I have just heard, that it's a good idea to choose initial weights of a neural network from the range $(\frac{-1}{\sqrt d} , \frac{1}{\sqrt d})$, where $d$ is the number of inputs to a given neuron. It is assumed, that the sets are normalized - mean 0, variance 1 (don't know if this matters).

Why is this a good idea?


7 Answers 7


I assume you are using logistic neurons, and that you are training by gradient descent/back-propagation.

The logistic function is close to flat for large positive or negative inputs. The derivative at an input of $2$ is about $1/10$, but at $10$ the derivative is about $1/22000$ . This means that if the input of a logistic neuron is $10$ then, for a given training signal, the neuron will learn about $2200$ times slower that if the input was $2$.

If you want the neuron to learn quickly, you either need to produce a huge training signal (such as with a cross-entropy loss function) or you want the derivative to be large. To make the derivative large, you set the initial weights so that you often get inputs in the range $[-4,4]$.

The initial weights you give might or might not work. It depends on how the inputs are normalized. If the inputs are normalized to have mean $0$ and standard deviation $1$, then a random sum of $d$ terms with weights uniform on $(\frac{-1}{\sqrt{d}},\frac{1}{\sqrt{d}})$ will have mean $0$ and variance $\frac{1}{3}$, independent of $d$. The probability that you get a sum outside of $[-4,4]$ is small. That means as you increase $d$, you are not causing the neurons to start out saturated so that they don't learn.

With inputs which are not normalized, those weights may not be effective at avoiding saturation.

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    $\begingroup$ So basically, one should always at least consider normalizing data.. It makes sense now. Could you explain why the std deviation will be 1/3 and how small is the probability of input sum outside the range <-4,4>? $\endgroup$
    – elmes
    Commented Jan 13, 2013 at 9:50
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    $\begingroup$ There are some basic properties of variance which imply this: If $X$ and $Y$ are independent, then $\text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y)$ and if $X$ and $Y$ are independent and have mean $0$, then $\text{Var}(X*Y) = \text{Var}(X)*\text{Var}(Y)$. $\endgroup$ Commented Jan 14, 2013 at 6:55
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    $\begingroup$ You can estimate the probability that a random variable is at least $12$ standard deviations away from the mean using the Chebyshev inequality. In practice this is not sharp, but the exact result depends on the distribution. $\endgroup$ Commented Jan 14, 2013 at 6:56
  • $\begingroup$ By the way, I miscalculated. The variance is $\frac{1}{3}$ so the standard deviation is $\sqrt{\frac13}$. $\endgroup$ Commented Jan 14, 2013 at 19:35
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    $\begingroup$ "The logistic function is close to flat for large positive or negative inputs. The derivative at an input of ..." Shouldn't the relevant subject be the derivative of the cost function of logistic regression? Whereby the input to the derivative of the cost function is already scaled by the logistic function to (0,1) regardless of the size of the weights and signals? $\endgroup$
    – Moobie
    Commented Nov 3, 2017 at 20:11

[1] addresses the question:

First, weights shouldn't be set to zeros in order to break the symmetry when backprogragating:

Biases can generally be initialized to zero but weights need to be initialized carefully to break the symmetry between hidden units of the same layer. Because different output units receive different gradient signals, this symmetry breaking issue does not concern the output weights (into the output units), which can therefore also be set to zero.

Some initialization strategies:

  • [2] and [3] recommend scaling by the inverse of the square root of the fan-in
  • Glorot and Bengio (2010) and the Deep Learning Tutorials use a combination of the fan-in and fan-out:
    • for hyperbolic tangent units: sample a Uniform(-r, r) with $r=\sqrt{\frac{6}{\text{fan-in}+\text{fan-out}}}$ (fan-in is the number of inputs of the unit).
    • for sigmoid units : sample a Uniform(-r, r) with $r=4 \sqrt{\frac{6}{\text{fan-in}+\text{fan-out}}}$ (fan-in is the number of inputs of the unit).
  • in the case of RBMs, a zero-mean Gaussian with a small standard deviation around 0.1 or 0.01 works well (Hinton, 2010) to initialize the weights.
  • Orthogonal random matrix initialization, i.e. W = np.random.randn(ndim, ndim); u, s, v = np.linalg.svd(W) then use u as your initialization matrix.

Also, unsupervised pre-training may help in some situations:

An important choice is whether one should use unsupervised pre-training (and which unsupervised feature learning algorithm to use) in order to initialize parameters. In most settings we have found unsupervised pre-training to help and very rarely to hurt, but of course that implies additional training time and additional hyper-parameters.

