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Why does logistic function use e rather than 2?

Why is Reconstruction in Autoencoders Using the Same Activation Function as Forward Activation, and not the Inverse?Sigmoid's stabilityWhy ReLU is better than the other activation functionsMachine Learning: Why do the error in cost function need to be squared?Why does Q-learning use an actor model and critic model?Why do we need the sigmoid function in logistic regression?Preprocessing and dropout in Autoencoders?Properly using activation functions of neural networkPurpose of backpropagation in neural networksPossible reasons for word2vec learning context words as most similar rather than words in similar contexts

.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;

7

1

$begingroup$

sigmoid function could be used as activation function in machine learning.

$$displaystyle S(x)=frac 11+e^-x=frac e^xe^x+1.$$

If substitute e with 2,

def sigmoid2(z):
 return 1/(1+2**(-z))
x = np.arange(-9,9,dtype=float)
y = sigmoid2(x)
plt.scatter(x,y)

the plot looks similar.

Why does the logistic function use $e$ rather than 2?

machine-learning deep-learning

share|improve this question

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

$endgroup$

add a comment | 

7

1

$begingroup$

sigmoid function could be used as activation function in machine learning.

$$displaystyle S(x)=frac 11+e^-x=frac e^xe^x+1.$$

If substitute e with 2,

def sigmoid2(z):
 return 1/(1+2**(-z))
x = np.arange(-9,9,dtype=float)
y = sigmoid2(x)
plt.scatter(x,y)

the plot looks similar.

Why does the logistic function use $e$ rather than 2?

machine-learning deep-learning

share|improve this question

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

$endgroup$

add a comment | 

$begingroup$

sigmoid function could be used as activation function in machine learning.

$$displaystyle S(x)=frac 11+e^-x=frac e^xe^x+1.$$

If substitute e with 2,

def sigmoid2(z):
 return 1/(1+2**(-z))
x = np.arange(-9,9,dtype=float)
y = sigmoid2(x)
plt.scatter(x,y)

the plot looks similar.

Why does the logistic function use $e$ rather than 2?

machine-learning deep-learning

share|improve this question

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

$endgroup$

def sigmoid2(z):
 return 1/(1+2**(-z))
x = np.arange(-9,9,dtype=float)
y = sigmoid2(x)
plt.scatter(x,y)

share|improve this question

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

share|improve this question

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

edited Jun 6 at 14:48

Ethan

1,00888 silver badges2929 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

asked Jun 6 at 7:55

baojieqhbaojieqh

8055 bronze badges

3 Answers
3

active

oldest

votes

11

$begingroup$

Since you are going to minimize later on the log likelihood, there is actually no big difference between $log 2^x=x * log2$ and $log e^x=x$. You see the difference is simply a constant.

Nevertheless one could argue to use $2^x$ instead of $e^x$ und also use $log_2$ instead of $log$ when it comes to the optimizing step. In fact it is possible to use $2^x$ and also many other functions, which show some desired properties.
Which are:

• $limlimits_x rightarrow inftyf(x)=1$


• $limlimits_x rightarrow -inftyf(x)=0$


• 
  $f(x) = -f(-x) + 1$, (symmetric in $(0, 0.5)$
  

Here is an example of suitable functions from wikipedia.

share|improve this answer

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

$endgroup$

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  I think it's also worth pointing out that one nice reason to use $e$ as the base is that the derivative of $sigma(x)=frac11+e^-x$ is $sigma'(x)=sigma(x)(1-sigma(x))$. Without doing the actual computation, I think if the base was different the formula would only differ by a constant again, but it's a nice property that is specific to $e$.
  $endgroup$
  – Calvin Godfrey
  Jun 6 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Same goes for $2^x$ when using $log_2$.
  $endgroup$
  – Andreas Look
  Jun 6 at 20:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndreasLook I'm not sure what you mean. If you use $2^-x$ then there's an extra factor of $ln(2)$ in the derivative (like Calvin Godfrey said).
  $endgroup$
  – sfmiller940
  Jun 12 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, check out binary logarithm. $log_2 (2^x)=x$.
  $endgroup$
  – Andreas Look
  Jun 13 at 19:32
  
  

  
  
