A mutual information equality












0












$begingroup$


$x$ and $y$ are two random variables and $I(u;v)$ is the mutual information between random variables $u$ and $v$. Does the following equality hold?
$$text{argmax}_a I(y;ax)=text{argmin}_a I(y;y-ax).$$






information-theory







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    $x$ and $y$ are two random variables and $I(u;v)$ is the mutual information between random variables $u$ and $v$. Does the following equality hold?
    $$text{argmax}_a I(y;ax)=text{argmin}_a I(y;y-ax).$$






    information-theory







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      $x$ and $y$ are two random variables and $I(u;v)$ is the mutual information between random variables $u$ and $v$. Does the following equality hold?
      $$text{argmax}_a I(y;ax)=text{argmin}_a I(y;y-ax).$$






      information-theory







      share|cite|improve this question









      $endgroup$




      $x$ and $y$ are two random variables and $I(u;v)$ is the mutual information between random variables $u$ and $v$. Does the following equality hold?
      $$text{argmax}_a I(y;ax)=text{argmin}_a I(y;y-ax).$$






      information-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 8 at 15:57









      HansHans

      4,98021032




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          1 Answer
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          oldest

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          $begingroup$

          It can be seen that for any $aneq 0$, $I(y;ax)=I(y;x)$ because $yrightarrow axrightarrow x$ and $yrightarrow xrightarrow ax$ are both Markov chain hence by the data processing inequality both $I(y;x)leq I(y;ax)$ and $I(y;x)geq I(y;ax)$ respectively.



          Since any $aneq 0$ is a maximizer of $I(y;ax)$, your question is ill defined.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
            $endgroup$
            – Hans
            Jan 8 at 16:48












          • $begingroup$
            Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
            $endgroup$
            – P. Quinton
            Jan 8 at 17:07










          • $begingroup$
            You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
            $endgroup$
            – Hans
            Jan 8 at 23:00












          • $begingroup$
            well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
            $endgroup$
            – P. Quinton
            Jan 9 at 6:28










          • $begingroup$
            I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
            $endgroup$
            – Hans
            Jan 9 at 17:45











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          1












          $begingroup$

          It can be seen that for any $aneq 0$, $I(y;ax)=I(y;x)$ because $yrightarrow axrightarrow x$ and $yrightarrow xrightarrow ax$ are both Markov chain hence by the data processing inequality both $I(y;x)leq I(y;ax)$ and $I(y;x)geq I(y;ax)$ respectively.



          Since any $aneq 0$ is a maximizer of $I(y;ax)$, your question is ill defined.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
            $endgroup$
            – Hans
            Jan 8 at 16:48












          • $begingroup$
            Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
            $endgroup$
            – P. Quinton
            Jan 8 at 17:07










          • $begingroup$
            You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
            $endgroup$
            – Hans
            Jan 8 at 23:00












          • $begingroup$
            well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
            $endgroup$
            – P. Quinton
            Jan 9 at 6:28










          • $begingroup$
            I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
            $endgroup$
            – Hans
            Jan 9 at 17:45
















          1












          $begingroup$

          It can be seen that for any $aneq 0$, $I(y;ax)=I(y;x)$ because $yrightarrow axrightarrow x$ and $yrightarrow xrightarrow ax$ are both Markov chain hence by the data processing inequality both $I(y;x)leq I(y;ax)$ and $I(y;x)geq I(y;ax)$ respectively.



          Since any $aneq 0$ is a maximizer of $I(y;ax)$, your question is ill defined.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
            $endgroup$
            – Hans
            Jan 8 at 16:48












          • $begingroup$
            Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
            $endgroup$
            – P. Quinton
            Jan 8 at 17:07










          • $begingroup$
            You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
            $endgroup$
            – Hans
            Jan 8 at 23:00












          • $begingroup$
            well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
            $endgroup$
            – P. Quinton
            Jan 9 at 6:28










          • $begingroup$
            I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
            $endgroup$
            – Hans
            Jan 9 at 17:45














          1












          1








          1





          $begingroup$

          It can be seen that for any $aneq 0$, $I(y;ax)=I(y;x)$ because $yrightarrow axrightarrow x$ and $yrightarrow xrightarrow ax$ are both Markov chain hence by the data processing inequality both $I(y;x)leq I(y;ax)$ and $I(y;x)geq I(y;ax)$ respectively.



