3
$\begingroup$

From Aghdam, H. H., & Heravi, E. J. (2017). Guide to Convolutional Neural Networks:

it is also desirable that the activation function approximates the identity mapping near origin. To explain this, we should consider the activation of a neuron. Formally, the activation of a neuron is given by G (wx T + b) where G is the activation function. Usually, the weight vector w and bias b are initialized with values close to zero by the gradient descend method. Consequently, wx T + b will be close to zero. If G approximates the identity function near zero, its gradient will be approximately equal to its input. In other words, δ G ≈ wx T + b ⇐⇒ wx T + b ≈ 0. In terms of the gradient descend, it is a strong gradient which helps the training algorithm to converge faster.

https://en.wikipedia.org/wiki/Activation_function also quotes them saying:

Approximates identity near the origin: When activation functions have this property, the neural network will learn efficiently when its weights are initialized with small random values. When the activation function does not approximate identity near the origin, special care must be used when initializing the weights.

but I don't understand why the gradient approximates the input near zero. I think it should be a constant, maybe w. Can someone please explain this step by step?

$\endgroup$

2 Answers 2

2
$\begingroup$

I'll break down the answer into 2 claims:

  1. If the network is initialized with near zero values ($w,b \approx 0$) then the gradient of each activation will also be approximately zero.
  2. Near-zero gradients are desirable, in terms of learning efficiency

The proof of the second claim is outside the scope of your question. In short, having bounded gradients at each layer is desirable, because gradients that are too small or too large ("exploding gradients", "vanishing gradients") are likely to interfere with the optimization procedure. Many works in the last year or two are around methods of solving this problem, probably the most famous one is the Batch Normalization scheme.

We'll prove the first claim. Denote as $x$ a neuron's input. Then its output is $G(w^T x)$, where $G$ is our non-linear activation function. For convenience, denote $z=w^T x$ and $y=G(z)$.

Since $w \approx 0$ we have that

$\frac{\partial z}{\partial x}=\frac{\partial (w^T x)}{\partial x}=w^T x\approx 0 $

And if we assume that $G$ is approximately the identity around zero,

$\frac{\partial G}{\partial x}=1$

Plugging these two results into the chain rule:

$\frac{\partial G}{\partial x}= \frac{\partial G}{\partial z} \cdot \frac{\partial z}{\partial x} \approx 0\cdot 1 = 0$

$\endgroup$
1
  • $\begingroup$ Did you mean $\frac{∂z}{∂x}=\frac{∂(w^Tx)}{∂x}=w^T$? So the gradient approximates the weights, but not the actual input? The authors i quoted say something different $\endgroup$
    – jj0
    Commented Jul 6, 2017 at 10:03
0
$\begingroup$

Depends on which input you are referring to.

  1. Gradient input to current layer. For $G(z)$, if $G$ is close to identity for near zero input, i.e., $G(z) = z$, and $G_z'(z) = 1$. That means, when you have a loss function L(w) backpropagating through the layer of $G(z)$, you have $$ L_z'(z) = L_G'(G)\cdot G_z'(z) = L_G'(G) $$ That is, activation layer directly passes through the gradient from upper layer (i.e., $L_G'(G)$, which can be regarded as the "input" to $G$ from backprop point of view). This is neat because it means the activation layer does not impact the gradient value in either exploding or vanishing.

  2. Forward propagation input. The actual gradient of $G$ w.r.t. the parameter $w$ is then, $$ G_w'(w) = G_z'(z)\cdot z_w'(w) = x $$ Here $x$ is the forward-propagate input to the layer and goes through the linear transformation and activation. When your input $x$ is small, that means your gradient of this layer is near zero, and the parameter optimization for the loss in this layer is close to a local minimum. This can save you lots of time wandering across the loss hyperplane with big gradient value.

  3. Gradient input to back-propagate to next layer. If you are further passing the gradient down to next layer, then you have, $$ G_x'(x) = G_z'(z)\cdot z_x'(x) = w $$ Here $w$ is your parameter input to your loss function, and it is the gradient value you input to the next layer. If it is near zero, it also helps the following layer optimization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.