word2vec數學分析

    本文沒有繁文縟節，純數學推導，建議先閱讀《word2vec中的數學原理詳解》

1、邏輯迴歸

    能夠閱讀《邏輯迴歸算法分析》理解邏輯迴歸。

    sigmoid函數： $\sigma(x) = \frac{1}{1 + e^{-x}}$

         $\sigma'(x) = \sigma(x)[1 - \sigma(x)]$

         $[log\sigma(x)]' = \frac{\sigma'(x)}{\sigma(x)} = 1 - \sigma(x)$

         $[log(1 - \sigma(x))]' = \frac{-\sigma'(x)}{1 - \sigma(x)} = - \sigma(x)$

    邏輯迴歸用於解決二分類問題，定義好極大對數似然函數，採用梯度上升的方法進行優化。事實上，word2vec的算法本質就是邏輯迴歸。

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2、CBOW

    根據上下文詞，預測當前詞，將預測偏差反向更新到每一個上下文詞上，以達到更準確的預測的目的。

    記：

    一、 $p^w$：從根節點出發，到達 $w$的路徑；

    二、 $l^w$：路徑 $p^w$包含節點的個數；

    三、 $p_1^w, p_2^w, \cdots, p_{l^w}^w$：路徑 $p^w$中第 $l^w$個節點，其中 $p_1^w$是根節點， $p_{l^w}^w$是 $w$對應的節點

    四、 $d_2^w, d_3^w, \cdots, d_{l^w}^w \ \epsilon \ \left \{ 0, 1 \right \}$：詞 $w$的哈夫曼編碼，它由 $l^w - 1$位編碼構成， $d_j^w$表示路徑 $p^w$第 $j$個節點對應的編碼（根節點不編碼）。

    五、 $\theta_1^w, \theta_2^w, \cdots, \theta_{l^w - 1}^w \ \epsilon \ \mathbb{R}^m$：路徑 $p^w$中非葉子節點對應的向量， $\theta_j^w$表示路徑 $p^w$第 $j$個非葉子節點對應的向量

    六、 $Label(p_j^w) = 1 - d_j^w, j = 2,3,\cdots,l^w$：表示路徑 $p^w$第 $j$個節點對應的分類標籤（根節點不分類）

一、Hierarchical Softmax

    極大似然： $\prod_{w \epsilon C}p(w|Context(w))$

    極大對數似然： $\pounds = \sum_{w \epsilon C} log \ p(w|Context(w))$

    條件機率： $p(w|Context(w)) = \prod_{j=2}^{l^w} p(d_j^w|X_w, \theta_{j-1}^w)$，其中：

         $p(d_j^w|X_w, \theta_{j-1}^w) = \left\{\begin{matrix}\sigma (X_w^T \theta_{j - 1}^w), \quad \quad \ \ d_j^w = 0 & \\ & \\ 1 - \sigma (X_w^T \theta_{j - 1}^w), \quad d_j^w = 1 & \end{matrix}\right.$，注意：在word2vec中的哈夫曼樹中，編碼0表示正類，編碼1表示負類。

         $X_w = \frac{\sum_{u \epsilon Context(w)} v(u)}{|Context(w)|}$

    寫成總體即： $p(d_j^w|X_w, \theta_{j-1}^w) = [\sigma (X_w^T \theta_{j - 1}^w)]^{1 - d_j^w} \cdot [1 - \sigma (X_w^T \theta_{j - 1}^w)]^{d_j^w}$，代入對數似然函數得：

     $\pounds = \sum _{w \epsilon C} log \prod_{j=2}^{l^w} p(d_j^w|X_w, \theta_{j-1}^w)$

        $= \sum _{w \epsilon C} log \prod_{j=2}^{l^w} \left \{ [\sigma (X_w^T \theta_{j - 1}^w)]^{1 - d_j^w} \cdot [1 - \sigma (X_w^T \theta_{j - 1}^w)]^{d_j^w} \right \}$

        $= \sum_{w \epsilon C} \sum_{j=2}^{l^w} \left \{ (1 - d_j^w) \cdot log [\sigma (X_w^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (X_w^T \theta_{j - 1}^w)] \right \}$

