Deep coral: Correlation alignment for deep domain adaptation. ECCV 2016. Domain Adaptation

Deep coral: Correlation alignment for deep domain adaptation. ECCV 2016. Domain Adaptation

** Sun, Baochen, and Kate Saenko. “Deep coral: Correlation alignment for deep domain adaptation.” ECCV. Springer, Cham, 2016. **

结构如图：

两个损失函数：

其中 L C L A S S mathcal{L}_{CLASS} LCLASS​为分类损失， L C O R A L mathcal{L}_{CORAL} LCORAL​:

C S C_S CS​和 C T C_T CT​为样本协方差 (二阶统计量)，反映了feature map上各位置的相关性。 其中D_S矩阵的每一行代表代表一个feature map。 C S C_S CS​和 C T C_T CT​的行数= C S C_S CS​和 C T C_T CT​的列数=feature map维度。

CORAL Loss系数为0时，训练过程中CORAL distance测量值变化情况:

上图证明了CORAL Loss的有效性。

原文表明：没有CORAL Loss直接fine-tuning容易对source domain的数据过拟合。

这里复习下样本方差为无偏估计量：

S 2 = ∑ ( x i − x ˉ ) 2 n S^2= frac{sum (x_i - bar{x} )^2 }{n} S2=n∑(xi​−xˉ)2​

证明:
E ( s 2 ) = 1 n − 1 E [ ∑ ( x i − x ˉ ) 2 ] = ∑ E ( x i 2 − 2 x ˉ x i + x ˉ 2 ) n − 1 = n E 2 ( x ) − 2 n E ( x ˉ 2 ) + n E ( x ˉ 2 ) n − 1 = n E 2 ( x ) − n E ( x ˉ 2 ) n − 1 = n E 2 ( x ) − n D ( x ˉ ) − n E 2 ( x ˉ ) n − 1 = n E 2 ( x ) − n D ( x ˉ ) − n E 2 ( x ) n − 1 = n E 2 ( x ) − n E 2 ( x ) − n D ( x ˉ ) n − 1 = n D ( x ) − n D ( x ) n n − 1 = D ( x ) begin{aligned} E(s^2)&= frac{1}{n-1} E[ sum(x_i -bar{x})^2 ] \ &=frac{sum E(x_i^2-2bar{x}x_i +bar{x}^2 ) }{n-1} \ &=frac{nE^2(x)-2nE(bar{x}^2)+nE(bar{x}^2)}{n-1} \ &=frac{nE^2(x)-nE(bar{x}^2)}{n-1} \ &=frac{nE^2(x)-nD(bar{x})-nE^2(bar{x}) }{n-1} \ &=frac{nE^2(x)-nD(bar{x})-nE^2(x) }{n-1} \ &=frac{nE^2(x)-nE^2(x) -nD(bar{x})}{n-1} \ &= frac{nD(x)-frac{nD(x)}{n}}{n-1}\ &=D(x) end{aligned} E(s2)​=n−11​E[∑(xi​−xˉ)2]=n−1∑E(xi2​−2xˉxi​+xˉ2)​=n−1nE2(x)−2nE(xˉ2)+nE(xˉ2)​=n−1nE2(x)−nE(xˉ2)​=n−1nE2(x)−nD(xˉ)−nE2(xˉ)​=n−1nE2(x)−nD(xˉ)−nE2(x)​=n−1nE2(x)−nE2(x)−nD(xˉ)​=n−1nD(x)−nnD(x)​​=D(x)​

注意:
E ( x ) ≠ x ˉ ∑ ( − 2 E ( x i ) x ˉ ) ≠ − 2 n E ( x ) E ( x ) = E ( x ˉ ) D ( x ˉ ) = E ( x ˉ 2 ) − E 2 ( x ˉ ) begin{aligned} E(x) &neq bar{x} \ sum (-2E(x_i) &bar{x}) neq -2nE(x) \ E(x) &= E(bar{x}) \ D(bar{x})&=E(bar{x}^2)-E^2(bar{x}) end{aligned} E(x)∑(−2E(xi​)E(x)D(xˉ)​̸​=xˉxˉ)̸​=−2nE(x)=E(xˉ)=E(xˉ2)−E2(xˉ)​

类似于 S 2 = ∑ ( x i − x ˉ ) 2 n S^2= frac{sum (x_i - bar{x} )^2 }{n} S2=n∑(xi​−xˉ)2​

样本协方差 = ∑ ( x i − x ˉ ) ( y i − y ˉ ) n − 1 = ∑ x i y i − n x ˉ y ˉ n − 1 = frac{sum (x_i-bar{x}) (y_i-bar{y}) }{n-1}= frac{ sum x_iy_i -n bar{x}bar{y} }{n-1} =n−1∑(xi​−xˉ)(yi​−yˉ​)​=n−1∑xi​yi​−nxˉyˉ​​