Proof $ I(X;Y) = H(X) - H(X|Y) $












0












$begingroup$


$I(X;Y) = H(X) - H(X|Y)$



proof



$ I(X;Y)=sum_{x,y}p(x,y) log frac{p(x,y)}{p(x)p(y)} $



$=sum_{x,y}p(x,y) log frac{p(x|y)}{p(x)}$



$= sum_{x,y}p(x,y) log p(x|y) + sum_{x,y}p(x,y) log p(x) $



$= - H(X|Y) + sum_{y} p(y|x) sum_x p(x) log p(x) $



Can you please help me get the term H(X)






probability probability-theory information-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    $I(X;Y) = H(X) - H(X|Y)$



    proof



    $ I(X;Y)=sum_{x,y}p(x,y) log frac{p(x,y)}{p(x)p(y)} $



    $=sum_{x,y}p(x,y) log frac{p(x|y)}{p(x)}$



    $= sum_{x,y}p(x,y) log p(x|y) + sum_{x,y}p(x,y) log p(x) $



    $= - H(X|Y) + sum_{y} p(y|x) sum_x p(x) log p(x) $



    Can you please help me get the term H(X)






    probability probability-theory information-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $I(X;Y) = H(X) - H(X|Y)$



      proof



      $ I(X;Y)=sum_{x,y}p(x,y) log frac{p(x,y)}{p(x)p(y)} $



      $=sum_{x,y}p(x,y) log frac{p(x|y)}{p(x)}$



      $= sum_{x,y}p(x,y) log p(x|y) + sum_{x,y}p(x,y) log p(x) $



      $= - H(X|Y) + sum_{y} p(y|x) sum_x p(x) log p(x) $



      Can you please help me get the term H(X)






      probability probability-theory information-theory







      share|cite|improve this question









      $endgroup$




      $I(X;Y) = H(X) - H(X|Y)$



      proof



      $ I(X;Y)=sum_{x,y}p(x,y) log frac{p(x,y)}{p(x)p(y)} $



      $=sum_{x,y}p(x,y) log frac{p(x|y)}{p(x)}$



      $= sum_{x,y}p(x,y) log p(x|y) + sum_{x,y}p(x,y) log p(x) $



      $= - H(X|Y) + sum_{y} p(y|x) sum_x p(x) log p(x) $



      Can you please help me get the term H(X)






      probability probability-theory information-theory




      probability probability-theory information-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 13 at 5:47









      Salwa MostafaSalwa Mostafa

      273




      273






















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          $begingroup$

          $$-sum_{x,y} p(x,y) log p(x)
          = - sum_x log p(x) sum_y p(x,y)
          = - sum_x (log p(x)) p(x)
          = H(X).$$






          share|cite|improve this answer









          $endgroup$













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            $begingroup$

            $$-sum_{x,y} p(x,y) log p(x)
            = - sum_x log p(x) sum_y p(x,y)
            = - sum_x (log p(x)) p(x)
            = H(X).$$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $$-sum_{x,y} p(x,y) log p(x)
              = - sum_x log p(x) sum_y p(x,y)
              = - sum_x (log p(x)) p(x)
              = H(X).$$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                $$-sum_{x,y} p(x,y) log p(x)
                = - sum_x log p(x) sum_y p(x,y)
                = - sum_x (log p(x)) p(x)
                = H(X).$$






                share|cite|improve this answer









                $endgroup$



                $$-sum_{x,y} p(x,y) log p(x)
                = - sum_x log p(x) sum_y p(x,y)
                = - sum_x (log p(x)) p(x)
                = H(X).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 13 at 5:58









                angryavianangryavian

                40.9k23380




                40.9k23380






























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