Some ANN libraries also have some interesting lists, e.g. Lasagne:

Constant([val]) Initialize weights with constant value.
Normal([std, mean]) Sample initial weights from the Gaussian distribution.
Uniform([range, std, mean]) Sample initial weights from the uniform distribution.
Glorot(initializer[, gain, c01b])   Glorot weight initialization.
GlorotNormal([gain, c01b])  Glorot with weights sampled from the Normal distribution.
GlorotUniform([gain, c01b]) Glorot with weights sampled from the Uniform distribution.
He(initializer[, gain, c01b])   He weight initialization.
HeNormal([gain, c01b])  He initializer with weights sampled from the Normal distribution.
HeUniform([gain, c01b]) He initializer with weights sampled from the Uniform distribution.
Orthogonal([gain])  Intialize weights as Orthogonal matrix.
Sparse([sparsity, std]) Initialize weights as sparse matrix.

[1] Bengio, Yoshua. "Practical recommendations for gradient-based training of deep architectures." Neural Networks: Tricks of the Trade. Springer Berlin Heidelberg, 2012. 437-478.

[2] LeCun, Y., Bottou, L., Orr, G. B., and Muller, K. (1998a). Efficient backprop. In Neural Networks, Tricks of the Trade.

[3] Glorot, Xavier, and Yoshua Bengio. "Understanding the difficulty of training deep feedforward neural networks." International conference on artificial intelligence and statistics. 2010.

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    $\begingroup$ I`d like to add two useful references: 1)Delving Deep into Rectifiers: Surpassing Human-Level Performance on ImageNet Classification -- about importance of activation-aware scaling arxiv.org/abs/1502.01852 2)Exact solutions to the nonlinear dynamics of learning in deep linear neural networks arxiv.org/abs/1312.6120 - orthonormal matrices are much better than just Gaussian noise $\endgroup$
    – old-ufo
    Commented Jan 26, 2016 at 11:31
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    $\begingroup$ An editor suggests the initializations for the sigmoid and the hyperbolic tangent should be switched to match the original paper. $\endgroup$ Commented Aug 15, 2016 at 11:20
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    $\begingroup$ Did you want to keep this edit, Frank? If not, you can roll it back. $\endgroup$ Commented Aug 16, 2016 at 13:51
  • $\begingroup$ I must be missing something. Where does it say in the Glorot and Bengio (2010) paper that they recommend to use 4 times the Equation 16 value when using logistic sigmoid activation functions? Equation 16 follows from using Equation 12 and the variance of a uniform distribution, but Equation 16 is derived assuming a symmetric activation with unit derivative at 0. Thus, e.g. a tanh activation function, but not a logistic activation function (non-symmetric). Further, they don't even test this proposed initialisation with logistic sigmoid; they only test it with tanh and softsign. $\endgroup$
    – Tommy L
    Commented Feb 28, 2017 at 8:54
  • $\begingroup$ Page 15 of ref [1] propose weights the way the answer has them. I.e. weights for sigmoid are 4x the weights for tanh. Ref [3] doesn't seem to mention these. $\endgroup$
    – Aelian
    Commented Jul 23, 2020 at 4:56

The following explanation is taken from the book: Neural Networks for Pattern Recognition by Christopher Bishop. Great book! Assume you have previously whitened the inputs to the input units, i.e. $$<x_{i}> = 0$$ and $$<x_{i}^{2}> = 1$$

The question is: how to best choose the weights?. The idea is to pick values of the weights at random following a distribution which helps the optimization process to converge to a meaningful solution.

You have for the activation of the units in the first layer, $$y = g(a) $$ where $$ a = \sum_{i=0}^{d}w_{i}x_{i}$$. Now, since you choose the weights independently from the inputs, $$<a> = \sum_{i=0}^{d}<w_{i}x_{i}> = \sum_{i=0}^{d}<w_{i}><x_{i}> = 0$$ and $$ <a^2> = \left<\left(\sum_{i=0}^{d}w_{i}x_{i}\right) \left(\sum_{i=0}^{d}w_{i}x_{i}\right)\right> = \sum_{i=0}^{d}<w_{i}^{2}><x_{i}^{2}> = \sigma^{2}d $$ where sigma is the variance of the distribution of weights. To derive this result you need to recall that weights are initialized independently from each other, i.e. $$<w_{i}w_{j}> = \delta_{ij}$$

