How does Naive Bayes classifier works for continuous variable?












2












$begingroup$


I know that for categorical features we just calculate the prior and likelihood probability assuming conditional independence between the features.



How does it work for continuous variables? How can we calculate likelihood probability for continuous variable?






machine-learning python naive-bayes-classifier







share|improve this question









$endgroup$

















    2












    $begingroup$


    I know that for categorical features we just calculate the prior and likelihood probability assuming conditional independence between the features.



    How does it work for continuous variables? How can we calculate likelihood probability for continuous variable?






    machine-learning python naive-bayes-classifier







    share|improve this question









    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I know that for categorical features we just calculate the prior and likelihood probability assuming conditional independence between the features.



      How does it work for continuous variables? How can we calculate likelihood probability for continuous variable?






      machine-learning python naive-bayes-classifier







      share|improve this question









      $endgroup$




      I know that for categorical features we just calculate the prior and likelihood probability assuming conditional independence between the features.



      How does it work for continuous variables? How can we calculate likelihood probability for continuous variable?






      machine-learning python naive-bayes-classifier




      machine-learning python naive-bayes-classifier






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked 17 hours ago









      user214user214

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      18216






















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          $begingroup$

          The difference boils down to "how we define $P(x_i|C_k)$?", where $x_i$ is a single feature, and $C_k$ is a class from a total of $K$ classes.



          Discrete



          In discrete case, $P(x_i|C_k)$ is represented by a table as follows:



          x_i   P(x_i|C_k)
          
a     0.5
          
b     0.2
          
c     0.3


          We have one of these tables for each feature-class pair $(i, k)$.



          Lets denote $i$-th feature of data point $n$ as $x_{n,i}$. Each row of this table can be estimated using



          $$hat{P}(x_i=a|C_k) = frac{sum_{n:n in C_k}{mathbb{1}_{x_{n,i}=a}}}{N_k}$$



          which divides the number of samples that have i-th feature equal to $a$ by total number of samples in class $C_k$ (of course from the training set).



          Continuous



          In continuous case, we either discretize the continuous interval of $x_i$ into bins ${b_1,..,b_m}$ and proceed the same as discrete case, or we assume a function like Gaussian (or any other one), as follows:



          $$P(x_i|C_k)=frac{1}{sqrt{2pi}sigma_{i,k}}e^{-(x_i-mu_{i,k})^2/2sigma_{i,k}^2}$$



          This way, for each feature-class pair $(i, k)$, $P(x_i|C_k)$ is represented with two parameters ${mu_{i,k}, sigma_{i, k}}$ instead of a table in discrete case. The estimation of the parameters is the same as fitting a Gaussian distribution to one dimensional data, that is:



          $$hat{mu}_{i,k} = frac{sum_{n:n in C_k}{x_{n,i}}}{N_k}, hat{sigma}^2_{i, k} = frac{sum_{n:n in C_k}{(x_{n,i} - hat{mu}_{i,k})^2}}{N_k-1}$$



          Instead of Gaussian, we can opt for a more complex function, even a neural network. In that case, we should look for a technique to fit the function to data just like what we did with Gaussian.



          The rest is independent of feature type



          Representing $P(C_k)$ is the same for both discrete and continuous cases and is estimated as $hat{P}(C_k)=N_k/N$.



          Finally, the classifier is $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}hat{P}(C_k)prod_{i}hat{P}(x_i=x_{n,i}|C_k)$$



          Or equivalently using log-probabilities,



          $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}mbox{log}hat{P}(C_k)+sum_{i}mbox{log}hat{P}(x_i=x_{n,i}|C_k)$$






          share|improve this answer











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            $begingroup$

            The difference boils down to "how we define $P(x_i|C_k)$?", where $x_i$ is a single feature, and $C_k$ is a class from a total of $K$ classes.



            Discrete



            In discrete case, $P(x_i|C_k)$ is represented by a table as follows:



            x_i   P(x_i|C_k)
            
a     0.5
            
b     0.2
            
c     0.3


            We have one of these tables for each feature-class pair $(i, k)$.



            Lets denote $i$-th feature of data point $n$ as $x_{n,i}$. Each row of this table can be estimated using



            $$hat{P}(x_i=a|C_k) = frac{sum_{n:n in C_k}{mathbb{1}_{x_{n,i}=a}}}{N_k}$$



            which divides the number of samples that have i-th feature equal to $a$ by total number of samples in class $C_k$ (of course from the training set).