    爲求導方便，記： $\pounds(w, j) = (1 - d_j^w) \cdot log [\sigma (X_w^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (X_w^T \theta_{j - 1}^w)]$

     $\pounds(w, j)$關於 $\theta_{j - 1}^w$的梯度：

         $\frac{\partial \pounds(w, j)}{\partial \theta_{j - 1}^w} = \frac{\partial }{\partial \theta_{j - 1}^w}\left \{ (1 - d_j^w) \cdot log [\sigma (X_w^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (X_w^T \theta_{j - 1}^w)] \right \}$

                 $= (1 - d_j^w)[1 - \sigma (X_w^T \theta_{j - 1}^w)]X_w - d_j^w [\sigma (X_w^T \theta_{j - 1}^w)]X_w$

                 $= \left \{ (1 - d_j^w)[1 - \sigma (X_w^T \theta_{j - 1}^w)] - d_j^w [\sigma (X_w^T \theta_{j - 1}^w)] \right \} X_w$

                 $= [1 - d_j^w - \sigma (X_w^T \theta_{j - 1}^w)] X_w$

    因而， $\theta_{j - 1}^w$的更新可寫爲：

         $\theta_{j - 1}^w \ := \ \theta_{j - 1}^w + \eta [1 - d_j^w - \sigma (X_w^T \theta_{j - 1}^w)] X_w$

    因爲在 $\pounds(w, j)$中 $\theta_{j - 1}^w$與 $X_w$是對稱的，因此 $\pounds(w, j)$關於 $X_w$的梯度：

         $\frac{\partial \pounds(w, j)}{\partial X_w} = [1 - d_j^w - \sigma (X_w^T \theta_{j - 1}^w)] \theta_{j - 1}^w$

     用 $\frac{\partial \pounds(w, j)}{\partial X_w}$來對上下文詞 $v(u), u \epsilon Context(w)$進行更新：

         $v(u) := v(u) + \eta \sum_{j=2}^{l^w}\frac{\partial \pounds(w, j)}{\partial X_w}$

    以樣本 $(Context(w), w)$爲例，訓練僞代碼以下：

     $e = 0$

     $X_w = \frac{\sum_{u \epsilon Context(w)} v(u)}{|Context(w)|}$

     $FOR \quad j = 2:l^w \quad DO$
     {
          $q = \sigma(X_w^T\theta_{j - 1}^w)$

          $g = \eta [1 - d_j^w - q]$

          $e := e + g\theta_{j - 1}^w$

          $\theta_{j - 1}^w := \theta_{j - 1}^w + gX_w$
     }

     $FOR \quad u \epsilon Context(w) \quad DO$
     {
          $v(u) := v(u) + e$
     }

    這裏有必要對以上僞代碼的含義作一個說明，固然，直接經過導數推導過程來理解也能夠，但導數的推導過程並無表達其真實的內在含義，下文中有相似的地方，再也不說明。

    一、 $\sigma(X_w^T\theta_{j - 1}^w)$：

        其含義是在已知上下文的前提下，在當前詞的哈夫曼路徑上作分類預測，根據路徑上的父節點 $\theta_{j - 1}^w$，預測其子節點 $\theta_{j}^w$，獲得的子節點的分類標籤。固然，這裏獲得的分類標籤是[0,1]之間的實數，而不是{0, 1}二分類，這個值與0，1之間的差距便是預測偏差。把 $\sigma(X_w^T\theta_{j - 1}^w)$理解成子節點 $\theta_{j}^w$正分類的機率也是能夠的。

    二、 $1 - d_j^w - q$：

         $1 - d_j^w$的含義是子節點 $\theta_{j}^w$的真實分類標籤， $1 - d_j^w - q$則是真實標籤與預測標籤之間的偏差。

    三、 $e := e + g\theta_{j - 1}^w$：

        這裏是一個關鍵點，回到咱們最開始的優化函數上，要求是的最大對數似然： $\pounds = \sum_{w \epsilon C} log \ p(w|Context(w))$，即求極大值，因此要用梯度上升的方法進行優化（機器學習中通常是梯度降低），因此e的更新是加法（梯度降低是減法）。

