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What loss function to use when labels are probabilities?

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What loss function to use when labels are probabilities?



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$begingroup$


What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].



It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.



Would something like MSE (after applying softmax) make sense, or is there a better loss function?










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    1












    $begingroup$


    What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].



    It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.



    Would something like MSE (after applying softmax) make sense, or is there a better loss function?










    share|improve this question







    New contributor




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







    $endgroup$















      1












      1








      1





      $begingroup$


      What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].



      It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.



      Would something like MSE (after applying softmax) make sense, or is there a better loss function?










      share|improve this question







      New contributor




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







      $endgroup$




      What loss function is most appropriate when training a model with target values that are probabilities? For example, I have a 3-output model with x=[some features] and y=[0.2, 0.3, 0.5].



      It seems like something like cross-entropy doesn't make sense here since it assumes that a single target is the correct label.



      Would something like MSE (after applying softmax) make sense, or is there a better loss function?







      neural-networks loss-functions probability-distribution






      share|improve this question







      New contributor




      Thomas Johnson 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 question







      New contributor




      Thomas Johnson 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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      asked 5 hours ago









      Thomas JohnsonThomas Johnson

      1083




      1083




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          $begingroup$

          Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.



          You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,



          $$H(p,q)=-sum_{xin X} p(x) log q(x).$$
          $ $



          Note that one-hot labels would mean that
          $$
          p(x) =
          begin{cases}
          1 & text{if }x text{ is the true label}\
          0 & text{otherwise}
          end{cases}$$



          which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:



          $$H(p,q) = -log q(x_{label})$$






          share|improve this answer









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            $begingroup$

            Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.



            You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,



            $$H(p,q)=-sum_{xin X} p(x) log q(x).$$
            $ $



            Note that one-hot labels would mean that
            $$
            p(x) =
            begin{cases}
            1 & text{if }x text{ is the true label}\
            0 & text{otherwise}
            end{cases}$$



            which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:



            $$H(p,q) = -log q(x_{label})$$






            share|improve this answer









            $endgroup$


















              1












              $begingroup$

              Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.



              You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,



              $$H(p,q)=-sum_{xin X} p(x) log q(x).$$
              $ $



              Note that one-hot labels would mean that
              $$
              p(x) =
              begin{cases}
              1 & text{if }x text{ is the true label}\
              0 & text{otherwise}
              end{cases}$$



              which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:



              $$H(p,q) = -log q(x_{label})$$






              share|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.



                You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,



                $$H(p,q)=-sum_{xin X} p(x) log q(x).$$
                $ $



                Note that one-hot labels would mean that
                $$
                p(x) =
                begin{cases}
                1 & text{if }x text{ is the true label}\
                0 & text{otherwise}
                end{cases}$$



                which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:



                $$H(p,q) = -log q(x_{label})$$






                share|improve this answer









                $endgroup$



                Actually, the cross-entropy loss function would be appropriate here, since it measures the "distance" between a distribution $q$ and the "true" distribution $p$.



                You are right, though, that using a loss function called "cross_entropy" in many APIs would be a mistake. This is because these functions, as you said, assume a one-hot label. You would need to use the general cross-entropy function,



                $$H(p,q)=-sum_{xin X} p(x) log q(x).$$
                $ $



                Note that one-hot labels would mean that
                $$
                p(x) =
                begin{cases}
                1 & text{if }x text{ is the true label}\
                0 & text{otherwise}
                end{cases}$$



                which causes the cross-entropy $H(p,q)$ to reduce to the form you're familiar with:



                $$H(p,q) = -log q(x_{label})$$







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 4 hours ago









                Philip RaeisghasemPhilip Raeisghasem

                963119




                963119






















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