Lower bound of the entropy over the probability distribution











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I am trying to show that:



$$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$



I thought about Jensen inequality or induction, but none of them provided me something.










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    up vote
    1
    down vote

    favorite












    I am trying to show that:



    $$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$



    I thought about Jensen inequality or induction, but none of them provided me something.










    share|cite|improve this question


























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I am trying to show that:



      $$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$



      I thought about Jensen inequality or induction, but none of them provided me something.










      share|cite|improve this question















      I am trying to show that:



      $$inf_{z: 1^Tz=1} sum_{i=1}^m {z_i log z_i} = -log m$$



      I thought about Jensen inequality or induction, but none of them provided me something.







      calculus inequality upper-lower-bounds jensen-inequality






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 28 at 2:28









      Alex Ravsky

      37.1k32078




      37.1k32078










      asked Nov 22 at 14:24









      pl-94

      1274




      1274






















          1 Answer
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          up vote
          1
          down vote



          accepted










          Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus



          $$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$



          and the minimum is attained when each $z_i$ equals $frac 1m$.






          share|cite|improve this answer























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            1 Answer
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            active

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            oldest

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            active

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            up vote
            1
            down vote



            accepted










            Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus



            $$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$



            and the minimum is attained when each $z_i$ equals $frac 1m$.






            share|cite|improve this answer



























              up vote
              1
              down vote



              accepted










              Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus



              $$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$



              and the minimum is attained when each $z_i$ equals $frac 1m$.






              share|cite|improve this answer

























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus



                $$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$



                and the minimum is attained when each $z_i$ equals $frac 1m$.






                share|cite|improve this answer














                Jensen’s inequality works straightforwardly. Since for each $i$, $log z_i$ is involved in the given inequality, we assume that $z_i>0$. Consider a function $f(x)=xlog x$ for $x>0$. Since $f’’(x)=frac 1{x}>0$, the function $f$ is convex. Thus



                $$frac{sum f(z_i)}mge f left(frac{sum z_i}mright)=fleft(frac 1mright)=frac 1mlogfrac 1m=-frac{log m}{m}.$$



                and the minimum is attained when each $z_i$ equals $frac 1m$.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 28 at 17:57









                pl-94

                1274




                1274










                answered Nov 28 at 2:28









                Alex Ravsky

                37.1k32078




                37.1k32078






























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