1
$\begingroup$

Consider the following network:

enter image description here

There are two weights, say $w_1$ and $w_2$, and two biases, $b_1$ and $b_2$. The hidden layer has a ReLU activation function $g^{(1)}$ and the output layer has a linear activation function $g^{(2)}$.

Say we have a dataset of $M$ datapoints, each of the form $(x^{[m]}, y^{[m]})$ (1D inputs/outputs)

The network's predictions are \begin{align*} \hat{y}^{[m]} = g^{(2)}\left(w_2 \left[g^{(1)} \left(w_1 x^{[m]} + b_1 \right)\right] + b_2\right) = g^{(2)}\left(w_2 {a_1^{(1)}}^{[m]}+ b_2\right) \,. \end{align*} Assuming a linear activation function in the output layer, this simplifies to \begin{align*} \hat{y}^{[m]} =w_2 \left[g^{(1)} \left(w_1 x^{[m]} + b_1 \right)\right] + b_2 = w_2 {a_1^{(1)}}^{[m]} + b_2 \,. \end{align*} The MSE loss is $$L_\text{MSE}=\frac{1}{M}\sum_{m=1}^M L_m = \frac{1}{M} \sum_{m=1}^M \left(\hat{y}^{[m]} - y^{[m]} \right)^2 \,.$$

The derivatives of the MSE loss with respect to the network parameters are: $$ \frac{\partial L_\text{MSE}}{\partial b_2} = \frac{1}{M} \sum_{m=1}^M \frac{\partial}{\partial b_2} \left[\left(\hat{y}^{[m]} - y^{[m]} \right)^2 \right] = \frac{1}{M} \sum_{m=1}^M 2\left(\hat{y}^{[m]} - y^{[m]} \right) \cdot 1 \, $$ $$\frac{\partial L_\text{MSE}}{\partial w_2} = \frac{1}{M} \sum_{m=1}^M \frac{\partial}{\partial w_2} \left[\left(\hat{y}^{[m]} - y^{[m]} \right)^2 \right] = \frac{1}{M} \sum_{m=1}^M 2\left(\hat{y}^{[m]} - y^{[m]} \right) {a_1^{(1)}}^{[m]} \,$$ $$\frac{\partial L_\text{MSE}}{\partial b_1} = \frac{1}{M} \sum_{m=1}^M \frac{\partial}{\partial b_1} \left[\left(\hat{y}^{[m]} - y^{[m]} \right)^2 \right] = \frac{1}{M} \sum_{m=1}^M 2\left(\hat{y}^{[m]} - y^{[m]} \right) w_2 \hspace{0.2em}g^{(1)^{'}}\left(w_1 x^{[m]} + b_1 \right) \,$$ $$ \frac{\partial L_\text{MSE}}{\partial w_1} = \frac{1}{M} \sum_{m=1}^M \frac{\partial}{\partial w_1} \left[\left(\hat{y}^{[m]} - y^{[m]} \right)^2 \right]= \frac{1}{M} \sum_{m=1}^M 2\left(\hat{y}^{[m]} - y^{[m]} \right) w_2 \hspace{0.2em}g^{(1)^{'}}\left(w_1 x^{[m]} + b_1 \right) x^{[m]} \,.$$

I'm trying to find a combination of parameters (weights and biases) and a dataset (of any size $M > 1$) such that $\displaystyle \frac{\partial L_\text{MSE}}{\partial b_2}$ and $\displaystyle\frac{\partial L_\text{MSE}}{\partial w_2}$ are 0 but $\displaystyle \frac{\partial L_\text{MSE}}{\partial b_1}$ and $\displaystyle \frac{\partial L_\text{MSE}}{\partial w_1}$ are not equal to 0.

However, I've been struggling to solve this problem and find a combination that works. Any help would be much appreciated.