  • $\begingroup$ Minor mistake: $<x_i^2> = 1$ instead of $0$. $\endgroup$
    – bayerj
    Commented Jan 13, 2013 at 21:30
  • $\begingroup$ This explains how you reach a ceratin $\sigma$ assuming you know the required $\alpha$. As I understand, $\alpha$ should be small to allow a big value of the sigmoid derivative, but not too small so that the deltas won't vanish. Is this true? If so - is it a good rule of thumb to say that $\alpha$ should be ~0.2? $\endgroup$
    – Uri
    Commented Apr 23, 2013 at 15:00
  • $\begingroup$ This is specially true for deep neural networks, where units tend to saturate quickly as you add layers. There are a number of papers dealing with that question. A good start point might be "Understanding the difficulty of training deep feedforward neural networks" by glorot and bengio $\endgroup$
    – jpmuc
    Commented Apr 24, 2013 at 7:10

Well just as an update, Delving Deep into Rectifiers: Surpassing Human-Level Performance n ImageNet Classification by He et al introduced an initialization specifically with initialization w = U([0,n]) * sqrt(2.0/n) where n is the number of inputs of your NN. I have seen this initialization used in many recent works (also with ReLU). They actually show how this starts to reduce the error rate much faster than the (-1/n, 1/n) that you mentioned. For the thorough explanation, see the paper but here's how fast it converges: The convergence of a 22-layer large model

  • $\begingroup$ Wow! Significant improvement for me. $\endgroup$ Commented Jun 8, 2017 at 14:20
  • $\begingroup$ Not for big numbers of inputs though... fails with MNIST. $\endgroup$ Commented Jun 12, 2017 at 16:37
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    $\begingroup$ Note that He-initialisation is specifically designed for (P)ReLUs and accounts for the fact that it is not symmetric (which is one of the assumptions in Xavier-initialisation). Don't be fooled by this graph out of context! $\endgroup$ Commented Aug 24, 2017 at 15:46

The idea is that you want to initialize the weights in a way that ensures good forward and backward data flow through the network. That is, you don't want the activations to be consistently shrinking or increasing as you progress through the network.

This image shows the activations of a 5 layer ReLU Multi-Layer Perceptron under 3 different initialization strategies after one pass of MNIST through the network.

Activations in a ReLU MLP with different initialization strategies

In all three cases weights are drawn from a zero-centered normal distribution which is determined by its standard deviation. You can see that if the initial weights are too small (the standard deviation is small) the activations get choked, and that if they are too large the activations explode. The middle value, that is approximately right can be found by setting the weights such that the variance of the activations and gradient updates stays approximately the same as you pass through the network.

I wrote a blog post about weight initialization that goes into more detail, but the basic idea is as follows.

If $x^{(i)}$ denotes the activations of the $i$-th layer, $n_i$ the size of the layer, and $w^{(i)}$ the weights connecting them to the $(i+1)$-st layer, then one can show that for activation functions $f$ with $f'(s) \approx 1$ we have

$$ \text{Var}(x^{(i+1)}) = n_i \text{Var}(x^{(i)}) \text{Var}(w^{(i)}) $$

In order to achieve $\text{Var}(x^{(i+1)}) = \text{Var}(x^{(i)})$ we therefore have to impose the condition

$$ \text{Var}( w^{(i)}) = \frac{1}{n_i}\,. $$

If we denote $\frac{\partial L}{\partial x_j^{(i)}}$ by $\Delta_j^{(i)}$, on the backward pass we similarly want

$$ \text{Var}(\Delta^{(i)} ) = n_{i+1} \text{Var}(\Delta^{(i+1)}) \text{Var}(w^{(i)})\,. $$

Unless $n_i = n_{i+1}$, we have to compromise between these two conditions, and a reasonable choice is the harmonic mean

$$ \text{Var}(w^{(i)}) = \frac{2}{n_i+n_{i+1}}\,. $$

If we sample weights from a normal distribution $N(0, \sigma) $ we satisfy this condition with $\sigma = \sqrt{\frac{2}{n_i + n_{i+1}}} $. For a uniform distribution $U(-a, a) $ we should take $a = \sqrt{\frac{6}{n_i+n_{i+1}}} $ since $\text{Var} \left( U(-a,a) \right) = a^2/3 $. We have thus arrived at Glorot initialization. This is the default initialization strategy for dense and 2D convolution layers in Keras, for instance.

