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Lei Mao

Machine Learning, Artificial Intelligence, Computer Science.

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Introduction

I have once come up with a question “how do we do back propagation through max-pooling layer?”. The short answer is “there is no gradient with respect to non-maximum values”.

Proof

Max-pooling is defined as

\[y = \max(x_1, x_2, \cdots, x_n)\]

where $y$ is the output and $x_i$ is the value of the neuron.


Alternatively, we could consider max-pooling layer as an affine layer without bias terms. The weight matrix in this affine layer is not trainable though.


Concretely, for the output $y$ after max-pooling, we have

\[y = \sum_{i=1}^{n} w_i x_i\]

where

\[w_i = \begin{cases} 1 & \text{if } x_i = \max(x_1, x_2, \cdots, x_n) \\ 0 & \text{otherwise} \end{cases}\]

The gradient to each neuron is

\[\begin{aligned} \frac{\partial y}{\partial x_i} &= w_i \\ &= \begin{cases} 1 & \text{if } x_i = \max(x_1, x_2, \cdots, x_n) \\ 0 & \text{otherwise} \end{cases} \end{aligned}\]

This simply means that the gradient to the neuron is 1 for the neuron with the maximum value. The gradients for all the other neurons are 0. The neurons whose gradients are 0 do not contribute to the gradients in the earlier neurons due to the chain rule.

Conclusions

There is no gradient with respect to non-maximum values.