$\endgroup$
3
  • $\begingroup$ Does this answer your question? Gradients of lower layers of NN when gradient of an upper layer is 0? $\endgroup$
    – Broele
    Aug 17 at 15:25
  • $\begingroup$ @Broele Hi, it doesn't quite, no. In that question, you say "If we consider the (accumulated) gradients of a batch, then your assumption is not true." I am trying to find an example of this, which shows the upper layer having 0 gradients but not the lower layers, hence my question. $\endgroup$
    – VJ123
    Aug 17 at 16:12
  • 1
    $\begingroup$ My mistake. I hope my answer helps. $\endgroup$
    – Broele
    Aug 17 at 21:58

1 Answer 1

2
$\begingroup$

There are simple examples. I will give one, here and later give an intuition how to find it.

Example

Set $w1=w2=1$ and $b1=b2=-1$.

with $M=3$, $x^{[m]}=(0,2,3)$ and $y^{[m]}=(-2,2,0)$

This leads to: $$\begin{align} a^{(1)^{[m]}} &= (0, 1, 2)\\ \hat{y}^{[m]} &= (-1, 0, 1)\\ (\hat{y}^{[m]} - y^{[m]}) &= (1,-2, 1)\\ \frac{\partial L_{\mathrm{MSE}}}{\partial b_2} &= \frac{2}{M}(1-2+1)=0\\ \frac{\partial L_{\mathrm{MSE}}}{\partial w_2} &= \frac{2}{M}(1\cdot 0-2\cdot 1+1\cdot 2)=0\\ \frac{\partial L_{\mathrm{MSE}}}{\partial b_1} &= \frac{2}{M}(1\cdot 0-2\cdot 1+1\cdot 1)=-\frac{2}{3}\\ \frac{\partial L_{\mathrm{MSE}}}{\partial w_1} &= \frac{2}{M}(1\cdot 0\cdot 0-2\cdot 1\cdot 2+1\cdot 1\cdot 3)=-\frac{2}{3}\\ \end{align}$$

How to find it
  1. First observation is, that for each sample every gradient is multiplied by $2(\hat{y}-y)$. So we can start by setting $y=\hat{y}-0.5$ to set this factor to $1$ (we will later change it)

  2. Now we need to find a setting ($w_1,w_2,b_1,b_2$ and samples $x^{[m]}$ so that there is a solution for the linear equation $$\lambda^{[m]}\cdot\left[\frac{\partial L_{\mathrm{MSE}}}{\partial w_1}, \frac{\partial L_{\mathrm{MSE}}}{\partial b_1}, \frac{\partial L_{\mathrm{MSE}}}{\partial w_2}, \frac{\partial L_{\mathrm{MSE}}}{\partial b_2}\right]=(r_1,r_2,0,0)$$ with $r_1\not=0\not=r_2$

  3. If $a^{(1)}=0$ for a sample, then every gradient, exept the one for $b_2$ is zero. This means we can reduce the linear equation above and ignore $\frac{\partial L_{\mathrm{MSE}}}{\partial b_2}$ for the moment. We later add a sample that moves the gradient to 0 ($x=0$ in our example).

  4. $w_1$ and $w_2$ are just linear factors and can mostly be ignored for building an example, so we set them to $1$.

  5. The value of $b_2$ doesn't matter for the gradients (if we can choose $y$ freely)

  6. This is trial and error. Try to find $x_1$, $x_2$, $b_1$, so that a linear combination of the gradients leads to $(r_1,r_2,0)$

  7. Add the sample for $b_2$ (see step 3) with a fitting weight.

  8. Change $y$ so that it fits to the weights of the linear combination that we found.

$\endgroup$
3
  • $\begingroup$ This is perfect, exactly what I was looking for! Very cool approach too, I definitely wouldn't have thought of that myself. Thanks a lot :) $\endgroup$
    – VJ123
    Aug 18 at 10:24
  • 1
    $\begingroup$ Note that with a different architecture (e.g. classification with sigmoid output and cross-entropy loss, or different activation functions) the approach might not work. I am even not sure that there exists an example in every case. $\endgroup$
    – Broele
    Aug 18 at 17:26
  • $\begingroup$ Okay thanks for clarifying. $\endgroup$
    – VJ123
    Aug 19 at 16:13

Your Answer

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

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