交叉熵损失(Cross Entropy)求导

比眉伴天荒 2023-06-01 06:22 9阅读 0赞

Cross Entropy是分类问题中常见的一种损失函数，我们在之前的文章提到过[二值交叉熵的证明][Link 1]和[交叉熵的作用][Link 2]，下面解释一下交叉熵损失的求导。  
首先一个模型的最后一层神经元的输出记为 f 0 . . . f i f\_\{0\}...f\_\{i\} f0...fi，  
输出经过softmax激活之后记为 p 0 . . . p i p\_\{0\}...p\_\{i\} p0...pi，那么：  
 p i = e f i ∑ k = 0 C − 1 e f k p\_\{i\} = \\frac\{e^\{f\_\{i\}\}\}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\} pi=∑k=0C−1efkefi  
类别的实际标签记为 y 0 . . . y i y\_\{0\}...y\_\{i\} y0...yi，那么交叉熵损失L为：  
 L = − ∑ i = 0 C − 1 y i l o g p i L = -\\sum\_\{i=0\}^\{C-1\} y\_\{i\}log^\{p\_\{i\}\} L=−i=0∑C−1yilogpi  
上式中的 l o g log log是一种简写，为了后续的求导方便，一般我们认为 l o g log log的底是 e e e，即 l o g log log为 l n ln ln。  
那么 L L L对第 i i i个神经元的输出 f i f\_\{i\} fi求偏导 ∂ L ∂ f i \\frac\{\\partial L\}\{\\partial f\_\{i\}\} ∂fi∂L:  
根据复合函数求导原则：  
 ∂ L ∂ f i = ∑ j = 0 C − 1 ∂ L j ∂ p j ∂ p j ∂ f i \\frac\{\\partial L\}\{\\partial f\_\{i\}\} = \\sum\_\{j=0\}^\{C-1\} \\frac\{\\partial L\_\{j\}\}\{\\partial p\_\{j\}\}\\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} ∂fi∂L=j=0∑C−1∂pj∂Lj∂fi∂pj  
在这里需要说明，在softmax中我们使用了下标 i i i和 k k k，在交叉熵中使用了下标 i i i，但是这里的两个 i i i并不等价，因为softmax的分母中包含了每个神经元的输出 f f f，也就是激活后所有的 p p p对任意的 f i f\_\{i\} fi求偏导都不为0，同时 L L L中又包含了所有的 p p p，所以为了避免重复我们需要为 p p p引入一个新的下标 j j j， j j j有 0... C − 1 0...C-1 0...C−1这C种情况。  
那么依次求导：

∂ L j ∂ p j = ∂ ( − y j l o g p j ) ∂ ( p j ) \\frac\{\\partial L\_\{j\}\}\{\\partial p\_\{j\}\}= \\frac\{\\partial (-y\_\{j\}log^\{p\_\{j\}\})\}\{\\partial (p\_\{j\})\} ∂pj∂Lj=∂(pj)∂(−yjlogpj)

由于默认一般我们认为 l o g log log的底是 e e e，即 l o g log log为 l n ln ln，所以：

∂ L j ∂ p j = ∂ ( − y j l o g p j ) ∂ ( p j ) = − y j p j \\frac\{\\partial L\_\{j\}\}\{\\partial p\_\{j\}\}= \\frac\{\\partial (-y\_\{j\}log^\{p\_\{j\}\})\}\{\\partial (p\_\{j\})\} =-\\frac\{y\_\{j\}\}\{p\_\{j\}\} ∂pj∂Lj=∂(pj)∂(−yjlogpj)=−pjyj

接着要求 ∂ p j ∂ f i \\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} ∂fi∂pj的值，在这里可以发现，每一个 p j p\_\{j\} pj中都包含 f i f\_\{i\} fi，所以 ∂ p j ∂ f i \\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} ∂fi∂pj都不是0，但是 j = i j=i j=i和 j ≠ i j \\neq i j=i的时候， ∂ p j ∂ f i \\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} ∂fi∂pj结果又不相同，所以这里需要分开讨论：

