Is applying dropout equivalent to zeroing output of random neurons in each mini-batch iteration and leaving rest of forward and backward steps in back-propagation unchanged? I'm implementing network from scratch in numpy.

  • 1
    $\begingroup$ Yes, although just to be super-duper-extra precise, bernoulli dropout is the same as zeroing-out random neurons (some people use other kinds of randomness and call it things like Gaussian dropout; see e.g. keras.io/api/layers/regularization_layers/gaussian_dropout) $\endgroup$ Nov 9, 2022 at 21:30
  • $\begingroup$ @Qbik please see edits to my reply below. $\endgroup$
    – hH1sG0n3
    Nov 10, 2022 at 9:07

1 Answer 1


Indeed. To be precise, the dropout operation will randomly zero some of the input tensor elements with probability $p$, and furthermore the rest of the non-dropped out outputs are scaled by a factor of $\frac{1}{1-p}$ during training.

For example, see how elements of each tensor in the input (top tensor in output) are zeroed in the output tensor (bottom tensor in output) using pytorch.

m = nn.Dropout(p=0.5)
input = torch.randn(3, 4)
output = m(input)

print(input, '\n', output)

>>> tensor([[-0.9698, -0.9397,  1.0711, -1.4557],
>>>        [-0.0249, -0.9614, -0.7848, -0.8345],
>>>        [ 0.9420,  0.6565,  0.4437, -0.2312]]) 
>>> tensor([[-0.0000, -0.0000,  2.1423, -0.0000],
>>>        [-0.0000, -0.0000, -1.5695, -1.6690],
>>>        [ 0.0000,  0.0000,  0.0000, -0.0000]])

EDIT: please note the post has been updated to reflect Todd Sewell's addition in the comments.

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    $\begingroup$ Note that non-dropped out elements are scaled by 1/(1-p) to compensate for the shift in average magnitude, so it's not just zeroing out some elements. $\endgroup$ Nov 9, 2022 at 20:12
  • $\begingroup$ That is very true, I omitted that info from the original pytorch docs for simplicity however in that way makes the post only half correct. Amended now to reflect your point. $\endgroup$
    – hH1sG0n3
    Nov 9, 2022 at 22:17

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