        當梯度爲正的時， $g\theta_{j - 1}^w > 0$，則 $e := e + g\theta_{j - 1}^w$相加後變大，將 $e$更新到 $v(u)$後讓 $X_w$變大， $X_w$變大後 $\sigma(X_w^T\theta_{j - 1}^w)$也就變大，也就是說預測的分類標籤越像1（正類），也能夠理解成預測爲正類的機率增大。爲何要讓 $\sigma(X_w^T\theta_{j - 1}^w)$增大呢？反過來思考，當梯度爲正的時， $(1 - d_j^w - q) > 0$，這時只有當 $d_j^w = 0$時其值纔可能爲正，而 $d_j^w = 0$表示正類，分類標籤爲1，因此優化時要讓 $\sigma(X_w^T\theta_{j - 1}^w)$趨近於1。

        一樣，當梯度爲負的時， $g\theta_{j - 1}^w < 0$，則 $e := e + g\theta_{j - 1}^w$相加後變小，將 $e$更新到 $v(u)$後讓 $X_w$變小， $X_w$變小後 $\sigma(X_w^T\theta_{j - 1}^w)$也就變小，也就是說預測的分類標籤越像0（負類），也能夠理解成預測爲正類的機率減少。一樣，反過來思考，當梯度爲負的時， $(1 - d_j^w - q) < 0$，這時只有當 $d_j^w = 1$時其值纔可能爲負，而 $d_j^w = 1$表示負類，分類標籤爲0，因此優化時要讓 $\sigma(X_w^T\theta_{j - 1}^w)$趨近於0。

    四、 $\theta_{j - 1}^w := \theta_{j - 1}^w + gX_w$：

        同上

    五、 $FOR \quad j = 2:l^w \quad DO$：

        該循環的含義是上下文預測的是葉子節點的詞（當前詞在葉子節點上），要通過該詞的哈夫曼路徑才能到達，因此要循環累計路徑上（除根節點）每一次分類的偏差。

    總結：根據上下文詞，遍歷當前詞的哈夫曼路徑，累計（除根節點之外）每一個節點的二分類偏差，將偏差反向更新到每一個上下文詞上（同時也會更新路徑上節點的輔助向量）。

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二、Negative Sampling

    對 $w$的負樣本子集 $NEG(w)$的每一個樣本，定義樣本標籤：

         $L^w(\tilde{w}) = \left\{\begin{matrix}1,\quad w = \tilde{w} & \\ & \\ 0,\quad w \neq \tilde{w} & \end{matrix}\right.$

    極大似然： $\prod_{w \epsilon C}p(w|Context(w))$

    極大對數似然： $\pounds = \sum_{w \epsilon C} log \ p(w|Context(w))$

    條件機率： $p(w|Context(w)) = \prod_{u \epsilon \left \{ w \right \} \cup NEG(w)}p(u|Context(w))$，其中：

         $p(u|Context(w)) = \left\{\begin{matrix}\sigma (X_w^T \theta^u), \quad \quad \ \ L^w(u) = 1 & \\ & \\ 1 - \sigma (X_w^T \theta^u), \quad L^w(u) = 0 & \end{matrix}\right.$

         $X_w = \frac{\sum_{u \epsilon Context(w)} v(u)}{|Context(w)|}$

    寫成總體即： $p(u|Context(w)) = [\sigma (X_w^T \theta^u)]^{L^w(u)} \cdot [1 - \sigma (X_w^T \theta^u)]^{1 - L^w(u)}$，代入對數似然函數得：

     $\pounds = \sum_{w \epsilon C} log \ \prod_{u \epsilon \left \{ w \right \} \cup NEG(w)}p(u|Context(w))$

        $= \sum_{w \epsilon C} log \ \prod_{u \epsilon \left \{ w \right \} \cup NEG(w)} [\sigma (X_w^T \theta^u)]^{L^w(u)} \cdot [1 - \sigma (X_w^T \theta^u)]^{1 - L^w(u)}$

