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

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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












$begingroup$


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.



enter image description here



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










share|improve this question











$endgroup$


















    7












    $begingroup$


    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.



    enter image description here



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










    share|improve this question











    $endgroup$














      7












      7








      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.



      enter image description here



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










      share|improve this question











      $endgroup$




      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.



      enter image description here



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







      machine-learning deep-learning






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Jun 6 at 14:48









      Ethan

      1,0088 silver badges29 bronze badges




      1,0088 silver badges29 bronze badges










      asked Jun 6 at 7:55









      baojieqhbaojieqh

      805 bronze badges




      805 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









          $endgroup$








          • 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



















          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











          $endgroup$








          • 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


















          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










          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.





          $endgroup$















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            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









            $endgroup$








            • 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
















            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









            $endgroup$








            • 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














            11












            11








            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









            $endgroup$



            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












            share|improve this answer



            share|improve this answer










            answered Jun 6 at 8:18









            Andreas LookAndreas Look

            6431 silver badge12 bronze badges




            6431 silver badge12 bronze badges







            • 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













            • 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








            8




            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




            $begingroup$
            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





            $begingroup$
            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





            $begingroup$
            @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





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














            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











            $endgroup$








            • 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















            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











            $endgroup$








            • 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













            4












            4








            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











            $endgroup$



            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














            share|improve this answer



            share|improve this answer








            edited Jun 6 at 15:00









            Ethan

            1,0088 silver badges29 bronze badges




            1,0088 silver badges29 bronze badges










            answered Jun 6 at 8:12









            Anonymous EmuAnonymous Emu

            1504 bronze badges




            1504 bronze badges







            • 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












            • 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







            2




            2




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




            $begingroup$
            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




            $begingroup$
            @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





            $begingroup$
            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




            $begingroup$
            @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




            $begingroup$
            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











            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










            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.





            $endgroup$

















              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










              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.





              $endgroup$















                0












                0








                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










                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.





                $endgroup$



                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










                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.








                share|improve this answer



                share|improve this answer








                edited Jun 16 at 13:03









                Stephen Rauch

                1,5436 gold badges13 silver badges30 bronze badges




                1,5436 gold badges13 silver badges30 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

                12 bronze badges




                12 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.




                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.





























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