I am using single observation to compute losses using neural network implementation in PyTorch. I am confused in a small detail of SGD. If I compute loss and do loss.backward(), I am accumulating gradients. If I do this on 100 observations and then run optimizer.step(), should I average out the gradients?

This is what I am doing as of now:

def compute_loss(training_data):
for data in training_data:
loss = F.mse_loss(data, data)
loss.backward()

def optimize(sample):
compute_loss(sample)
optimizer.step()


Should it be rather:

def compute_loss(training_data):
for data in training_data:
loss = F.mse_loss(data, data)
loss.backward(torch.Tensor([1.0/len(training_data)]))

• I was assuming that learning-rate is not dependent on the batch size. Given everything else is same I think I should average out gradients or gradients will cumulate over samples. Jul 24, 2018 at 20:23

The following assumes a loss function $$f$$ that's expressed as a sum, not an average. Expressing the loss as an average means that the scaling $$\frac{1}{n}$$ is "baked in" and no further action is needed. In particular, note that F.mse_loss uses reduction="mean" by default, so in the case of OP's code, no further modification is necessary to achieve an average of gradients. Indeed, rescaling the gradients and using reduction="mean" does not accomplish the desired result and amounts to a reduction in the learning rate by a factor of $$\frac{1}{n}$$.
Suppose that $$G = \sum_{i=1}^n \nabla f(x_i)$$ is the sum of the gradients for some minibatch with $$n$$ samples. The SGD update with learning rate (step size) $$r$$ is $$x^{(t+1)} = x^{(t)}- r G.$$
Now suppose that you use the mean of the gradients instead. This will change the update. If we use learning rate $$\tilde{r}$$, we have $$x^{(t+1)} = x^{(t)}- \frac{\tilde{r}}{n} G.$$ These expressions can be made to be equal by re-scaling either $$r$$ or $$\tilde{r}$$. So in that sense, the distinction between the mean and the sum is unimportant because $$r$$ is chosen by the researcher in either case, and choosing a good $$r$$ for the sum has an equivalent, rescaled $$\tilde{r}$$ for the mean.