Mixing
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
asked Jan 8 at 15:57
HansHans
4,99021032
$endgroup$
add a comment |
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
asked Jan 8 at 15:57
HansHans
4,99021032
$endgroup$
add a comment |
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
asked Jan 8 at 15:57
HansHans
4,99021032
$endgroup$
$x$ 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
asked Jan 8 at 15:57
HansHans
4,99021032
asked Jan 8 at 15:57
HansHans
4,99021032
asked Jan 8 at 15:57
HansHans
4,99021032
asked Jan 8 at 15:57
HansHans
4,99021032
asked Jan 8 at 15:57
HansHans
4,99021032
4,99021032
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
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.
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
$endgroup$
$begingroup$
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
add a comment |
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1 Answer
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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.
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
$endgroup$
$begingroup$
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
add a comment |
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.
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
$endgroup$
$begingroup$
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
add a comment |
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.
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
$endgroup$
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.
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
edited Jan 8 at 17:09
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
answered Jan 8 at 16:25
P. QuintonP. Quinton
1,7911213
1,7911213
$begingroup$
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
add a comment |
$begingroup$
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
add a comment |
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