An Information-Theoretic View for Deep Learning

24 Apr 2018  ·  Jingwei Zhang, Tongliang Liu, DaCheng Tao ·

Deep learning has transformed computer vision, natural language processing, and speech recognition\cite{badrinarayanan2017segnet, dong2016image, ren2017faster, ji20133d}. However, two critical questions remain obscure: (1) why do deep neural networks generalize better than shallow networks; and (2) does it always hold that a deeper network leads to better performance? Specifically, letting $L$ be the number of convolutional and pooling layers in a deep neural network, and $n$ be the size of the training sample, we derive an upper bound on the expected generalization error for this network, i.e., \begin{eqnarray*} \mathbb{E}[R(W)-R_S(W)] \leq \exp{\left(-\frac{L}{2}\log{\frac{1}{\eta}}\right)}\sqrt{\frac{2\sigma^2}{n}I(S,W) } \end{eqnarray*} where $\sigma >0$ is a constant depending on the loss function, $0<\eta<1$ is a constant depending on the information loss for each convolutional or pooling layer, and $I(S, W)$ is the mutual information between the training sample $S$ and the output hypothesis $W$. This upper bound shows that as the number of convolutional and pooling layers $L$ increases in the network, the expected generalization error will decrease exponentially to zero. Layers with strict information loss, such as the convolutional layers, reduce the generalization error for the whole network; this answers the first question. However, algorithms with zero expected generalization error does not imply a small test error or $\mathbb{E}[R(W)]$. This is because $\mathbb{E}[R_S(W)]$ is large when the information for fitting the data is lost as the number of layers increases. This suggests that the claim `the deeper the better' is conditioned on a small training error or $\mathbb{E}[R_S(W)]$. Finally, we show that deep learning satisfies a weak notion of stability and the sample complexity of deep neural networks will decrease as $L$ increases.

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