*  首先 j = i j=i j=i时：  
     ∂ p j ∂ f i = ∂ p i ∂ f i = ∂ e f i ∑ k = 0 C − 1 e f k ∂ f i \\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} = \\frac\{\\partial p\_\{i\}\}\{\\partial f\_\{i\}\} = \\frac\{\\partial \\frac\{e^\{f\_\{i\}\}\}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\}\}\{\\partial f\_\{i\}\} ∂fi∂pj=∂fi∂pi=∂fi∂∑k=0C−1efkefi  
     = ( e f i ) ′ ∑ k = 0 C − 1 e f k − e f i ( ∑ k = 0 C − 1 e f k ) ′ ( ∑ k = 0 C − 1 e f k ) 2 = \\frac\{ (e^\{f\_\{i\}\})' \\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\} - e^\{f\_\{i\}\}(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})' \}\{(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})^\{2\}\} =(∑k=0C−1efk)2(efi)′∑k=0C−1efk−efi(∑k=0C−1efk)′  
     = e f i ∑ k = 0 C − 1 e f k − ( e f i ) 2 ( ∑ k = 0 C − 1 e f k ) 2 = e f i ∑ k = 0 C − 1 e f k − ( e f i ∑ k = 0 C − 1 e f k ) 2 = \\frac\{ e^\{f\_\{i\}\}\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\} - (e^\{f\_\{i\}\})^2 \}\{(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})^\{2\}\}= \\frac\{ e^\{f\_\{i\}\} \}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\} - (\\frac\{ e^\{f\_\{i\}\} \}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\})^2 =(∑k=0C−1efk)2efi∑k=0C−1efk−(efi)2=∑k=0C−1efkefi−(∑k=0C−1efkefi)2  
     = p i − ( p i ) 2 = p i ( 1 − p i ) = p\_\{i\}-(p\{i\})^2 = p\_\{i\}(1-p\_\{i\}) =pi−(pi)2=pi(1−pi)
 *  然后 j ≠ i j\\neq i j=i时：  
     ∂ p j ∂ f i = ∂ e f j ∑ k = 0 C − 1 e f k ∂ f i \\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\}= \\frac\{\\partial \\frac\{e^\{f\_\{j\}\}\}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\}\}\{\\partial f\_\{i\}\} ∂fi∂pj=∂fi∂∑k=0C−1efkefj  
     = ( e f j ) ′ ∑ k = 0 C − 1 e f k − e f j ( ∑ k = 0 C − 1 e f k ) ′ ( ∑ k = 0 C − 1 e f k ) 2 = \\frac\{ (e^\{f\_\{j\}\})' \\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\} - e^\{f\_\{j\}\}(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})' \}\{(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})^\{2\}\} =(∑k=0C−1efk)2(efj)′∑k=0C−1efk−efj(∑k=0C−1efk)′  
     = − e f i e f j ( ∑ k = 0 C − 1 e f k ) 2 = − e f i ∑ k = 0 C − 1 e f k e f j ∑ k = 0 C − 1 e f k = \\frac\{ - e^\{f\_\{i\}\} e^\{f\_\{j\}\} \}\{(\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\})^\{2\}\} = - \\frac\{ e^\{f\_\{i\}\} \}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\} \\frac\{ e^\{f\_\{j\}\} \}\{\\sum\_\{k=0\}^\{C-1\} e^\{f\_\{k\}\}\} =(∑k=0C−1efk)2−efiefj=−∑k=0C−1efkefi∑k=0C−1efkefj  
     = − p i p j = -p\_\{i\}p\_\{j\} =−pipj

对于最后的偏导数，需要把上述两个部分加起来：  
 ∂ L ∂ f i = ∑ j = i C − 1 ∂ L j ∂ p j ∂ p j ∂ f i + ∑ j ≠ i C − 1 ∂ L j ∂ p j ∂ p j ∂ f i \\frac\{\\partial L\}\{\\partial f\_\{i\}\} = \\sum\_\{j=i\}^\{C-1\} \\frac\{\\partial L\_\{j\}\}\{\\partial p\_\{j\}\}\\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} + \\sum\_\{j\\neq i\}^\{C-1\} \\frac\{\\partial L\_\{j\}\}\{\\partial p\_\{j\}\}\\frac\{\\partial p\_\{j\}\}\{\\partial f\_\{i\}\} ∂fi∂L=j=i∑C−1∂pj∂Lj∂fi∂pj\+j=i∑C−1∂pj∂Lj∂fi∂pj  
 = − y i p i p i ( 1 − p i ) + ∑ j ≠ i C − 1 − p i p j ( − y j p j ) =-\\frac\{y\_\{i\}\}\{p\_\{i\}\}p\_\{i\}(1-p\_\{i\}) + \\sum\_\{j\\neq i\}^\{C-1\}-p\_\{i\}p\_\{j\}(-\\frac\{y\_\{j\}\}\{p\_\{j\}\}) =−piyipi(1−pi)\+j=i∑C−1−pipj(−pjyj)  
 = − y i ( 1 − p i ) + ∑ j ≠ i C − 1 p i y j =-y\_\{i\}(1-p\_\{i\}) + \\sum\_\{j\\neq i\}^\{C-1\}p\_\{i\}y\_\{j\} =−yi(1−pi)\+j=i∑C−1piyj  
 = y i p i − y i + ∑ j ≠ i C − 1 p i y j =y\_\{i\}p\_\{i\}-y\_\{i\} + \\sum\_\{j\\neq i\}^\{C-1\}p\_\{i\}y\_\{j\} =yipi−yi\+j=i∑C−1piyj

在上式中， j ≠ i j\\neq i j=i的情况中刚好缺了 j = i j=i j=i，所以可以继续改写为：  
 = ∑ j = 0 C − 1 p i y j − y i =\\sum\_\{j=0\}^\{C-1\}p\_\{i\}y\_\{j\} - y\_\{i\} =j=0∑C−1piyj−yi  
 = p i ∑ j = 0 C − 1 y j − y i =p\_\{i\}\\sum\_\{j=0\}^\{C-1\}y\_\{j\} - y\_\{i\} =pij=0∑C−1yj−yi  
而 ∑ j = 0 C − 1 y j = 1 \\sum\_\{j=0\}^\{C-1\}y\_\{j\} = 1 ∑j=0C−1yj=1，所以：  
 = p i ∑ j = 0 C − 1 y j − y i = p i − y i =p\_\{i\}\\sum\_\{j=0\}^\{C-1\}y\_\{j\} - y\_\{i\} = p\_\{i\}-y\_\{i\} =pij=0∑C−1yj−yi=pi−yi

[Link 1]: https://blog.csdn.net/chaipp0607/article/details/78082154
[Link 2]: https://blog.csdn.net/chaipp0607/article/details/73392175