Symmetric function and product of two functions

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Question: [amended referring to the comments below]

In reading up on a nice result among generative mechanisms for outcomes that are described by power law distributions, I came across this paper, where the central claim is that given a function $f(x, y)$ that governs the link formation probability between two nodes, one can derive the conditions under which a scale-free network structure emerges.

The authors assert that $f(x,y)$ is a function that is symmetric with respect to its arguments and consider the case when $f(x, y) = g(x)h(y)$. Based on this, they conclude that $g(x) equiv h(x)$? I am trying to trace the arguments to support this assertion.

Attempt:

I started by assuming that $g(x) neq h(x)$ for any $x$. So, let $g(x) = lambda h(x), lambda in mathbbR$. Now, $f(x, y) = lambda h(x)h(y)$ and $f(y, x) = lambda h(y) h(x)$. This means that $f(x, y) = f(y, x)$ irrespective of $lambda$, i.e., we are unable to conclude that the original assumption of unequal $g$ and $h$ is true unless $lambda = 1$.

Any other approach to prove the question?

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edited Jul 17 at 5:01

asked Jul 17 at 4:12

buzaku

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Question: [amended referring to the comments below]

In reading up on a nice result among generative mechanisms for outcomes that are described by power law distributions, I came across this paper, where the central claim is that given a function $f(x, y)$ that governs the link formation probability between two nodes, one can derive the conditions under which a scale-free network structure emerges.

The authors assert that $f(x,y)$ is a function that is symmetric with respect to its arguments and consider the case when $f(x, y) = g(x)h(y)$. Based on this, they conclude that $g(x) equiv h(x)$? I am trying to trace the arguments to support this assertion.

Attempt:

I started by assuming that $g(x) neq h(x)$ for any $x$. So, let $g(x) = lambda h(x), lambda in mathbbR$. Now, $f(x, y) = lambda h(x)h(y)$ and $f(y, x) = lambda h(y) h(x)$. This means that $f(x, y) = f(y, x)$ irrespective of $lambda$, i.e., we are unable to conclude that the original assumption of unequal $g$ and $h$ is true unless $lambda = 1$.

Any other approach to prove the question?

functions

share|cite|improve this question

edited Jul 17 at 5:01

asked Jul 17 at 4:12

buzaku

3091212

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Question: [amended referring to the comments below]

In reading up on a nice result among generative mechanisms for outcomes that are described by power law distributions, I came across this paper, where the central claim is that given a function $f(x, y)$ that governs the link formation probability between two nodes, one can derive the conditions under which a scale-free network structure emerges.

The authors assert that $f(x,y)$ is a function that is symmetric with respect to its arguments and consider the case when $f(x, y) = g(x)h(y)$. Based on this, they conclude that $g(x) equiv h(x)$? I am trying to trace the arguments to support this assertion.

Attempt:

I started by assuming that $g(x) neq h(x)$ for any $x$. So, let $g(x) = lambda h(x), lambda in mathbbR$. Now, $f(x, y) = lambda h(x)h(y)$ and $f(y, x) = lambda h(y) h(x)$. This means that $f(x, y) = f(y, x)$ irrespective of $lambda$, i.e., we are unable to conclude that the original assumption of unequal $g$ and $h$ is true unless $lambda = 1$.

Any other approach to prove the question?

functions

share|cite|improve this question

edited Jul 17 at 5:01

asked Jul 17 at 4:12

buzaku

3091212

Question: [amended referring to the comments below]

In reading up on a nice result among generative mechanisms for outcomes that are described by power law distributions, I came across this paper, where the central claim is that given a function $f(x, y)$ that governs the link formation probability between two nodes, one can derive the conditions under which a scale-free network structure emerges.

The authors assert that $f(x,y)$ is a function that is symmetric with respect to its arguments and consider the case when $f(x, y) = g(x)h(y)$. Based on this, they conclude that $g(x) equiv h(x)$? I am trying to trace the arguments to support this assertion.

Attempt:

I started by assuming that $g(x) neq h(x)$ for any $x$. So, let $g(x) = lambda h(x), lambda in mathbbR$. Now, $f(x, y) = lambda h(x)h(y)$ and $f(y, x) = lambda h(y) h(x)$. This means that $f(x, y) = f(y, x)$ irrespective of $lambda$, i.e., we are unable to conclude that the original assumption of unequal $g$ and $h$ is true unless $lambda = 1$.

Any other approach to prove the question?

functions

share|cite|improve this question

edited Jul 17 at 5:01

asked Jul 17 at 4:12

buzaku

3091212

share|cite|improve this question

share|cite|improve this question

edited Jul 17 at 5:01

edited Jul 17 at 5:01

edited Jul 17 at 5:01

asked Jul 17 at 4:12

buzaku

3091212

asked Jul 17 at 4:12

buzaku

3091212

asked Jul 17 at 4:12

buzaku

3091212

3091212

add a comment | 

add a comment | 

1 Answer
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This is not true at all! For example, the function $f(x,y) equiv 2$ is symmetric, obviously, and can also be written as $f(x,y) = g(x)h(y)$ where $g(x) equiv 2$ and $h(x) equiv 1$.