  
  

add a comment | 

4

$begingroup$

So there are many functions that look sigmoid including the 2 you mentioned, but there are reasons why $e$ is special. The main reason it that the logistic function was originally used to model population growth. And populations, much like interest, can compound over time. Thus, the $e$ becomes a very natural object for this reason. In addition, for theoretical reasons concerning the canonical link function of a glm the logistic is one of the theoretically simplest objects to work with which makes it easy to prove things with.

share|improve this answer

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for your answer. what does "canonical link function of a glm" mean?
  $endgroup$
  – baojieqh
  Jun 6 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @baojieqh For all generalized linear models, one needs to specify a member of the exponential family of distributions. These distributions all share a property where they can be written in such a way so that a function of the scale parameter of the distribution sits "by itself" in an exponent (and the function is only a function of the scale parameter). This function is what people refer to as the canonical link function. For the bernoulli/binomial distribution, where the scale parameter is p, it turns out that this function is ln(p/(1-p)) which is the logit link function.
  $endgroup$
  – aranglol
  Jun 6 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hence, the canonical link function for the logistic regression, which assumes a Bernoulli distribution for each row, is the logit link. There are other more theoretical properties as well that make the canonical link function desirable. But it is technically not necessary to use it, you could use the probit for example.
  $endgroup$
  – aranglol
  Jun 6 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aranglol thanks for you comments, would you please take a look at this link math.stackexchange.com/q/3253634/656371
  $endgroup$
  – baojieqh
  Jun 7 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This seems to be just a hand-waving appeal to the claim that "$e$ is special", without giving any justification about why $e$ is special. Really, the only specialness is the convenience that $tfracddxa^x=a^xln a$, which means that $tfracddxe^x=e^x$.
  $endgroup$
  – David Richerby
  Jun 7 at 9:21
  

  
  
  
  

add a comment | 

0

$begingroup$

It comes from the basic assumption of the model that there exists a continuous/latent/unobservable $Y^*$ that relates somehow to the observed values of $Y$. The model further assumes that $Y=1$ if the signal of $Y^*$ is above some threshold, and otherwise $Y=0$. The third and last assumption is that the underlying distribution of $Y*$ is the logistic distribution. Once you have these assumptions, it is only a matter of algebra to derive the model.

You can read more details at my blog.

share|improve this answer

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges

New contributor

Yossi Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

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11

$begingroup$

Since you are going to minimize later on the log likelihood, there is actually no big difference between $log 2^x=x * log2$ and $log e^x=x$. You see the difference is simply a constant.

Nevertheless one could argue to use $2^x$ instead of $e^x$ und also use $log_2$ instead of $log$ when it comes to the optimizing step. In fact it is possible to use $2^x$ and also many other functions, which show some desired properties.
Which are:

• $limlimits_x rightarrow inftyf(x)=1$


• $limlimits_x rightarrow -inftyf(x)=0$


• 
  $f(x) = -f(-x) + 1$, (symmetric in $(0, 0.5)$
  

Here is an example of suitable functions from wikipedia.

share|improve this answer

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

$endgroup$

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  I think it's also worth pointing out that one nice reason to use $e$ as the base is that the derivative of $sigma(x)=frac11+e^-x$ is $sigma'(x)=sigma(x)(1-sigma(x))$. Without doing the actual computation, I think if the base was different the formula would only differ by a constant again, but it's a nice property that is specific to $e$.
  $endgroup$
  – Calvin Godfrey
  Jun 6 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Same goes for $2^x$ when using $log_2$.
  $endgroup$
  – Andreas Look
  Jun 6 at 20:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndreasLook I'm not sure what you mean. If you use $2^-x$ then there's an extra factor of $ln(2)$ in the derivative (like Calvin Godfrey said).
  $endgroup$
  – sfmiller940
  Jun 12 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, check out binary logarithm. $log_2 (2^x)=x$.
  $endgroup$
  – Andreas Look
  Jun 13 at 19:32
  
  

  
  
  
  

add a comment | 

11

$begingroup$

Since you are going to minimize later on the log likelihood, there is actually no big difference between $log 2^x=x * log2$ and $log e^x=x$. You see the difference is simply a constant.