        $= \sum_{w \epsilon C} \sum_{u \epsilon \left \{ w \right \} \cup NEG(w)} \left \{ L^w(u) \cdot log [\sigma (X_w^T \theta^u)] + [1 - L^w(u)] \cdot log [1 - \sigma (X_w^T \theta^u)] \right \}$

    爲求導方便，記： $\pounds(w, u) = L^w(u) \cdot log [\sigma (X_w^T \theta^u)] + [1 - L^w(u)] \cdot log [1 - \sigma (X_w^T \theta^u)]$

     $\pounds(w, u)$關於 $\theta^u$的梯度：

         $\frac{\partial \pounds(w, u)}{\partial \theta^u} = \frac{\partial }{\partial \theta^u}\left \{ L^w(u) \cdot log [\sigma (X_w^T \theta^u)] + [1 - L^w(u)] \cdot log [1 - \sigma (X_w^T \theta^u)] \right \}$

                 $= L^w(u) \cdot [1 - \sigma (X_w^T \theta^u)]X_w - [1 - L^w(u)] \cdot [\sigma (X_w^T \theta^u)] X_w$

                 $= \left \{ L^w(u) \cdot [1 - \sigma (X_w^T \theta^u)] - [1 - L^w(u)] \cdot [\sigma (X_w^T \theta^u)] \right \} X_w$

                 $= [L^w(u) - \sigma (X_w^T \theta^u) ] X_w$

    因而， $\theta^u$的更新可寫爲：

         $\theta^u \ := \ \theta^u + \eta [L^w(u) - \sigma (X_w^T \theta^u) ] X_w$

    因爲在 $\pounds(w, u)$中 $\theta^u$與 $X_w$是對稱的，因此 $\pounds(w, u)$關於 $X_w$的梯度：

         $\frac{\partial \pounds(w, u)}{\partial X_w} = [L^w(u) - \sigma (X_w^T \theta^u) ] \theta^u$

     用 $\frac{\partial \pounds(w, u)}{\partial X_w}$來對上下文詞 $v(u), u \epsilon Context(w)$進行更新：

         $v(u) := v(u) + \eta \sum_{u \epsilon \left \{ w \right \} \cup NEG(w)}\frac{\partial \pounds(w, u)}{\partial X_w}$

    以樣本 $(Context(w), w)$爲例，訓練僞代碼以下：

     $e = 0$

     $X_w = \frac{\sum_{u \epsilon Context(w)} v(u)}{|Context(w)|}$

     $FOR \quad u \epsilon \left \{ w \right \} \cup NEG(w) \quad DO$
     {
          $q = \sigma(X_w^T\theta^u)$

          $g = \eta [L^w(u) - q]$

          $e := e + g\theta^u$

          $\theta^u := \theta^u + gX_w$
       }

     $FOR \quad u \epsilon Context(w) \quad DO$
     {
          $v(u) := v(u) + e$
     }

    總結：根據上下文詞，對當前詞作一次負採樣（包括當前詞，當前詞是正樣本），遍歷每一個樣本，累計上下文對每一個樣本的預測偏差，將偏差反向更新到每一個上下文詞上（同時也會更新樣本向量）。

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3、Skip-gram

    根據當前詞，預測上下文詞，將預測偏差反向更新到當前詞上，以達到更準確的預測的目的。但word2vec並無按這個思路訓練，而是依然按照CBOW的思路，用上下文中的每一個詞（注意這裏的區別，CBOW是合併了上下文，即 $\sum_{u \epsilon Context(w)}v(u)$），對當前詞進行預測，再將預測偏差反向更新到該上下文詞上。

一、Hierarchical Softmax

    極大似然： $\prod_{w \epsilon C}p(Context(w)|w)$

    極大對數似然： $\pounds = \sum_{w \epsilon C} log \ p(Context(w)|w)$

    條件機率： $p(Context(w)|w) = \prod_{u \epsilon Context(w)} p(u|w)$，其中：

         $p(u|w) = \prod_{j=2}^{l^u} p(d_j^u|v(w), \theta_{j-1}^u)$

         $p(d_j^u|v(w), \theta_{j-1}^u) = \left\{\begin{matrix}\sigma (v(w)^T \theta_{j - 1}^u), \quad \quad \ \ \ d_j^u = 0 & \\ & \\ 1 - \sigma (v(w)^T \theta_{j - 1}^u), \quad d_j^u = 1 & \end{matrix}\right.$

