I have a question about how the averaging works when doing mini-batch gradient descent.

I think I now understood the general gradient descent algorithm, but only for online learning. When doing mini-batch gradient descent, do I have to:

  • forward propagate

  • calculate error

  • calculate all gradients

...repeatedly over all samples in the batch, and then average all gradients and apply the weight change?

I thought it would work that way, but recently I have read somewhere that you basically only average the error of each example in the batch, and then calculate the gradients at the end of each batch. That left me wondering though, because, the activations of which sample in the mini-batch am I supposed to use to calculate the gradients at the end of every batch?

It would be nice if somebody could explain what exactly happens during mini-batch gradient descent, and what actually gets calculated and averaged.


2 Answers 2



First of all, it's completely normal that you are confused because nobody really explains this well and accurately enough. Here's my partial attempt to do that. So, this answer doesn't completely answer the original question. In fact, I leave some unanswered questions at the end (that I will eventually answer).

The gradient is a linear operator

The gradient operator $\nabla$ is a linear operator, because, for some $f : \mathbb{R} \rightarrow \mathbb{R} $ and $g: \mathbb{R} \rightarrow \mathbb{R}$, the following two conditions hold.

  • $\nabla(f + g)(x) = (\nabla f)(x) + (\nabla g)(x),\; \forall x \in \mathbb{R}$
  • $\nabla(kf)(x) = k(\nabla f)(x),\; \forall k, x \in \mathbb{R}$

In other words, the restriction, in this case, is that the functions are evaluated at the same point $x$ in the domain. This is a very important restriction to understand the answer to your question below!

The linearity of the gradient directly follows from the linearity of the derivative. See a simple proof here.


For example, let $f(x) = x^2$, $g(x) = x^3$ and $h(x) = f(x) + g(x) = x^2 + x^3$, then $\frac{dh}{dx} = \frac{d (x^2 + x^3)}{d x} = \frac{d x^2}{d x} + \frac{d x^3}{d x} = \frac{d f}{d x} + \frac{d g}{d x} = 2x + 3x$.

Note that both $f$ and $g$ are not linear functions (i.e. straight-lines), so the linearity of the gradients is not just applicable in the case of straight-lines.

Straight-lines are not necessarily linear maps

Before proceeding, I want to note that there are at least two notions of linearity.

  1. There's the notion of a linear map (or linear operator), i.e. which is the definition above (i.e. the gradient operator is a linear operator because it satisfies the two conditions, i.e. it preserves addition and scalar multiplication).

  2. There's the notion of a straight-line function: $f(x) = c*x + k$. A function can be a straight-line and not be a linear map. For example, $f(x) = x+1$ is a straight-line but it doesn't satisfy the conditions above. More precisely, in general, $f(x+y) \neq f(x) + f(y)$, and you can easily verify that this is the case if $x = 2$ and $y=3$ (i.e. $f(2+3) = 6$, $f(2) = 3$, $f(3) = 4$, but $f(2) + f(3) = 7 \neq f(2+3)$.

Neural networks

A neural network is a composition of (typically) non-linear functions (let's ignore the case of linear functions), which can thus be represented as $$y'_{\theta}= f^{L}_{\theta_L} \circ f^{L-1}_{\theta_{L-1}} \circ \dots \circ f_{\theta_1},$$ where

  • $f^{l}_{\theta_l}$ is the $i$th layer of your neural network and it computes a non-linear function
  • ${\theta_l}$ is a vector of parameters associated with the $l$th layer
  • $L$ is the number of layers,
  • $y'_{\theta}$ is your neural network,
  • $\theta$ is a vector containing all parameters of the neural network
  • $y'_{\theta}(x)$ is the output of your neural network
  • $\circ $ means the composition of functions

Given that $f^l_{\theta}$ are non-linear, $y'_{\theta}$ is also a non-linear function of the input $x$. This notion of linearity is the second one above (i.e. $y'_{\theta}$ is not a straight-line). In fact, neural networks are typically composed of sigmoids, ReLUs, and hyperbolic tangents, which are not straight-lines.

