Will the mutual information equal to $2$ when H(X) and H(Y)=$0$?

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$P(X,Y)=$beginbmatrix
P(x_1,y_1) & P(x_1,y_2)\
P(x_2,y_1) & P(x_2,y_2)
endbmatrix =



beginbmatrix
0.54 & 0.06\
0.06 & 0.34
endbmatrix



And i calculate the $H(x)=-1log_21=0$ and $H(y)=-1log_21=0$,and $H(X|Y)=H(Y|X)=-2log_22=-2$ ,so $I(X,Y)=H(X)-H(X|Y)=0-(-2)=2$,but according the explanation of mutual information in wiki https://en.wikipedia.org/wiki/Conditional_probability ,the picture is as below,it seems that it is impossible that the mutual information equal to $2$ when H(X) and H(Y)=$0$ ,is my calculation wrong?or it does possible that mutual information equal to $2$ when H(X) and H(Y)=$0$?
enter image description here




probability information-theory




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    up vote
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    down vote

    favorite












    $P(X,Y)=$beginbmatrix
    P(x_1,y_1) & P(x_1,y_2)\
    P(x_2,y_1) & P(x_2,y_2)
    endbmatrix =



    beginbmatrix
    0.54 & 0.06\
    0.06 & 0.34
    endbmatrix



    And i calculate the $H(x)=-1log_21=0$ and $H(y)=-1log_21=0$,and $H(X|Y)=H(Y|X)=-2log_22=-2$ ,so $I(X,Y)=H(X)-H(X|Y)=0-(-2)=2$,but according the explanation of mutual information in wiki https://en.wikipedia.org/wiki/Conditional_probability ,the picture is as below,it seems that it is impossible that the mutual information equal to $2$ when H(X) and H(Y)=$0$ ,is my calculation wrong?or it does possible that mutual information equal to $2$ when H(X) and H(Y)=$0$?
    enter image description here




    probability information-theory




    share|cite|improve this question





















      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      $P(X,Y)=$beginbmatrix
      P(x_1,y_1) & P(x_1,y_2)\
      P(x_2,y_1) & P(x_2,y_2)
      endbmatrix =



      beginbmatrix
      0.54 & 0.06\
      0.06 & 0.34
      endbmatrix



      And i calculate the $H(x)=-1log_21=0$ and $H(y)=-1log_21=0$,and $H(X|Y)=H(Y|X)=-2log_22=-2$ ,so $I(X,Y)=H(X)-H(X|Y)=0-(-2)=2$,but according the explanation of mutual information in wiki https://en.wikipedia.org/wiki/Conditional_probability ,the picture is as below,it seems that it is impossible that the mutual information equal to $2$ when H(X) and H(Y)=$0$ ,is my calculation wrong?or it does possible that mutual information equal to $2$ when H(X) and H(Y)=$0$?
      enter image description here




      probability information-theory




      share|cite|improve this question











      $P(X,Y)=$beginbmatrix
      P(x_1,y_1) & P(x_1,y_2)\
      P(x_2,y_1) & P(x_2,y_2)
      endbmatrix =



      beginbmatrix
      0.54 & 0.06\
      0.06 & 0.34
      endbmatrix



      And i calculate the $H(x)=-1log_21=0$ and $H(y)=-1log_21=0$,and $H(X|Y)=H(Y|X)=-2log_22=-2$ ,so $I(X,Y)=H(X)-H(X|Y)=0-(-2)=2$,but according the explanation of mutual information in wiki https://en.wikipedia.org/wiki/Conditional_probability ,the picture is as below,it seems that it is impossible that the mutual information equal to $2$ when H(X) and H(Y)=$0$ ,is my calculation wrong?or it does possible that mutual information equal to $2$ when H(X) and H(Y)=$0$?
      enter image description here





      probability information-theory





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      share|cite|improve this question




      share|cite|improve this question









      asked Jul 29 at 13:45









      Shine Sun

      1258




      1258




















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          I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$
          H(x)=-(0.6)log(0.6)-(0.4)log(0.4),
          $$
          and similarly for $H(y)$. Neither of these is zero.



          Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic,
          $$
          H(x|y)=H(y|x)=-(0.54)log(0.9)-(0.06)log(0.1)-(0.06)log(0.15)-(0.34)log(0.85).
          $$






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            I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$
            H(x)=-(0.6)log(0.6)-(0.4)log(0.4),
            $$
            and similarly for $H(y)$. Neither of these is zero.



            Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic,
            $$
            H(x|y)=H(y|x)=-(0.54)log(0.9)-(0.06)log(0.1)-(0.06)log(0.15)-(0.34)log(0.85).
            $$






            share|cite|improve this answer



























              up vote
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              down vote













              I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$
              H(x)=-(0.6)log(0.6)-(0.4)log(0.4),
              $$
              and similarly for $H(y)$. Neither of these is zero.



              Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic,
              $$
              H(x|y)=H(y|x)=-(0.54)log(0.9)-(0.06)log(0.1)-(0.06)log(0.15)-(0.34)log(0.85).
              $$






              share|cite|improve this answer

























                up vote
                0
                down vote










                up vote
                0
                down vote









                I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$
                H(x)=-(0.6)log(0.6)-(0.4)log(0.4),
                $$
                and similarly for $H(y)$. Neither of these is zero.



                Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic,
                $$
                H(x|y)=H(y|x)=-(0.54)log(0.9)-(0.06)log(0.1)-(0.06)log(0.15)-(0.34)log(0.85).
                $$






                share|cite|improve this answer















                I would interpret the $P$ matrix in the statement of the problem as telling me that the random variable $x$ has the value $x_1$ with probability $0.54+0.06=0.6$ and has the value $x_2$ with probability $0.6+0.34=0.4$. So $$
                H(x)=-(0.6)log(0.6)-(0.4)log(0.4),
                $$
                and similarly for $H(y)$. Neither of these is zero.



                Your calculation of $H(x|y)$ and $H(y|x)$ also looks wrong (nor do I understand why the variables $x$ and $y$ became $X$ and $Y$ half way through the question).Unless I've messed up the arithmetic,
                $$
                H(x|y)=H(y|x)=-(0.54)log(0.9)-(0.06)log(0.1)-(0.06)log(0.15)-(0.34)log(0.85).
                $$







                share|cite|improve this answer















                share|cite|improve this answer



                share|cite|improve this answer








                edited Jul 29 at 15:16


























                answered Jul 29 at 15:07









                Andreas Blass

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