    寫成總體即： $p(d_j^u|v(w), \theta_{j-1}^u) = [\sigma (v(w)^T \theta_{j - 1}^u)]^{1 - d_j^u} \cdot [1 - \sigma (v(w)^T \theta_{j - 1}^u)]^{d_j^u}$，代入對數似然函數得：

     $\pounds = \sum _{w \epsilon C} log \prod_{u \epsilon Context(w)} \prod_{j=2}^{l^u} p(d_j^u|v(w), \theta_{j-1}^u)$

        $= \sum _{w \epsilon C} log \prod_{u \epsilon Context(w)} \prod_{j=2}^{l^u} [\sigma (v(w)^T \theta_{j - 1}^u)]^{1 - d_j^u} \cdot [1 - \sigma (v(w)^T \theta_{j - 1}^u)]^{d_j^u}$

        $= \sum _{w \epsilon C} \sum_{u \epsilon Context(w)} \sum_{j=2}^{l^u} \left \{ (1 - d_j^u) \cdot log \ [\sigma (v(w)^T \theta_{j - 1}^u)] + d_j^u \cdot log [1 - \sigma (v(w)^T \theta_{j - 1}^u)] \right \}$

    爲求導方便，記： $\pounds(w, u, j) = (1 - d_j^u) \cdot log \ [\sigma (v(w)^T \theta_{j - 1}^u)] + d_j^u \cdot log [1 - \sigma (v(w)^T \theta_{j - 1}^u)]$

     $\pounds(w, u, j)$關於 $\theta_{j - 1}^u$的梯度：

         $\frac{\partial \pounds(w, u, j)}{\partial \theta_{j - 1}^u} = \frac{\partial }{\partial \theta_{j - 1}^u}\left \{ (1 - d_j^u) \cdot log \ [\sigma (v(w)^T \theta_{j - 1}^u)] + d_j^u \cdot log [1 - \sigma (v(w)^T \theta_{j - 1}^u)] \right \}$

                 $= (1 - d_j^u)[1 - \sigma (v(w)^T \theta_{j - 1}^u)]v(w) - d_j^u [\sigma (v(w)^T \theta_{j - 1}^u)]v(w)$

                 $= \left \{ (1 - d_j^u)[1 - \sigma (v(w)^T \theta_{j - 1}^u)] - d_j^u [\sigma (v(w)^T \theta_{j - 1}^u)] \right \}v(w)$

                 $= [1 - d_j^u - \sigma (v(w)^T \theta_{j - 1}^u)] v(w)$

    因而， $\theta_{j - 1}^u$的更新可寫爲：

         $\theta_{j - 1}^u \ := \ \theta_{j - 1}^u + \eta [1 - d_j^u - \sigma (v(w)^T \theta_{j - 1}^u)] v(w)$

    因爲在 $\pounds(w, u, j)$中 $\theta_{j - 1}^u$與 $v(w)$是對稱的，因此 $\pounds(w, u, j)$關於 $v(w)$的梯度：

         $\frac{\partial \pounds(w, u, j)}{\partial v(w)} = [1 - d_j^u - \sigma (v(w)^T \theta_{j - 1}^u)] \theta_{j - 1}^u$

     用 $\frac{\partial \pounds(w, u, j)}{\partial v(w)}$來對當前詞 $v(w)$進行更新：

         $v(w) := v(w) + \eta \sum_{u \epsilon Context(w)} \sum_{j=2}^{l^u}\frac{\partial \pounds(w, u, j)}{\partial v(w)}$

    以樣本 $(w, Context(w))$爲例，訓練僞代碼以下：

     $e = 0$

     $FOR \quad u \epsilon Context(w) \quad DO$
     {
          $FOR \quad j = 2:l^u \quad DO$
          {
               $q = \sigma(v(w)^T\theta_{j - 1}^u)$