Nevertheless one could argue to use $2^x$ instead of $e^x$ und also use $log_2$ instead of $log$ when it comes to the optimizing step. In fact it is possible to use $2^x$ and also many other functions, which show some desired properties.
Which are:

• $limlimits_x rightarrow inftyf(x)=1$


• $limlimits_x rightarrow -inftyf(x)=0$


• 
  $f(x) = -f(-x) + 1$, (symmetric in $(0, 0.5)$
  

Here is an example of suitable functions from wikipedia.

share|improve this answer

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

$endgroup$

• 
  
  
  

  
  8
  

  
  
  
  
  
  

  
  $begingroup$
  I think it's also worth pointing out that one nice reason to use $e$ as the base is that the derivative of $sigma(x)=frac11+e^-x$ is $sigma'(x)=sigma(x)(1-sigma(x))$. Without doing the actual computation, I think if the base was different the formula would only differ by a constant again, but it's a nice property that is specific to $e$.
  $endgroup$
  – Calvin Godfrey
  Jun 6 at 16:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Same goes for $2^x$ when using $log_2$.
  $endgroup$
  – Andreas Look
  Jun 6 at 20:44
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @AndreasLook I'm not sure what you mean. If you use $2^-x$ then there's an extra factor of $ln(2)$ in the derivative (like Calvin Godfrey said).
  $endgroup$
  – sfmiller940
  Jun 12 at 20:08
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, check out binary logarithm. $log_2 (2^x)=x$.
  $endgroup$
  – Andreas Look
  Jun 13 at 19:32
  
  

  
  
  
  

add a comment | 

$begingroup$

Since you are going to minimize later on the log likelihood, there is actually no big difference between $log 2^x=x * log2$ and $log e^x=x$. You see the difference is simply a constant.

Nevertheless one could argue to use $2^x$ instead of $e^x$ und also use $log_2$ instead of $log$ when it comes to the optimizing step. In fact it is possible to use $2^x$ and also many other functions, which show some desired properties.
Which are:

• $limlimits_x rightarrow inftyf(x)=1$


• $limlimits_x rightarrow -inftyf(x)=0$


• 
  $f(x) = -f(-x) + 1$, (symmetric in $(0, 0.5)$
  

Here is an example of suitable functions from wikipedia.

share|improve this answer

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

$endgroup$

share|improve this answer

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

answered Jun 6 at 8:18

Andreas LookAndreas Look

64311 silver badge1212 bronze badges

4

$begingroup$

So there are many functions that look sigmoid including the 2 you mentioned, but there are reasons why $e$ is special. The main reason it that the logistic function was originally used to model population growth. And populations, much like interest, can compound over time. Thus, the $e$ becomes a very natural object for this reason. In addition, for theoretical reasons concerning the canonical link function of a glm the logistic is one of the theoretically simplest objects to work with which makes it easy to prove things with.

share|improve this answer

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for your answer. what does "canonical link function of a glm" mean?
  $endgroup$
  – baojieqh
  Jun 6 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @baojieqh For all generalized linear models, one needs to specify a member of the exponential family of distributions. These distributions all share a property where they can be written in such a way so that a function of the scale parameter of the distribution sits "by itself" in an exponent (and the function is only a function of the scale parameter). This function is what people refer to as the canonical link function. For the bernoulli/binomial distribution, where the scale parameter is p, it turns out that this function is ln(p/(1-p)) which is the logit link function.
  $endgroup$
  – aranglol
  Jun 6 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hence, the canonical link function for the logistic regression, which assumes a Bernoulli distribution for each row, is the logit link. There are other more theoretical properties as well that make the canonical link function desirable. But it is technically not necessary to use it, you could use the probit for example.
  $endgroup$
  – aranglol
  Jun 6 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aranglol thanks for you comments, would you please take a look at this link math.stackexchange.com/q/3253634/656371
  $endgroup$
  – baojieqh
  Jun 7 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This seems to be just a hand-waving appeal to the claim that "$e$ is special", without giving any justification about why $e$ is special. Really, the only specialness is the convenience that $tfracddxa^x=a^xln a$, which means that $tfracddxe^x=e^x$.
  $endgroup$
  – David Richerby
  Jun 7 at 9:21
  

  
  