Glorot initialization works pretty well for trivial and $ \tanh $ activations, but doesn't do as well for $ \text{ReLU} $. Luckily, since $ f(s) = \text{ReLU}(s) $ just zeroes out negative inputs, it roughly removes half the variance and this is easily amended by multiplying one of our conditions above by two:

$$ \text{Var}(w^{(i)}) = \frac{2}{n_i}\,. $$


One other technique that alleviates the problem of weight initialization is Batch Normalization. It acts to standardize the mean and variance of each unit in order to stabilize learning as described in the original paper. In practice, networks that use Batch Normalization (BN) are significantly more robust to bad initialization. BN works as follows: $$ \mu_B = \frac{1}{m}\sum_{i=1}^{M}x_i~~~and~~~ \sigma_{B}^{2} = \frac{1}{m}\sum_{i=1}^{m}(x_i - \mu_B)^{2} \\ \hat{x}_i = \frac{x_i - \mu_B}{\sqrt{\sigma_{B}^{2} + \epsilon}}~~~and~~~BN(x_i) = \gamma \hat{x}_i + \beta $$ We compute empirical mean and variance for each mini-batch, then we standardize the input $x_i$ and form the output $BN(x_i)$ by scaling $\hat{x}_i$ by $\gamma$ and adding $\beta$ both of which are learned during training.

BN introduces two extra parameters ($\gamma$ and $\beta$) per activation that allow the $\hat{x}_i$ to have any mean and standard deviation. The reason for that is normalizing $x_i$ can reduce its expressive power. This new parameterization has better learning dynamics: in the old parameterization the mean of $x_i$ was determined by a complicated interaction between parameters of all preceding layers - so small changes to the network parameters amplify as the network becomes deeper. In the new parameterization the mean of $\hat{x}_i$ is determined by $\beta$ that we learn along with $\gamma$ during training. Thus, Batch Normalization stabilizes learning.

As a result, Batch Normalization enables faster training by using much higher learning rates and alleviates the problem of bad initialization. BN also makes it possible to use saturating non-linearities by preventing the network from getting stuck in saturation modes. In summary, Batch Normalization is a differentiable transform that introduces normalized activations into the network. In practice, a BN layer can be inserted immediately after a fully connected layer.


There are two distinct ideas in this heuristic:

  1. Initialize the weights to be small - in addition to Douglas Zare excellent answer about sigmoid activations, the problem is more general. Even when the gradients are of "good" magnitude (e.g., using ReLU activations) training is hampered with big weights. Think about 2 neurons whose real weights should be $(3, -2)$. If you initialize them close to $0$, the maximal "distance" the weights have to traverse is roughly $3.6$ (Euclidean distance; 5 in Manhattan distance). While if you initialize them e.g. from $U(-3,3)$ you run into the risk that in the worst case the initial weights will be set to $(-3,3)$, in that case the distance the weights have to traverse is roughly $7.8$ (Euclidean; 11 Manhatten).

  2. Keep the variance of each weight $\propto \frac{1}{d}$ - the input to the next layer will thus have a variance $\propto 1$ (as it is a sum of $d$ neurons times their respective weights). Why do we want this? We want to keep the magnitude of the inputs to the layers the same. We don't want that inputs from a layer with a lot of hidden units will be much bigger than inputs from a layer with fewer hidden units. If we add a lot of inputs, we want the weights to be relatively smaller in magnitude, and if we're adding fewer inputs we want them to be larger.

The $\frac{1}{3}$ variance constant in the $U(-\frac{1}{\sqrt d}, \frac{1}{\sqrt d})$ heuristic is actually problematic. To keep information flowing we would like that $\mathbb V[a_l] = \mathbb V[a_{l-1}]$- i.e. that the variance of the activations inputs / outputs stay more or less the same across layers, and similarly for backprop that $\mathbb V[\frac{\partial \mathcal L}{\partial z_l}] = \mathbb V[\frac{\partial \mathcal L}{\partial z_{l+1}}]$ i.e., that the variance of the backprop derivatives will be more or less the same. The $\frac{1}{3}$ factor get's in our way. You can see this in the following graphs (taken from the original Xavier Glorot init paper):

enter image description here

Here you can see that the activations follow $\mathbb V[a_l] \approx \frac{1}{3}\mathbb V[a_{l-1}] $. And:

enter image description here

Here you can see that the derivatives follow $\mathbb V[\frac{\partial \mathcal L}{\partial z_l}] \approx \frac{1}{3}\mathbb V[\frac{\partial \mathcal L}{\partial z_{l+1}}]$

In both cases - the narrow parts are bad: in the forward pass (activations) it means each neuron is basically calculating the same thing, and also that we are not really taking advantage of the activation function non-linearity ($\tanh$ were used in this network); in the backprop it means we are not really learning in the early layers.

Xavier Glorot init fixed this by changing the distribution to $U(-\frac{\sqrt 3}{\sqrt n}, \frac{\sqrt 3}{\sqrt n}) $which eliminates the 1/3 factor. Also, since we care about both the previous layer number of neurons (for the forward prop) and the next layer number of neurons (for backprop) - Xavier Glorot init uses the harmonic mean between them as a compromise. Using this, the same network now looks like:

enter image description here

If you want to learn more, check out my YouTube videos on the topic: Part 1 and Part 2.


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