Why is the de-facto standard sigmoid function, $\frac{1}{1+e^{-x}}$, so popular in (non-deep) neural-networks and logistic regression?

Why don't we use many of the other derivable functions, with faster computation time or slower decay (so vanishing gradient occurs less). Few examples are on Wikipedia about sigmoid functions. One of my favorites with slow decay and fast calculation is $\frac{x}{1+|x|}$.


The question is different to Comprehensive list of activation functions in neural networks with pros/cons as I'm only interested in the 'why' and only for the sigmoid.

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    $\begingroup$ Note the logistic sigmoid is a special case of the softmax function, and see my answer to this question: stats.stackexchange.com/questions/145272/… $\endgroup$
    – Neil G
    Commented Jul 24, 2015 at 11:31
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    $\begingroup$ There are other functions like probit or cloglog that are commonly used, see: stats.stackexchange.com/questions/20523/… $\endgroup$
    – Tim
    Commented Jul 24, 2015 at 11:54
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    $\begingroup$ @user777 I am not sure if it is a duplicate since the thread you refer to does not really answer the why question. $\endgroup$
    – Tim
    Commented Jul 24, 2015 at 14:17
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    $\begingroup$ I hate to disagree with so many distinguished community members who voted to close this as a duplicate, but I am persuaded that the apparent duplicate does not address the "why" and so I have voted to reopen this question. $\endgroup$
    – whuber
    Commented Jul 25, 2015 at 14:00
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    $\begingroup$ Just want to add that this paper by Michael I. Jordan titled "Why the logistic function? A tutorial discussion on probabilities and neural networks" provides a satisfying mathematical reason. $\endgroup$
    – mhdadk
    Commented Nov 19, 2020 at 0:24

4 Answers 4


Quoting myself from this answer to a different question:

In section 4.2 of Pattern Recognition and Machine Learning (Springer 2006), Bishop shows that the logit arises naturally as the form of the posterior probability distribution in a Bayesian treatment of two-class classification. He then goes on to show that the same holds for discretely distributed features, as well as a subset of the family of exponential distributions. For multi-class classification the logit generalizes to the normalized exponential or softmax function.

This explains why this sigmoid is used in logistic regression.

Regarding neural networks, this blog post explains how different nonlinearities including the logit / softmax and the probit used in neural networks can be given a statistical interpretation and thereby a motivation. The underlying idea is that a multi-layered neural network can be regarded as a hierarchy of generalized linear models; according to this, activation functions are link functions, which in turn correspond to different distributional assumptions.

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    $\begingroup$ Great! So when we are using sigmoids in a network, we can say we are implicitly assuming that the network "models" probabilities of various events (in the internal layers or in the output). This can be a sensible model inside a network even for squared error (allowing for the output neuron a different activation function). Never thought of this intuition before, thanks! $\endgroup$ Commented Jul 27, 2015 at 0:49
  • $\begingroup$ Historically, not so. My best summary of a messy history is that logit entered statistical science largely because functional forms used to predict change over time (populations expected to follow logistic curves) looked about right when adapted and adopted as link functions [anachronistic use there!] for binary responses; and they are easy to manipulate with simple calculus, which expressions in absolute values aren't. But naturally the simplest logical justification for such functions is interesting and crucial, and your answer addresses that. $\endgroup$
    – Nick Cox
    Commented Dec 11, 2017 at 15:26
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    $\begingroup$ I've read through the sections in both Bishop books (2006 and 1995) and I'm still not convinced that the sigmoid is essential here, although I certainly get the motivation with the logit. What if I write down the same cross-entropy loss function based on the 2-class Poisson assumption, but then use a different activation function instead of sigmoid? For instance, this similar but not quite as nice one defined piecewise: g(x) = 1/(2-2x) if x <0, 1 - 1/(2+2x) for x>0, g(0) = 0.5. Now the max likelihood equation looks different, but if we minimize it don't we still get probabilities as outputs? $\endgroup$
    – eraoul
    Commented Dec 19, 2018 at 21:48
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    $\begingroup$ if Bischop would have taken $a = \frac{p(x, C_1)}{\sqrt{(1 + p(x, C_1)) p(x, C_2)}}$, the "naturally arising" function would be $\frac{a}{\sqrt{1 + a^2}}$, wouldn't it? $\endgroup$ Commented Sep 20, 2019 at 8:05

I have asked myself this question for months. The answers on CrossValidated and Quora all list nice properties of the logistic sigmoid function, but it all seems like we cleverly guessed this function. What I missed was the justification for choosing it. I finally found one in section of the "Deep Learning" book by Bengio (2016). In my own words:

In short, we want the logarithm of the model's output to be suitable for gradient-based optimization of the log-likelihood of the training data.


  • We want a linear model, but we can't use $z = w^T x + b$ directly as $z \in (-\infty, +\infty)$.
  • For classification, it makes sense to assume the Bernoulli distribution and model its parameter $\theta$ in $P(Y=1) = \theta$.
  • So, we need to map $z$ from $(-\infty, +\infty)$ to $[0, 1]$ to do classification.

Why the logistic sigmoid function?

Cutting off $z$ with $P(Y=1|z) = max\{0, min\{1, z\}\}$ yields a zero gradient for $z$ outside of $[0, 1]$. We need a strong gradient whenever the model's prediction is wrong, because we solve logistic regression with gradient descent. For logistic regression, there is no closed form solution.