Sum of squared errors

Now, for simplicity, let's consider the sum of squared error (SSE) as the loss function of your neural network, which is defined as

$$ \mathcal{L}_{\theta}(\mathbf{x}, \mathbf{y}) = \sum_{i=1}^N \mathcal{S}_{\theta}(\mathbf{x}_i, \mathbf{y}_i) = \sum_{i=1}^N (\mathbf{y}_i - y'_{\theta}(\mathbf{x}_i))^2 $$ where

  • $\mathbf{x} \in \mathbb{R}$ and $\mathbf{y} \in \mathbb{R}$ are vectors of inputs and labels, respectively
  • $\mathbf{y}_i$ is the label for the $i$th input $\mathbf{x}_i$
  • $\mathcal{S}_{\theta}(\mathbf{x}_i, \mathbf{y}_i) = (\mathbf{y}_i - y'_{\theta}(\mathbf{x}_i))^2$

Sum of gradients vs gradient of a sum

Given the gradient is a linear operator, one could think that computing the sum of the gradients is equal to the gradient of the sums.

However, in our case, we are summing $\mathcal{S}_{\theta}(\mathbf{x}_i, \mathbf{y}_i)$ and, in general, $\mathbf{x}_i \neq \mathbf{x}_j$, for $i \neq j$. So, essentially, the SSE is the sum of the same function, i.e. $S_{\theta}$, evaluated at different points of the domain. However, the definition of a linear map applies when the functions are evaluated at the same point in the domain, as I said above.

So, in general, in the case of neural networks with SSE, the gradient of the sum may not be equal to the sum of gradients, i.e. the definition of the linear operator for the gradient doesn't apply here because we are evaluating every squared error at different points of their domains.

Stochastic gradient descent

The idea of stochastic gradient descent is to approximate the true gradient (i.e. the gradient that would be computed with all training examples) with a noisy gradient (which is an approximation of the true gradient).

How does the noisy gradient approximate the true gradient?

In the case of mini-batch ($M \leq N$, where $M$ is the size of the mini-batch and $N$ is the total number of training examples), this is actually a sum of the gradients, one for each example in the mini-batch.

The papers Bayesian Learning via Stochastic Gradient Langevin Dynamics (equation 1) or Auto-Encoding Variational Bayes (in section 2.2) use this type of approximation. See also these slides.


To give you some intuition of why we sum the gradients of the error of each input point $\mathbf{x}_i$, let's consider the case $M=1$, which is often referred to as the (actual) stochastic gradient descent algorithm.

Let's assume we uniformly sample an arbitrary tuple $(\mathbf{x}_j, \mathbf{y}_j)$ from the dataset $\mathcal{D} = \{ (\mathbf{x}_i, \mathbf{y}_i) \}_{i=1}^N$.

Formally, we want to show that

\begin{align} \nabla_{\theta} \mathcal{L}_{\theta}(\mathbf{x}, \mathbf{y}) &= \mathbb{E}_{(\mathbf{x}_j, \mathbf{y}_j) \sim \mathbb{U}}\left[ \nabla_{\theta} \mathcal{S}_{\theta} \right] \label{1} \tag{1} \end{align}


  • $\nabla_{\theta} \mathcal{S}_{\theta}$ is the gradient of $\mathcal{S}_{\theta}$ with respect to the parameters $\theta$

  • $\mathbb{E}_{(\mathbf{x}_j, \mathbf{y}_j) \sim \mathbb{U}}$ is the expectation with respect to the random variable associated with a sample $(\mathbf{x}_j, \mathbf{y}_j)$ from the uniform distribution $\mathbb{U}$

Under some conditions (see this), we can exchange the expectation and gradient operators, so \ref{1} becomes \begin{align} \nabla_{\theta} \mathcal{L}_{\theta}(\mathbf{x}, \mathbf{y}) &= \nabla_{\theta} \mathbb{E}_{(\mathbf{x}_j, \mathbf{y}_j) \sim \mathbb{U}}\left[ \mathcal{S}_{\theta} \right] \label{2} \tag{2} \end{align} Given that we uniformly sample, the probability of sampling an arbitrary $(\mathbf{x}_j, \mathbf{y}_j)$ is $\frac{1}{N}$. So, equation \ref{2} becomes \begin{align} \nabla_{\theta} \mathcal{L}_{\theta} (\mathbf{x}, \mathbf{y}) &= \nabla_{\theta} \sum_{i=1}^N \frac{1}{N} \mathcal{S}_{\theta}(\mathbf{x}_i, \mathbf{y}_i) \\ &= \nabla_{\theta} \frac{1}{N} \sum_{i=1}^N \mathcal{S}_{\theta}(\mathbf{x}_i, \mathbf{y}_i) \end{align}

Note that $\frac{1}{N}$ is a constant with respect to the summation variable $i$ and so it can be taken out of the summation.

