$G(x,y)G(y,z)$ independant of $y$ $implies$ Most general function form $G(x,y) = rH(x)/H(y)$ , $r$ a const.
I am reading Probability Theory, The Logic of Science by E.T. Jaynes; and encountered the following statement in his derivation of the product/chain rule (of probability) from basic principles:
$$(1),, frac partial {partial y}left(G(x,y)G(y,z)right)=0 implies G(x,y)=rfrac{H(x)}{H(y)}$$
for some constant $r$ without any loss of generality.
I am having trouble showing this. This question was already asked here by user56834, but received no answers.
After some 10-15 hours, I've only been able to arrive at:
$$(1) implies G(x,y)G(y,x) = K$$ for some constant $K$.
This is highly suggestibe, but I can't seem to prove that the final (fraction of same function) form is necessary.
Can some one help?
functional-equations
asked Nov 25 '18 at 16:44
3dot
83
add a comment |
I am reading Probability Theory, The Logic of Science by E.T. Jaynes; and encountered the following statement in his derivation of the product/chain rule (of probability) from basic principles:
$$(1),, frac partial {partial y}left(G(x,y)G(y,z)right)=0 implies G(x,y)=rfrac{H(x)}{H(y)}$$
for some constant $r$ without any loss of generality.
I am having trouble showing this. This question was already asked here by user56834, but received no answers.
After some 10-15 hours, I've only been able to arrive at:
$$(1) implies G(x,y)G(y,x) = K$$ for some constant $K$.
This is highly suggestibe, but I can't seem to prove that the final (fraction of same function) form is necessary.
Can some one help?
functional-equations
asked Nov 25 '18 at 16:44
3dot
83
MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
1
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05
add a comment |
I am reading Probability Theory, The Logic of Science by E.T. Jaynes; and encountered the following statement in his derivation of the product/chain rule (of probability) from basic principles:
$$(1),, frac partial {partial y}left(G(x,y)G(y,z)right)=0 implies G(x,y)=rfrac{H(x)}{H(y)}$$
for some constant $r$ without any loss of generality.
I am having trouble showing this. This question was already asked here by user56834, but received no answers.
After some 10-15 hours, I've only been able to arrive at:
$$(1) implies G(x,y)G(y,x) = K$$ for some constant $K$.
This is highly suggestibe, but I can't seem to prove that the final (fraction of same function) form is necessary.
Can some one help?
functional-equations
asked Nov 25 '18 at 16:44
3dot
83
I am reading Probability Theory, The Logic of Science by E.T. Jaynes; and encountered the following statement in his derivation of the product/chain rule (of probability) from basic principles:
$$(1),, frac partial {partial y}left(G(x,y)G(y,z)right)=0 implies G(x,y)=rfrac{H(x)}{H(y)}$$
for some constant $r$ without any loss of generality.
I am having trouble showing this. This question was already asked here by user56834, but received no answers.
After some 10-15 hours, I've only been able to arrive at:
$$(1) implies G(x,y)G(y,x) = K$$ for some constant $K$.
This is highly suggestibe, but I can't seem to prove that the final (fraction of same function) form is necessary.
Can some one help?
functional-equations
functional-equations
asked Nov 25 '18 at 16:44
3dot
83
asked Nov 25 '18 at 16:44
3dot
83
edited Nov 25 '18 at 17:05
asked Nov 25 '18 at 16:44
3dot
83
asked Nov 25 '18 at 16:44
3dot
83
asked Nov 25 '18 at 16:44
3dot
83
83
MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
1
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05
add a comment |
MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
1
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05
MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
1
1
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05
add a comment |
1 Answer
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Ignoring questions of division by 0, you can write $$frac{G(x,y)}{G(x,1)}frac{G(y,z)}{G(1,z)}=1$$ and hence each of the ratios $G(x,y)/{G(x,1)}$ and $G(y,z)/{G(1,z)}$ is independent of $x$ and of $z$. Hence $G(x,y)/G(x,1)$ is a function of $y$ alone; call it $h(y)$. So we know $G(x,y)=g(x)h(y),$ using the shorthand $g(x)=G(x,1)$. So now we know $g(x)h(y)g(y)h(z),$ which equals $G(x,y)G(y,z),$ is independent of $y$. That is, $h(y)g(y)$ is a constant, so $h(y)=k/g(y)$, for some $k$, and so on.