            Continuous



            In continuous case, we either discretize the continuous interval of $x_i$ into bins ${b_1,..,b_m}$ and proceed the same as discrete case, or we assume a function like Gaussian (or any other one), as follows:



            $$P(x_i|C_k)=frac{1}{sqrt{2pi}sigma_{i,k}}e^{-(x_i-mu_{i,k})^2/2sigma_{i,k}^2}$$



            This way, for each feature-class pair $(i, k)$, $P(x_i|C_k)$ is represented with two parameters ${mu_{i,k}, sigma_{i, k}}$ instead of a table in discrete case. The estimation of the parameters is the same as fitting a Gaussian distribution to one dimensional data, that is:



            $$hat{mu}_{i,k} = frac{sum_{n:n in C_k}{x_{n,i}}}{N_k}, hat{sigma}^2_{i, k} = frac{sum_{n:n in C_k}{(x_{n,i} - hat{mu}_{i,k})^2}}{N_k-1}$$



            Instead of Gaussian, we can opt for a more complex function, even a neural network. In that case, we should look for a technique to fit the function to data just like what we did with Gaussian.



            The rest is independent of feature type



            Representing $P(C_k)$ is the same for both discrete and continuous cases and is estimated as $hat{P}(C_k)=N_k/N$.



            Finally, the classifier is $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}hat{P}(C_k)prod_{i}hat{P}(x_i=x_{n,i}|C_k)$$



            Or equivalently using log-probabilities,



            $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}mbox{log}hat{P}(C_k)+sum_{i}mbox{log}hat{P}(x_i=x_{n,i}|C_k)$$






            share|improve this answer











            $endgroup$


















              2












              $begingroup$

              The difference boils down to "how we define $P(x_i|C_k)$?", where $x_i$ is a single feature, and $C_k$ is a class from a total of $K$ classes.



              Discrete



              In discrete case, $P(x_i|C_k)$ is represented by a table as follows:



              x_i   P(x_i|C_k)
              
a     0.5
              
b     0.2
              
c     0.3


              We have one of these tables for each feature-class pair $(i, k)$.



              Lets denote $i$-th feature of data point $n$ as $x_{n,i}$. Each row of this table can be estimated using



              $$hat{P}(x_i=a|C_k) = frac{sum_{n:n in C_k}{mathbb{1}_{x_{n,i}=a}}}{N_k}$$



              which divides the number of samples that have i-th feature equal to $a$ by total number of samples in class $C_k$ (of course from the training set).



              Continuous



              In continuous case, we either discretize the continuous interval of $x_i$ into bins ${b_1,..,b_m}$ and proceed the same as discrete case, or we assume a function like Gaussian (or any other one), as follows:



              $$P(x_i|C_k)=frac{1}{sqrt{2pi}sigma_{i,k}}e^{-(x_i-mu_{i,k})^2/2sigma_{i,k}^2}$$



              This way, for each feature-class pair $(i, k)$, $P(x_i|C_k)$ is represented with two parameters ${mu_{i,k}, sigma_{i, k}}$ instead of a table in discrete case. The estimation of the parameters is the same as fitting a Gaussian distribution to one dimensional data, that is:



              $$hat{mu}_{i,k} = frac{sum_{n:n in C_k}{x_{n,i}}}{N_k}, hat{sigma}^2_{i, k} = frac{sum_{n:n in C_k}{(x_{n,i} - hat{mu}_{i,k})^2}}{N_k-1}$$



              Instead of Gaussian, we can opt for a more complex function, even a neural network. In that case, we should look for a technique to fit the function to data just like what we did with Gaussian.



              The rest is independent of feature type



              Representing $P(C_k)$ is the same for both discrete and continuous cases and is estimated as $hat{P}(C_k)=N_k/N$.



              Finally, the classifier is $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}hat{P}(C_k)prod_{i}hat{P}(x_i=x_{n,i}|C_k)$$



              Or equivalently using log-probabilities,



              $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}mbox{log}hat{P}(C_k)+sum_{i}mbox{log}hat{P}(x_i=x_{n,i}|C_k)$$






              share|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                The difference boils down to "how we define $P(x_i|C_k)$?", where $x_i$ is a single feature, and $C_k$ is a class from a total of $K$ classes.



                Discrete



                In discrete case, $P(x_i|C_k)$ is represented by a table as follows:



                x_i   P(x_i|C_k)
                
a     0.5
                
b     0.2
                
c     0.3


                We have one of these tables for each feature-class pair $(i, k)$.