  
  

add a comment | 

4

$begingroup$

So there are many functions that look sigmoid including the 2 you mentioned, but there are reasons why $e$ is special. The main reason it that the logistic function was originally used to model population growth. And populations, much like interest, can compound over time. Thus, the $e$ becomes a very natural object for this reason. In addition, for theoretical reasons concerning the canonical link function of a glm the logistic is one of the theoretically simplest objects to work with which makes it easy to prove things with.

share|improve this answer

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  thanks for your answer. what does "canonical link function of a glm" mean?
  $endgroup$
  – baojieqh
  Jun 6 at 9:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @baojieqh For all generalized linear models, one needs to specify a member of the exponential family of distributions. These distributions all share a property where they can be written in such a way so that a function of the scale parameter of the distribution sits "by itself" in an exponent (and the function is only a function of the scale parameter). This function is what people refer to as the canonical link function. For the bernoulli/binomial distribution, where the scale parameter is p, it turns out that this function is ln(p/(1-p)) which is the logit link function.
  $endgroup$
  – aranglol
  Jun 6 at 23:55
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Hence, the canonical link function for the logistic regression, which assumes a Bernoulli distribution for each row, is the logit link. There are other more theoretical properties as well that make the canonical link function desirable. But it is technically not necessary to use it, you could use the probit for example.
  $endgroup$
  – aranglol
  Jun 6 at 23:58
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @aranglol thanks for you comments, would you please take a look at this link math.stackexchange.com/q/3253634/656371
  $endgroup$
  – baojieqh
  Jun 7 at 0:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This seems to be just a hand-waving appeal to the claim that "$e$ is special", without giving any justification about why $e$ is special. Really, the only specialness is the convenience that $tfracddxa^x=a^xln a$, which means that $tfracddxe^x=e^x$.
  $endgroup$
  – David Richerby
  Jun 7 at 9:21
  

  
  
  
  

add a comment | 

$begingroup$

So there are many functions that look sigmoid including the 2 you mentioned, but there are reasons why $e$ is special. The main reason it that the logistic function was originally used to model population growth. And populations, much like interest, can compound over time. Thus, the $e$ becomes a very natural object for this reason. In addition, for theoretical reasons concerning the canonical link function of a glm the logistic is one of the theoretically simplest objects to work with which makes it easy to prove things with.

share|improve this answer

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

$endgroup$

share|improve this answer

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

edited Jun 6 at 15:00

Ethan

1,00888 silver badges2929 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

answered Jun 6 at 8:12

Anonymous EmuAnonymous Emu

15044 bronze badges

0

$begingroup$

It comes from the basic assumption of the model that there exists a continuous/latent/unobservable $Y^*$ that relates somehow to the observed values of $Y$. The model further assumes that $Y=1$ if the signal of $Y^*$ is above some threshold, and otherwise $Y=0$. The third and last assumption is that the underlying distribution of $Y*$ is the logistic distribution. Once you have these assumptions, it is only a matter of algebra to derive the model.

You can read more details at my blog.

share|improve this answer

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges

New contributor

Yossi Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

It comes from the basic assumption of the model that there exists a continuous/latent/unobservable $Y^*$ that relates somehow to the observed values of $Y$. The model further assumes that $Y=1$ if the signal of $Y^*$ is above some threshold, and otherwise $Y=0$. The third and last assumption is that the underlying distribution of $Y*$ is the logistic distribution. Once you have these assumptions, it is only a matter of algebra to derive the model.

You can read more details at my blog.

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edited Jun 16 at 13:03

Stephen Rauch♦

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answered Jun 16 at 11:50

Yossi LevyYossi Levy

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$begingroup$

It comes from the basic assumption of the model that there exists a continuous/latent/unobservable $Y^*$ that relates somehow to the observed values of $Y$. The model further assumes that $Y=1$ if the signal of $Y^*$ is above some threshold, and otherwise $Y=0$. The third and last assumption is that the underlying distribution of $Y*$ is the logistic distribution. Once you have these assumptions, it is only a matter of algebra to derive the model.

You can read more details at my blog.

share|improve this answer

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges

New contributor

Yossi Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|improve this answer

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges

New contributor

Yossi Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

edited Jun 16 at 13:03

Stephen Rauch♦

1,54366 gold badges1313 silver badges3030 bronze badges

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges

New contributor

Yossi Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered Jun 16 at 11:50

Yossi LevyYossi Levy

122 bronze badges