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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:

$$(1)\,\, \frac \partial {\partial y}\left(G(x,y)G(y,z)\right)=0 \implies G(x,y)=r\frac{H(x)}{H(y)}$$ 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?

3dot
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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.

kimchi lover
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  • 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