This shows that the gradient with respect to $\theta$ of the loss function $\mathcal{L}_{\theta}$ that includes all training examples is equivalent, in expectation, to the gradient of $\mathcal{S}_{\theta}$ (the loss function of one training example).


  1. How can we extend the previous proof to the case $1 < M \leq N$?

  2. Which conditions need exactly to be satisfied so that we can exchange the gradient and the expectation operators? And are they satisfied in the case of typical loss functions, or sometimes they aren't (but in which cases)?

  3. What is the relationship between the proof above and the linearity of the gradient?

    • In the proof above, we are dealing with expectations and probabilities!
  4. What would the gradient of a sum of errors represent? Can we still use it in place of the sum of gradients?


do I have to:

  • forward propagate

  • calculate error

  • calculate all gradients

  • ...repeatedly over all samples in the batch, and then average all gradients and apply the weight change?

Yes, that is correct. You can save a bit of memory by summing gradients as you go. Once you have calculated the gradients for one example for the weights of one layer, then you do not re-use the individual gradients again, so you can just keep a sum. Alternatively for speed, you can calculate a minibatch of gradients in parallel, as each example is independent - which is a major part of why GPU acceleration is so effective in neural network training.

It is critical to getting correct results that you calculate the gradient of the loss function with respect to each example input/output pair separately. Once you have done that, you can average the gradients across a batch or mini-batch to estimate a true gradient for the dataset which can be used to take a gradient descent step.

recently I have read somewhere that you basically only average the error of each example in the batch, and then calculate the gradients at the end of each batch.

Without a reference it is hard to tell whether this is an error in the "somewhere", or you have misunderstood, or there is a specific context.

If by "error" you mean the literal difference $\hat{y}_i - y_i$, where $\hat{y}_i$ is your estimate for data input $i$ and $y_i$ is the ground-truth training value, then that is the gradient for many loss functions and activation function pairs. For instance, it is the error gradient for mean square error and linear output. Some texts loosely refer to this as the "error", and talk about backpropagating "the error", but actually it is a gradient.

In addition, if the article was referring to linear regression, logistic regression or softmax regression, everything else is linear - in those specific models then you can just "average the error" and use that as the gradient.

In general, however, the statement is incorrect because a neural network with one or more hidden layers has many non-linearities that will give different results when calculating average first then backpropagating vs taking backpropagating first the averaging - that is $f'(\mu(Y))$ vs $\mu(f'(Y))$ where $f'$ is the derivative of the transfer function and $\mu$ is the mean for the batch (i.e. $\mu(Y) = \frac{1}{N}\sum_{i=1}^{N} y_i$ and $Y$ represents all the $y_i$ in a given batch of size $N$)

When $y_i = f(x_i) = ax_i +b$ i.e. the transfer function is linear, then $f'(\mu(Y)) = \mu(f'(Y)) = \frac{a}{N}\sum_{i=1}^N x_i$, but almost all useful loss functions and all transfer functions except some output layers in neural networks are non-linear. For those, $f'(\mu(Y)) \neq \mu(f'(Y))$.

A simple example would show this, if we start a small minibatch back propagation with the loss function (as opposed to its gradient).

Say you had the following data for regression:

  x    y

  1    2
  1    4

You want a model that can regress to least mean squared error $y$ when given an input $x = 1$. The best model should predict $3$ in that case.

If your model has converged, the average MSE of the dataset is $1$. Using that would make your model move away from convergence and it will perform worse.

If you first take the gradients, then average those, you will calculate $0$. A simple gradient update step using that value will make no change, leaving the model in the optimal position.

This issue occurs on every hidden layer in a neural network, so in general you cannot simply resolve the loss function gradient and start with the average error gradient at the output. You would still hit the inequality $f'(\mu(Y)) \neq \mu(f'(Y))$ on each nonlinearly.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – nbro
    Commented Apr 19, 2020 at 17:23

You must log in to answer this question.