               $g = \eta [1 - d_j^u - q]$

               $e := e + g\theta_{j - 1}^u$

               $\theta_{j - 1}^u := \theta_{j - 1}^u + gv(w)$
          }
     }

     $v(w) := v(w) + e$

    值得注意的是，word2vec並非按上面的流程進行訓練的，而依然按CBOW的思路，對每個上下文詞，預測當前詞，分析以下：

    極大似然： $\prod_{w \epsilon C}\prod_{u \epsilon Context(w)}p(w|u)$

    極大對數似然： $\pounds = \sum_{w \epsilon C}\sum_{u \epsilon Context(w)}log \ p(w|u)$

    條件機率： $p(w|u) = \prod_{j=2}^{l^w} p(d_j^w|v(u), \theta_{j-1}^w)$，其中：

         $p(d_j^w|v(u), \theta_{j-1}^w) = \left\{\begin{matrix}\sigma (v(u)^T \theta_{j - 1}^w), \quad \quad \ \ d_j^w = 0 & \\ & \\ 1 - \sigma (v(u)^T \theta_{j - 1}^w), \quad d_j^w = 1 & \end{matrix}\right.$

    寫成總體即： $p(d_j^w|v(u), \theta_{j-1}^w) = [\sigma (v(u)^T \theta_{j - 1}^w)]^{1 - d_j^w} \cdot [1 - \sigma (v(u)^T \theta_{j - 1}^w)]^{d_j^w}$，代入對數似然函數得：

     $\pounds = \sum_{w \epsilon C}\sum_{u \epsilon Context(w)} log \prod_{j=2}^{l^w} p(d_j^w|v(u), \theta_{j-1}^w)$

        $= \sum_{w \epsilon C}\sum_{u \epsilon Context(w)} log \prod_{j=2}^{l^w} [\sigma (v(u)^T \theta_{j - 1}^w)]^{1 - d_j^w} \cdot [1 - \sigma (v(u)^T \theta_{j - 1}^w)]^{d_j^w}$

        $= \sum_{w \epsilon C}\sum_{u \epsilon Context(w)}\sum_{j=2}^{l^w} \left \{ (1 - d_j^w) \cdot log [\sigma (v(u)^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (v(u)^T \theta_{j - 1}^w)] \right \}$

    爲求導方便，記： $\pounds(w, u, j) = (1 - d_j^w) \cdot log [\sigma (v(u)^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (v(u)^T \theta_{j - 1}^w)]$

     $\pounds(w, u, j)$關於 $\theta_{j - 1}^w$的梯度：

         $\frac{\partial \pounds(w, u, j)}{\partial \theta_{j - 1}^w} = \frac{\partial }{\partial \theta_{j - 1}^w}\left \{ (1 - d_j^w) \cdot log [\sigma (v(u)^T \theta_{j - 1}^w)] + d_j^w \cdot log [1 - \sigma (v(u)^T \theta_{j - 1}^w)] \right \}$

                 $= (1 - d_j^w)[1 - \sigma (v(u)^T \theta_{j - 1}^w)]v(u) - d_j^w [\sigma (v(u)^T \theta_{j - 1}^w)]v(u)$

                 $= \left \{ (1 - d_j^w)[1 - \sigma (v(u)^T \theta_{j - 1}^w)] - d_j^w [\sigma (v(u)^T \theta_{j - 1}^w)] \right \} v(u)$

                 $= [1 - d_j^w - \sigma (v(u)^T \theta_{j - 1}^w)] v(u)$

    因而， $\theta_{j - 1}^w$的更新可寫爲：

         $\theta_{j - 1}^w \ := \ \theta_{j - 1}^w + \eta [1 - d_j^w - \sigma (v(u)^T \theta_{j - 1}^w)] v(u)$

    因爲在 $\pounds(w, u, j)$中 $\theta_{j - 1}^w$與 $v(u)$是對稱的，因此 $\pounds(w, u, j)$關於 $v(u)$的梯度：

