This question not related to a specific method or technique, rather there is a broader concept that I'm struggling to see clearly.


In machine learning, we have loss functions that we're trying to minimize. Gradient descent is a general solution to minimizing the output of these loss functions. I understand the basic idea there, but it's the details that I'm getting stuck on.

Suppose that I have some loss function $J_\Theta(x)$, where I have some $x$ input, where $\Theta$ is just some matrix of arbitrary parameters.

In many examples, I see $\Theta$ written something like:

$\Theta = \begin{bmatrix} \hat{x_0} \\ \hat{x_1} \\ \hat{x_2} \\ \vdots \\ \end{bmatrix} $

and each row of $\Theta$ is actually a vector.

Note that some concrete examples that I've run across include updating word embeddings in word2vec, or updating a softmax layer. I'm happy to elaborate if my explanation is too abstract.

In the examples of gradient descent that I've seen, typically the derivative of $J$ is taken w.r.t each row of $\Theta$, not individual elements.

So something like $\frac{dJ}{d\hat{x_0}}$, where each vector in $\Theta$ is updated according to the output of this gradient function.

Now for my point of confusion:

Suppose the parameters are initialized to zero, which is sometimes a thing. Wouldn't the updates be the same for each element of $\hat{x_i}$ in $\Theta$ during the update step? Wouldn't that lead to each vector having the same numbers for each element? I know that's the wrong conclusion, but I'm not able to see how each dimension would, in the end, become different values. I assume (hope) I'm missing something simple.


1 Answer 1


I reached the same conclusion for a general case below. However, in practice, at least in the case of neural networks, weights are initialized randomly; initializing to the same value is avoided.

A general problematic case

Suppose data is $x=(x_0, x_1)$, where $x_0$ is the vector of features and $x_1$ is the outcome. Let the loss function be:

$$J_{\theta}(x) = (x_1 - f_{\theta}(x_0))^2$$

Only one data point $x$ is considered for simplicity (for a set of points, conclusion would be the same since the argument applies to each term in the summation).

Each parameter $\hat{x}_i$ could be a vector or a single parameter. Gradient of loss would be:

$$\nabla_{\hat{x}_i}J_{\theta}(x) = 2\nabla_{\hat{x}_i}f_{\theta}(x_0)(x_1 - f_{\theta}(x_0))$$

According to this gradient, if two parameters $i$ and $j$ have symmetric roles in $f$, i.e. $$\nabla_{\hat{x}_i}f_{\theta} = \nabla_{\hat{x}_j}f_{\theta},$$ their corresponding loss gradient will also be the same since the components $\nabla f_{\theta}(x_0)$, $x_1$, and $f_{\theta}(x_0)$ are all the same. A concrete example would be a neural network with equal weights and mean squared loss function. All weights between two specific layers would have the same role in the network, thus they will remain equal after each update. However, in practice, weights of neural networks are initialized randomly which breaks this role symmetry between the weights.

A specific counterexample

For example, consider 2D data $x=(x_0, x_1)$ and two 1D parameters $\theta=[\hat{x}_0, \hat{x}_1]$, and let the loss function be: $$J_{\theta}(x)= \hat{x}_0 x_0 + \hat{x}_1 x_1,$$ for a batch of one point $x$ (this is for simplicity to avoid a summation over points in the batch).

The gradient w.r.t. parameters is: $$\frac{\partial J_{\theta}(x)}{\partial \hat{x}_0} = x_0,\mbox{and }\frac{\partial J_{\theta}(x)}{\partial \hat{x}_1} = x_1,$$

This illustrates the dependency of gradient on data $x=(x_0, x_1)$. Now, if both parameters are zero, i.e. $\hat{x}_0=\hat{x}_1=0$, the gradient is still different and non zero. More specifically, suppose learning rate is $\lambda$, the next values for parameters would be: $$\hat{x}'_0 = \hat{x}_0 - \lambda\frac{\partial J_{\theta}(x)}{\partial \hat{x}_0} = 0 - \lambda x_0 \neq 0 - \lambda x_1 = \hat{x}_1 -\lambda\frac{\partial J_{\theta}(x)}{\partial \hat{x}_1} = \hat{x}'_1$$

But, what if $x_0=x_1$?
In this case, parameters always remain the same if we always use specific data point $x$ to update the parameters. However, this case is pathological (unlikely). Because, to let this equality keep going, any other data point $y$ that we peak must satisfy $y_0=y_1$ too. So in this example, the problem is unlikely to happen.

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    $\begingroup$ Thanks for the response! I'll take a closer look at this when I've got some time tonight! $\endgroup$ Mar 13, 2019 at 17:52
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    $\begingroup$ In your example, $\hat{x_0}$ and $\hat{x_1}$ are both 1D. But I'm wondering what happens when they aren't. Suppose they are now 2D vectors. If I understand correctly, even when $x_0$ and $x_1$ are different, $\hat{x}'_0$ and $\hat{x}'_1$ would be different from one another, but the numbers within the vectors have the same operations being performed on them (in your example, $0 - \lambda x_0$ and $0 - \lambda x_1$, respectively) $\endgroup$ Mar 14, 2019 at 11:56
  • $\begingroup$ @wheresmycookie your points led me to come up with a general problematic case, which means the problem is not that unlikely that I thought. $\endgroup$
    – Esmailian
    Mar 14, 2019 at 13:10

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