The logistic function has the nice property of asymptoting a constant gradient when the model's prediction is wrong, given that we use Maximum Likelihood Estimation to fit the model. This is shown below:

For numerical benefits, Maximum Likelihood Estimation can be done by minimizing the negative log-likelihood of the training data. So, our cost function is:

$$ \begin{align} J(w, b) &= \frac{1}{m} \sum_{i=1}^m -\log P(Y = y_i | x_i; w, b) \\ &= \frac{1}{m} \sum_{i=1}^m - \big(y_i \log P(Y=1 | z) + (y_i-1)\log P(Y=0 | z)\big) \end{align}$$

Since $P(Y=0 | z) = 1-P(Y=1|z)$, we can focus on the $Y=1$ case. So, the question is how to model $P(Y=1 | z)$ given that we have $z = w^T x + b$.

The obvious requirements for the function $f$ mapping $z$ to $P(Y=1 | z)$ are:

  • $\forall z \in \mathbb{R}: f(z) \in [0, 1]$
  • $f(0) = 0.5$
  • $f$ should be rotationally symmetrical w.r.t. $(0, 0.5)$, i.e. $f(-x) = 1-f(x)$, so that flipping the signs of the classes has no effect on the cost function.
  • $f$ should be non-decreasing, continuous and differentiable.

These requirements are all fulfilled by rescaling sigmoid functions. Both $f(z) = \frac{1}{1 + e^{-z}}$ and $f(z) = 0.5 + 0.5 \frac{z}{1+|z|}$ fulfill them. However, sigmoid functions differ with respect to their behavior during gradient-based optimization of the log-likelihood. We can see the difference by plugging the logistic function $f(z) = \frac{1}{1 + e^{-z}}$ into our cost function.

Saturation for $Y=1$

For $P(Y=1|z) = \frac{1}{1 + e^{-z}}$ and $Y=1$, the cost of a single misclassified sample (i.e. $m=1$) is:

$$ \begin{align} J(z) &= -\log(P(Y=1|z)) \\ &= -\log(\frac{1}{1 + e^{-z}}) \\ &= -\log(\frac{e^z}{1+e^z}) \\ &= -z + \log(1 + e^z) \end{align} $$

We can see that there is a linear component $-z$. Now, we can look at two cases:

  • When $z$ is large, the model's prediction was correct, since $Y=1$. In the cost function, the $\log(1 + e^z)$ term asymptotes to $z$ for large $z$. Thus, it roughly cancels the $-z$ out leading to a roughly zero cost for this sample and a weak gradient. That makes sense, as the model is already predicting the correct class.
  • When $z$ is small (but $|z|$ is large), the model's prediction was not correct, since $Y=1$. In the cost function, the $\log(1 + e^z)$ term asymptotes to $0$ for small $z$. Thus, the overall cost for this sample is roughly $-z$, meaning the gradient w.r.t. $z$ is roughly $-1$. This makes it easy for the model to correct its wrong prediction based on the constant gradient it receives. Even for very small $z$, there is no saturation going on, which would cause vanishing gradients.

Saturation for $Y=0$

Above, we focussed on the $Y=1$ case. For $Y=0$, the cost function behaves analogously, providing strong gradients only when the model's prediction is wrong.

This is the cost function $J(z)$ for $Y=1$:

enter image description here

It is the horizontally flipped softplus function. For $Y=0$, it is the softplus function.


You mentioned the alternatives to the logistic sigmoid function, for example $\frac{z}{1+|z|}$. Normalized to $[0,1]$, this would mean that we model $P(Y=1|z) = 0.5 + 0.5 \frac{z}{1+|z|}$.

During MLE, the cost function for $Y=1$ would then be

$J(z) = - \log (0.5 + 0.5 \frac{z}{1+|z|})$,

which looks like this:

enter image description here

You can see, that the gradient of the cost function gets weaker and weaker for $z \rightarrow - \infty$.

  • $\begingroup$ What do you mean when you write "when the model is wrong" ? $\endgroup$ Commented Aug 8, 2018 at 12:46
  • $\begingroup$ @GabrielRomon I mean when the model's prediction is wrong. So for a training sample $(x_i, y_i)$, we would have for example $z = 5$, i.e. our prediction is class 1, but $y_i = 0$. $\endgroup$ Commented Feb 11, 2019 at 15:33

One reason this function might seem more "natural" than others is that it happens to be the inverse of the canonical parameter of the Bernoulli distribution: \begin{align} f(y) &= p^y (1 - p)^{1 - y} \\ &= (1 - p) \exp \left \{ y \log \left ( \frac{p}{1 - p} \right ) \right \} . \end{align} (The function of $p$ within the exponent is called the canonical parameter.)

Maybe a more compelling justification comes from information theory, where the sigmoid function can be derived as a maximum entropy model. Roughly speaking, the sigmoid function assumes minimal structure and reflects our general state of ignorance about the underlying model.

  • $\begingroup$ Good justification for logistic regression. The funny thing that we keep using this for squared error too... $\endgroup$ Commented Jul 24, 2015 at 17:51

Since the original question mentioned the decaying gradient problem, I'd just like to add that, for intermediate layers (where you don't need to interpret activations as class probabilities or regression outputs), other nonlinearities are often preferred over sigmoidal functions. The most prominent are rectifier functions (as in ReLUs), which are linear over the positive domain and zero over the negative. One of their advantages is that they're less subject to the decaying gradient problem, because the derivative is constant over the positive domain. ReLUs have become popular to the point that sigmoids probably can't be called the de-facto standard anymore.

Glorot et al. (2011). Deep sparse rectifier neural networks

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    $\begingroup$ Yep. I think the reason why the logistic function was so popular was due to its importation from statistics. Relu is the most popular in a lot of fields nowadays. $\endgroup$ Commented Mar 31, 2017 at 9:31

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