          Since any $aneq 0$ is a maximizer of $I(y;ax)$, your question is ill defined.






          share|cite|improve this answer











          $endgroup$



          It can be seen that for any $aneq 0$, $I(y;ax)=I(y;x)$ because $yrightarrow axrightarrow x$ and $yrightarrow xrightarrow ax$ are both Markov chain hence by the data processing inequality both $I(y;x)leq I(y;ax)$ and $I(y;x)geq I(y;ax)$ respectively.



          Since any $aneq 0$ is a maximizer of $I(y;ax)$, your question is ill defined.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 8 at 17:09

























          answered Jan 8 at 16:25









          P. QuintonP. Quinton

          1,7261213




          1,7261213












          • $begingroup$
            Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
            $endgroup$
            – Hans
            Jan 8 at 16:48












          • $begingroup$
            Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
            $endgroup$
            – P. Quinton
            Jan 8 at 17:07










          • $begingroup$
            You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
            $endgroup$
            – Hans
            Jan 8 at 23:00












          • $begingroup$
            well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
            $endgroup$
            – P. Quinton
            Jan 9 at 6:28










          • $begingroup$
            I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
            $endgroup$
            – Hans
            Jan 9 at 17:45


















          • $begingroup$
            Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
            $endgroup$
            – Hans
            Jan 8 at 16:48












          • $begingroup$
            Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
            $endgroup$
            – P. Quinton
            Jan 8 at 17:07










          • $begingroup$
            You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
            $endgroup$
            – Hans
            Jan 8 at 23:00












          • $begingroup$
            well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
            $endgroup$
            – P. Quinton
            Jan 9 at 6:28










          • $begingroup$
            I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
            $endgroup$
            – Hans
            Jan 9 at 17:45
















          $begingroup$
          Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
          $endgroup$
          – Hans
          Jan 8 at 16:48






          $begingroup$
          Could there be a misunderstanding, say, as shown in the expression $I(y;x)>0=mathrm{argmin}_a I(y;y-ax)$? I am asking about argmin which the argument $a$ for which a minimization is achieved, not min.
          $endgroup$
          – Hans
          Jan 8 at 16:48














          $begingroup$
          Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
          $endgroup$
          – P. Quinton
          Jan 8 at 17:07




          $begingroup$
          Yes, Sorry about that, but then you have to define $mathrm{argmax}$ more carefully since any $aneq 0$ is a maximizer of $I(y;ax)$ hence yes, it could be true depending on how you do this definition
          $endgroup$
          – P. Quinton
          Jan 8 at 17:07












          $begingroup$
          You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
          $endgroup$
          – Hans
          Jan 8 at 23:00






          $begingroup$
          You are right. Indeed, from the definition $I(x;y)$ is invariant under any injective one variable transformation, i.e. $I(x;y)=I(f(x),g(y))$ where $f$ and $g$ measurable injective functions. More importantly, you may be interested in taking a look at the question of my main concern stats.stackexchange.com/q/386101/44368.
          $endgroup$
          – Hans
          Jan 8 at 23:00














          $begingroup$
          well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
          $endgroup$
          – P. Quinton
          Jan 9 at 6:28




          $begingroup$
          well, what you say is true if and only if $f$ and $g$ are almost surely one to one. About your main question, I don't understand what "z distributed normally conditioned on x" means, I would say you may want $x$ and $z$ independ and $zsim mathcal N(0,sigma^2)$ or something of the sort. I would say without proof that if $x$ is also gaussian then your minimization is exactly the same as the usual regression so it may be of interest in other cases. I'm pretty sure this has been done before but I do not have any references.
          $endgroup$
          – P. Quinton
          Jan 9 at 6:28












          $begingroup$
          I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
          $endgroup$
          – Hans
          Jan 9 at 17:45




          $begingroup$
          I said "injective... transformation". That means one-to-one. "Almost surely", sure it makes it more complete, but is more like icing on the cake. So we are saying the same thing. Regarding my main question, there may be some misunderstanding there. The normality is to introduce the subject by giving the background. The new concept using mutual information comes after the phrase "now dropping the normality". Would you please reread the question more carefully? Thank you.
          $endgroup$
          – Hans
          Jan 9 at 17:45


















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