It is true that $g(x)h(y) = g(y)h(x)$ for all $x,y$ by symmetry. Only if, we furthermore suppose there is some $X$ such that $g(X) = h(X) neq 0$, then for all $Y$, $g(Y) = h(Y)$ by cancellation. However, this may not always happen, as the example above shows.

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
  – buzaku
  Jul 17 at 4:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
  – dxiv
  Jul 17 at 4:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
  – buzaku
  Jul 17 at 5:03
  

  
  
  
  

add a comment | 

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This is not true at all! For example, the function $f(x,y) equiv 2$ is symmetric, obviously, and can also be written as $f(x,y) = g(x)h(y)$ where $g(x) equiv 2$ and $h(x) equiv 1$.

It is true that $g(x)h(y) = g(y)h(x)$ for all $x,y$ by symmetry. Only if, we furthermore suppose there is some $X$ such that $g(X) = h(X) neq 0$, then for all $Y$, $g(Y) = h(Y)$ by cancellation. However, this may not always happen, as the example above shows.

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
  – buzaku
  Jul 17 at 4:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
  – dxiv
  Jul 17 at 4:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
  – buzaku
  Jul 17 at 5:03
  

  
  
  
  

add a comment | 

up vote
2
down vote

This is not true at all! For example, the function $f(x,y) equiv 2$ is symmetric, obviously, and can also be written as $f(x,y) = g(x)h(y)$ where $g(x) equiv 2$ and $h(x) equiv 1$.

It is true that $g(x)h(y) = g(y)h(x)$ for all $x,y$ by symmetry. Only if, we furthermore suppose there is some $X$ such that $g(X) = h(X) neq 0$, then for all $Y$, $g(Y) = h(Y)$ by cancellation. However, this may not always happen, as the example above shows.

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
  – buzaku
  Jul 17 at 4:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
  – dxiv
  Jul 17 at 4:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
  – buzaku
  Jul 17 at 5:03
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

This is not true at all! For example, the function $f(x,y) equiv 2$ is symmetric, obviously, and can also be written as $f(x,y) = g(x)h(y)$ where $g(x) equiv 2$ and $h(x) equiv 1$.

It is true that $g(x)h(y) = g(y)h(x)$ for all $x,y$ by symmetry. Only if, we furthermore suppose there is some $X$ such that $g(X) = h(X) neq 0$, then for all $Y$, $g(Y) = h(Y)$ by cancellation. However, this may not always happen, as the example above shows.

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

This is not true at all! For example, the function $f(x,y) equiv 2$ is symmetric, obviously, and can also be written as $f(x,y) = g(x)h(y)$ where $g(x) equiv 2$ and $h(x) equiv 1$.

It is true that $g(x)h(y) = g(y)h(x)$ for all $x,y$ by symmetry. Only if, we furthermore suppose there is some $X$ such that $g(X) = h(X) neq 0$, then for all $Y$, $g(Y) = h(Y)$ by cancellation. However, this may not always happen, as the example above shows.

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

share|cite|improve this answer

share|cite|improve this answer

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

answered Jul 17 at 4:29

астон вілла олоф мэллбэрг

32k22463

32k22463

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
  – buzaku
  Jul 17 at 4:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
  – dxiv
  Jul 17 at 4:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
  – buzaku
  Jul 17 at 5:03
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
  – buzaku
  Jul 17 at 4:35
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
  – dxiv
  Jul 17 at 4:52
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
  – buzaku
  Jul 17 at 5:03
  

  
  
  
  

You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
– buzaku
Jul 17 at 4:35

You are right! Actually, this appeared as a part of this paper - arxiv.org/pdf/cond-mat/0309659.pdf. In page 2, where the authors explicitly state - 'We start with the special case of f(x, y) = g(x)h(y) where both the direct and inverse problems can be analytically solved. Because of the symmetry of f(x, y) with respect to its arguments one has g(x) ≡ h(x), so that f(x, y) = g(x)g(y)'.
– buzaku
Jul 17 at 4:35

@buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
– dxiv
Jul 17 at 4:52

@buzaku Would be better if you added that context to the posted question. It follows from the answer above that $g,h$ differ at most by a multiplicative constant. If there is some additional scaling constraint/assumption (such that for example $g,h$ are probability density functions) then it would indeed follow that $gequiv h$, though I don't know that's necessarily the case in the linked paper.
– dxiv
Jul 17 at 4:52

amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
– buzaku
Jul 17 at 5:03

amended the question. I re-read the paper but cant trace $g$ and $h$ to be pdfs. I guess the key thing like you point out is that they differ at most by a multiplicative constant. I will work out if this makes any difference to the results in the paper.
– buzaku
Jul 17 at 5:03

add a comment | 

 

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