                Lets denote $i$-th feature of data point $n$ as $x_{n,i}$. Each row of this table can be estimated using



                $$hat{P}(x_i=a|C_k) = frac{sum_{n:n in C_k}{mathbb{1}_{x_{n,i}=a}}}{N_k}$$



                which divides the number of samples that have i-th feature equal to $a$ by total number of samples in class $C_k$ (of course from the training set).



                Continuous



                In continuous case, we either discretize the continuous interval of $x_i$ into bins ${b_1,..,b_m}$ and proceed the same as discrete case, or we assume a function like Gaussian (or any other one), as follows:



                $$P(x_i|C_k)=frac{1}{sqrt{2pi}sigma_{i,k}}e^{-(x_i-mu_{i,k})^2/2sigma_{i,k}^2}$$



                This way, for each feature-class pair $(i, k)$, $P(x_i|C_k)$ is represented with two parameters ${mu_{i,k}, sigma_{i, k}}$ instead of a table in discrete case. The estimation of the parameters is the same as fitting a Gaussian distribution to one dimensional data, that is:



                $$hat{mu}_{i,k} = frac{sum_{n:n in C_k}{x_{n,i}}}{N_k}, hat{sigma}^2_{i, k} = frac{sum_{n:n in C_k}{(x_{n,i} - hat{mu}_{i,k})^2}}{N_k-1}$$



                Instead of Gaussian, we can opt for a more complex function, even a neural network. In that case, we should look for a technique to fit the function to data just like what we did with Gaussian.



                The rest is independent of feature type



                Representing $P(C_k)$ is the same for both discrete and continuous cases and is estimated as $hat{P}(C_k)=N_k/N$.



                Finally, the classifier is $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}hat{P}(C_k)prod_{i}hat{P}(x_i=x_{n,i}|C_k)$$



                Or equivalently using log-probabilities,



                $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}mbox{log}hat{P}(C_k)+sum_{i}mbox{log}hat{P}(x_i=x_{n,i}|C_k)$$






                share|improve this answer











                $endgroup$



                The difference boils down to "how we define $P(x_i|C_k)$?", where $x_i$ is a single feature, and $C_k$ is a class from a total of $K$ classes.



                Discrete



                In discrete case, $P(x_i|C_k)$ is represented by a table as follows:



                x_i   P(x_i|C_k)
                
a     0.5
                
b     0.2
                
c     0.3


                We have one of these tables for each feature-class pair $(i, k)$.



                Lets denote $i$-th feature of data point $n$ as $x_{n,i}$. Each row of this table can be estimated using



                $$hat{P}(x_i=a|C_k) = frac{sum_{n:n in C_k}{mathbb{1}_{x_{n,i}=a}}}{N_k}$$



                which divides the number of samples that have i-th feature equal to $a$ by total number of samples in class $C_k$ (of course from the training set).



                Continuous



                In continuous case, we either discretize the continuous interval of $x_i$ into bins ${b_1,..,b_m}$ and proceed the same as discrete case, or we assume a function like Gaussian (or any other one), as follows:



                $$P(x_i|C_k)=frac{1}{sqrt{2pi}sigma_{i,k}}e^{-(x_i-mu_{i,k})^2/2sigma_{i,k}^2}$$



                This way, for each feature-class pair $(i, k)$, $P(x_i|C_k)$ is represented with two parameters ${mu_{i,k}, sigma_{i, k}}$ instead of a table in discrete case. The estimation of the parameters is the same as fitting a Gaussian distribution to one dimensional data, that is:



                $$hat{mu}_{i,k} = frac{sum_{n:n in C_k}{x_{n,i}}}{N_k}, hat{sigma}^2_{i, k} = frac{sum_{n:n in C_k}{(x_{n,i} - hat{mu}_{i,k})^2}}{N_k-1}$$



                Instead of Gaussian, we can opt for a more complex function, even a neural network. In that case, we should look for a technique to fit the function to data just like what we did with Gaussian.



                The rest is independent of feature type



                Representing $P(C_k)$ is the same for both discrete and continuous cases and is estimated as $hat{P}(C_k)=N_k/N$.



                Finally, the classifier is $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}hat{P}(C_k)prod_{i}hat{P}(x_i=x_{n,i}|C_k)$$



                Or equivalently using log-probabilities,



                $$C(x_n) = underset{k in {1,..,K}}{mbox{argmax }}mbox{log}hat{P}(C_k)+sum_{i}mbox{log}hat{P}(x_i=x_{n,i}|C_k)$$







                share|improve this answer














                share|improve this answer



                share|improve this answer








                edited 13 hours ago

























                answered 15 hours ago









                EsmailianEsmailian

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                5966






























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