A hint for the entropy problem-entropy of one discrete variable is greater than the entropy of another one












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I need a hint on how to start solving the following problem.



Entropy of a discrete variable X is $H(X) = −sum_{xin {x:P(X=x)>0}}P(X=x)logP(X=x)$. Let $f:R → R$ be any function.



a) Show that entropy of a discrete variable X is greater than or equal to entropy of a discrete variable f(X).



b) Show that equality occurs if and only if function f is injective on {x : P(X = x) > 0}






probability entropy







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    $begingroup$


    I need a hint on how to start solving the following problem.



    Entropy of a discrete variable X is $H(X) = −sum_{xin {x:P(X=x)>0}}P(X=x)logP(X=x)$. Let $f:R → R$ be any function.



    a) Show that entropy of a discrete variable X is greater than or equal to entropy of a discrete variable f(X).



    b) Show that equality occurs if and only if function f is injective on {x : P(X = x) > 0}






    probability entropy







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      I need a hint on how to start solving the following problem.



      Entropy of a discrete variable X is $H(X) = −sum_{xin {x:P(X=x)>0}}P(X=x)logP(X=x)$. Let $f:R → R$ be any function.



      a) Show that entropy of a discrete variable X is greater than or equal to entropy of a discrete variable f(X).



      b) Show that equality occurs if and only if function f is injective on {x : P(X = x) > 0}






      probability entropy







      share|cite|improve this question









      $endgroup$




      I need a hint on how to start solving the following problem.



      Entropy of a discrete variable X is $H(X) = −sum_{xin {x:P(X=x)>0}}P(X=x)logP(X=x)$. Let $f:R → R$ be any function.



      a) Show that entropy of a discrete variable X is greater than or equal to entropy of a discrete variable f(X).



      b) Show that equality occurs if and only if function f is injective on {x : P(X = x) > 0}






      probability entropy




      probability entropy






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 9 at 22:08









      Novak DjokovicNovak Djokovic

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      44328






















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          $$begin{align}&-sum_{y : P(f(X) = y) > 0} P(Y=y) log P(Y = y)\
          &= -sum_{y : P(f(X) = y) > 0} left{left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right)right}.end{align}$$
          It then suffices to show
          $-left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right) le - sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)$
          for each $y$,
          since then the above would be bounded by
          $$le - sum_{y : P(f(X)=y)>0}sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)
          = - sum_{x : P(X=x)>0} P(X=x) log P(X=x).$$



          The unverified claim reduces to showing that the function $g(x) = - x log x$ is sub-additive, i.e. $g(a+b) le g(a) + g(b)$.






          share|cite|improve this answer











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            $begingroup$

            $$begin{align}&-sum_{y : P(f(X) = y) > 0} P(Y=y) log P(Y = y)\
            &= -sum_{y : P(f(X) = y) > 0} left{left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right)right}.end{align}$$
            It then suffices to show
            $-left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right) le - sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)$
            for each $y$,
            since then the above would be bounded by
            $$le - sum_{y : P(f(X)=y)>0}sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)
            = - sum_{x : P(X=x)>0} P(X=x) log P(X=x).$$



            The unverified claim reduces to showing that the function $g(x) = - x log x$ is sub-additive, i.e. $g(a+b) le g(a) + g(b)$.






            share|cite|improve this answer











            $endgroup$


















              0












              $begingroup$

              $$begin{align}&-sum_{y : P(f(X) = y) > 0} P(Y=y) log P(Y = y)\
              &= -sum_{y : P(f(X) = y) > 0} left{left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right)right}.end{align}$$
              It then suffices to show
              $-left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right) le - sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)$
              for each $y$,
              since then the above would be bounded by
              $$le - sum_{y : P(f(X)=y)>0}sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)
              = - sum_{x : P(X=x)>0} P(X=x) log P(X=x).$$



              The unverified claim reduces to showing that the function $g(x) = - x log x$ is sub-additive, i.e. $g(a+b) le g(a) + g(b)$.






              share|cite|improve this answer











              $endgroup$
















                0












                0








                0





                $begingroup$

                $$begin{align}&-sum_{y : P(f(X) = y) > 0} P(Y=y) log P(Y = y)\
                &= -sum_{y : P(f(X) = y) > 0} left{left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right)right}.end{align}$$
                It then suffices to show
                $-left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right) le - sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)$
                for each $y$,
                since then the above would be bounded by
                $$le - sum_{y : P(f(X)=y)>0}sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)
                = - sum_{x : P(X=x)>0} P(X=x) log P(X=x).$$



                The unverified claim reduces to showing that the function $g(x) = - x log x$ is sub-additive, i.e. $g(a+b) le g(a) + g(b)$.






                share|cite|improve this answer











                $endgroup$



                $$begin{align}&-sum_{y : P(f(X) = y) > 0} P(Y=y) log P(Y = y)\
                &= -sum_{y : P(f(X) = y) > 0} left{left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right)right}.end{align}$$
                It then suffices to show
                $-left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X = x)right) log left(sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x)right) le - sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)$
                for each $y$,
                since then the above would be bounded by
                $$le - sum_{y : P(f(X)=y)>0}sum_{x in f^{-1}(y) : P(X=x) > 0} P(X=x) log P(X=x)
                = - sum_{x : P(X=x)>0} P(X=x) log P(X=x).$$



                The unverified claim reduces to showing that the function $g(x) = - x log x$ is sub-additive, i.e. $g(a+b) le g(a) + g(b)$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Jan 9 at 22:44

























                answered Jan 9 at 22:37









                angryavianangryavian

                40.7k23380




                40.7k23380






























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