 0.22em;"></span></span><span class="base"><span class="strut" style="height: 2em; vertical-align: -0.68em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 1.32em;"><span style="top: -2.31em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">4</span></span></span><span style="top: -3.23em;"><span class="pstrut" style="height: 3em;"></span><span class="frac-line" style="border-bottom-width: 0.04em;"></span></span><span style="top: -3.67em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">1</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.68em;"></span></span></span></span><span class="mclose nulldelimiter"></span></span><span class="mspace" style="margin-right: 0.27em;"></span><span class="mrel">=</span><span class="mspace" style="margin-right: 0.27em;"></span></span><span class="base"><span class="strut" style="height: 2em; vertical-align: -0.68em;"></span><span class="mord"><span class="mopen nulldelimiter"></span><span class="mfrac"><span class="vlist-t vlist-t2"><span class="vlist-r"><span class="vlist" style="height: 1.32em;"><span style="top: -2.31em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">2</span><span class="mord">5</span><span class="mord">6</span></span></span><span style="top: -3.23em;"><span class="pstrut" style="height: 3em;"></span><span class="frac-line" style="border-bottom-width: 0.04em;"></span></span><span style="top: -3.67em;"><span class="pstrut" style="height: 3em;"></span><span class="mord"><span class="mord">2</span><span class="mord">7</span></span></span></span><span class="vlist-s"></span></span><span class="vlist-r"><span class="vlist" style="height: 0.68em;"></span></span></span></span><span class="mclose nulldelimiter"></span></span></span></span></span></span></span><br> 现在明确了这个游戏的规则,那么我们分析一下如下几个箱子与赢得奖金的情况,分别计算一下获胜概率<br> <img alt="在这里插入图片描述" src="https://img-blog.csdnimg/20200423165753935.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1NjI0MDMw,size_16,color_FFFFFF,t_70"><br> <img alt="在这里插入图片描述" src="https://img-blog.csdnimg/20200423165836890.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1NjI0MDMw,size_16,color_FFFFFF,t_70"><br> <img alt="在这里插入图片描述" src="https://img-blog.csdnimg/20200423165853642.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1NjI0MDMw,size_16,color_FFFFFF,t_70"><br> (每个图的左边是这个箱子的红球与蓝球初始条件,右侧是获胜所要求的序列,大家自己算算这个概率和图上写的一样吗?)<br> <img alt="在这里插入图片描述" src="https://img-blog.csdnimg/20200423170009974.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1NjI0MDMw,size_16,color_FFFFFF,t_70"><br> 我们来分析一下这三种状态的游戏,第一次由于箱子内全是红球、该箱子的随机性很弱,也就是说还没摸我就知道结果了。这种状态带来的信息量很小;第三个箱子随机性很强,也就是说,对于结果我们充满未知。这个状态的信息量很大。我们算出的这个概率越接近1时候,这个信息熵应该越接近0;算出的这个概率越小的时候,信息熵反而应该越大。我们需要找到这么一个公式来满足这一点。(暂停思考一下,有哪些公式可以满足)<br> <strong>该可以很轻松的想到,用刚才P(winning)的概率取倒数可以吗?</strong> 答案是,只考虑刚才的问题,是可以的。但取倒数这个方式还是存在一些问题,比如第三个箱子取0.0625的倒数,算出来的这个信息量为16,如果这个P(winning)概率算出结果很小时,取倒数后就会变得非常大,所以我们认为这个乘法的规则并不是很好(换言之,我们希望曲线平滑一些)。香农发现log函数可以很好的解决这个问题。 因为log(a*b) = log(a)+log(b);而恰巧像抽箱子的独立事件P(AB) = P(A)*P(B)。【原因不止于此,后面我们还会再详细讨论】。<br> 所以香农取了一个-log(x)这么一个函数来表示某一状态的信息量大小,因为x是概率事件取[0,1],所以log(x)是一个递增的恒负的值,取一个负号,-log(x)是一个恒正、递减的函数,正好符合我们的预期。<br> <img alt="在这里插入图片描述" src="https://img-blog.csdnimg/2020042410083693.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1NjI0MDMw,size_16,color_FFFFFF,t_70"></p>
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