         $\frac{\partial \pounds(w, u, j)}{\partial v(u)} = [1 - d_j^w - \sigma (v(u)^T \theta_{j - 1}^w)] \theta_{j - 1}^w$

     用 $\frac{\partial \pounds(w, u, j)}{\partial v(u)}$來對上下文詞 $v(u), u \epsilon Context(w)$進行更新：

         $v(u) := v(u) + \eta \sum_{j=2}^{l^w}\frac{\partial \pounds(w, u, j)}{\partial v(u)}$

    以樣本 $(w, Context(w))$爲例，訓練僞代碼以下：

     $FOR \quad u \epsilon Context(w) \quad DO$
     {
          $e = 0$

          $FOR \quad j = 2:l^w \quad DO$
          {
               $q = \sigma(v(u)^T\theta_{j - 1}^w)$

               $g = \eta [1 - d_j^w - q]$

               $e := e + g\theta_{j - 1}^w$

               $\theta_{j - 1}^w := \theta_{j - 1}^w + gv(u)$
          }

          $v(u) := v(u) + e$
     }

    總結：根據上下文詞（用該上下文詞來預測當前詞），遍歷當前詞的哈夫曼路徑，累計（除根節點之外）每一個節點的二分類偏差，將偏差反向更新到該上下文詞上（同時也會更新路徑上節點的輔助向量）。

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二、Negative Sampling

    對 $w$的負樣本子集 $NEG(w)$的每一個樣本，定義樣本標籤：

         $L^w(\tilde{w}) = \left\{\begin{matrix}1,\quad w = \tilde{w} & \\ & \\ 0,\quad w \neq \tilde{w} & \end{matrix}\right.$

    極大似然： $\prod_{w \epsilon C}p(Context(w)|w)$

    極大對數似然： $\pounds = \sum_{w \epsilon C} log \ p(Context(w)|w)$

    條件機率： $p(Context(w)|w) = \prod_{u \epsilon Context(w)} p(u|w)$，其中：

         $p(u|w) = \prod_{z \epsilon \left \{ u \right \} \cup NEG(u)} p(z|w)$

         $p(z|w) = \left\{\begin{matrix}\sigma (v(w)^T \theta^z), \quad \quad \ \ L^u(z) = 1 & \\ & \\ 1 - \sigma (v(w)^T \theta^z), \quad L^u(z) = 0 & \end{matrix}\right.$

    寫成總體即： $p(z|w) = [\sigma (v(w)^T \theta^z)]^{L^u(z)} \cdot [1 - \sigma (v(w)^T \theta^z)]^{1 - L^u(z)}$，代入對數似然函數得：

     $\pounds = \sum _{w \epsilon C} log \prod_{u \epsilon Context(w)} \prod_{z \epsilon \left \{ u \right \} \cup NEG(u)} p(z|w)$

        $= \sum _{w \epsilon C} log \prod_{u \epsilon Context(w)} \prod_{z \epsilon \left \{ u \right \} \cup NEG(u)}[\sigma (v(w)^T \theta^z)]^{L^u(z)} \cdot [1 - \sigma (v(w)^T \theta^z)]^{1 - L^u(z)}$

        $= \sum _{w \epsilon C} \sum_{u \epsilon Context(w)} \sum_{z \epsilon \left \{ u \right \} \cup NEG(u)} \left \{ L^u(z) \cdot log [\sigma (v(w)^T \theta^z)] + [1 - L^u(z)] \cdot log [1 - \sigma (v(w)^T \theta^z)] \right \}$

    爲求導方便，記： $\pounds(w, u, z) = L^u(z) \cdot log [\sigma (v(w)^T \theta^z)] + [1 - L^u(z)] \cdot log [1 - \sigma (v(w)^T \theta^z)]$

     $\pounds(w, u, z)$關於 $\theta^z$的梯度：

         $\frac{\partial \pounds(w, u, z)}{\partial \theta^z} = \frac{\partial }{\partial \theta^z}\left \{ L^u(z) \cdot log [\sigma (v(w)^T \theta^z)] + [1 - L^u(z)] \cdot log [1 - \sigma (v(w)^T \theta^z)] \right \}$