# A和B的推导

$$A(\delta|w)$$是$$L(w+\delta) - L(w)$$的下界：

令：\
$$f\_i$$：$$f\_i(x,y)$$

## $$A(\delta|w)$$的推导

证明：

$$
L(w+\delta) - L(w)
\= \sum\_{x,y}\tilde P(x,y)\sum\_{i=1}^n\delta\_if\_i - \sum\_x\tilde P(x) \log \frac{Z\_{w+\delta}(x)}{Z\_w(x)}
$$

当a>0时，$$-\log a \ge 1-a$$，得：

$$
(1) \ge \sum\_{x,y}\tilde P(x,y)\sum\_{i=1}^n\delta\_if\_i + \sum\_x\tilde P(x)(1-\frac{Z\_{w+\delta}(x)}{Z\_w(x)})
\= \sum\_{x,y}\tilde P(x,y)\sum\_{i=1}^n\delta\_if\_i + \sum\_x\tilde P(x) - \sum\_x\tilde P(x)\frac{Z\_{w+\delta}(x)}{Z\_w(x)})
$$

根据$$\tilde P(x)$$可知$$\sum\_x\tilde P(x)=1$$，得：

$$
(2) = \sum\_{x,y}\tilde P(x,y)\sum\_{i=1}^n\delta\_if\_i + 1 - \sum\_x\tilde P(x)\frac{Z\_{w+\delta}(x)}{Z\_w(x)})
$$

根据【？】，得：

$$
(3) = \sum\_{x,y}\tilde P(x,y)\sum\_{i=1}^n\delta\_if\_i(x,y) + 1 - \sum\_x\tilde P(x)\sum\_yP\_w(y|x)\exp \sum\_{i=1}^n\delta\_if\_i(x,y)
$$

即$$A(\delta|w)$$

## $$B(\delta|w)$$的推导

**一次只优化其中一个变量**$$\delta\_i$$**，而固定其它变量**$$\delta\_j,i \neq j$$，得：\
【？】是一只优化一个$$w\_i$$还是一个$$\delta\_i$$？ 【？】如果是只优化一个$$w\_i$$，为什么不能直接假设其它$$\delta\_j=0$$？\
【？】如果是只优化一个$$\delta\_i$$，为什么算法6.1步骤2-(b)只更新一个$$w\_i$$？

后面的推导不难，只是这一块没想通，就不往下记了。


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