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
add a comment |
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Ignoring questions of division by 0, you can write $$frac{G(x,y)}{G(x,1)}frac{G(y,z)}{G(1,z)}=1$$ and hence each of the ratios $G(x,y)/{G(x,1)}$ and $G(y,z)/{G(1,z)}$ is independent of $x$ and of $z$. Hence $G(x,y)/G(x,1)$ is a function of $y$ alone; call it $h(y)$. So we know $G(x,y)=g(x)h(y),$ using the shorthand $g(x)=G(x,1)$. So now we know $g(x)h(y)g(y)h(z),$ which equals $G(x,y)G(y,z),$ is independent of $y$. That is, $h(y)g(y)$ is a constant, so $h(y)=k/g(y)$, for some $k$, and so on.
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
add a comment |
Ignoring questions of division by 0, you can write $$frac{G(x,y)}{G(x,1)}frac{G(y,z)}{G(1,z)}=1$$ and hence each of the ratios $G(x,y)/{G(x,1)}$ and $G(y,z)/{G(1,z)}$ is independent of $x$ and of $z$. Hence $G(x,y)/G(x,1)$ is a function of $y$ alone; call it $h(y)$. So we know $G(x,y)=g(x)h(y),$ using the shorthand $g(x)=G(x,1)$. So now we know $g(x)h(y)g(y)h(z),$ which equals $G(x,y)G(y,z),$ is independent of $y$. That is, $h(y)g(y)$ is a constant, so $h(y)=k/g(y)$, for some $k$, and so on.
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
add a comment |
Ignoring questions of division by 0, you can write $$frac{G(x,y)}{G(x,1)}frac{G(y,z)}{G(1,z)}=1$$ and hence each of the ratios $G(x,y)/{G(x,1)}$ and $G(y,z)/{G(1,z)}$ is independent of $x$ and of $z$. Hence $G(x,y)/G(x,1)$ is a function of $y$ alone; call it $h(y)$. So we know $G(x,y)=g(x)h(y),$ using the shorthand $g(x)=G(x,1)$. So now we know $g(x)h(y)g(y)h(z),$ which equals $G(x,y)G(y,z),$ is independent of $y$. That is, $h(y)g(y)$ is a constant, so $h(y)=k/g(y)$, for some $k$, and so on.
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
Ignoring questions of division by 0, you can write $$frac{G(x,y)}{G(x,1)}frac{G(y,z)}{G(1,z)}=1$$ and hence each of the ratios $G(x,y)/{G(x,1)}$ and $G(y,z)/{G(1,z)}$ is independent of $x$ and of $z$. Hence $G(x,y)/G(x,1)$ is a function of $y$ alone; call it $h(y)$. So we know $G(x,y)=g(x)h(y),$ using the shorthand $g(x)=G(x,1)$. So now we know $g(x)h(y)g(y)h(z),$ which equals $G(x,y)G(y,z),$ is independent of $y$. That is, $h(y)g(y)$ is a constant, so $h(y)=k/g(y)$, for some $k$, and so on.
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
answered Nov 25 '18 at 21:38
kimchi lover
9,65131128
9,65131128
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
add a comment |
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
Thank you! I was about to start trying to show that for $U(x,y) = ln(G(x,y))$, $U_1(x,y) = -U_2(y,x)$ for all $x, y$, and that $U_1,_2(x,y) = 0$, but your approach is so much more elegant!
– 3dot
Nov 25 '18 at 22:05
add a comment |
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MathJax works in the title section, don't you know?
– Shaun
Nov 25 '18 at 16:58
1
changed it, thanks.
– 3dot
Nov 25 '18 at 17:05