How do we make this integration rigorous?

Question

This is from Jaynes, Probability Theory: The Logic of Science, pp 27-28.

We have a function $F$ which is $\mathbb{R}^2 \rightarrow \mathbb{R}$, and we set $v = F(y,z)$. We discover that

$$ F_1(y,z) = { \partial F \over \partial y } = { H(v) \over H(y) } \\ F_2(y,z) = { \partial F \over \partial z } = r{ H(v) \over H(z) } $$

where $H$ is arbitrary, but can't change sign in the region of interest. (Specifically, $H$ is such that $F_2(y,z) / F_1(y,z)$ takes the form $r H(y) / H(z) $.)

We later discover $r = 1$, so I'm going to ignore that for clarity. Then, since $\mathrm d v = F_1 \mathrm d y + F_2 \mathrm d z $, we get

$$ { \mathrm d v \over H(v) } = { \mathrm d y \over H(y) } + { \mathrm d z \over H(z) } $$

So far, so good. Now we define

$$ w(x) = \exp\left( \int^x { \mathrm d x \over H(x) } \right) $$

and it follows that $w(v) = w(y) w(z)$.

I can kinda sorta see how this happens, but not really. Apparently,

$$ \int { \mathrm d v \over H(v) } = \int \left( { \mathrm d y \over H(y) } + { \mathrm d z \over H(z) } \right) = \int { \mathrm d y \over H(y) } + \int{ \mathrm d z \over H(z) } $$

And then we just apply $e^{a+b} = e^a e^b$. This makes some sense notationally, but I'm not familiar with the rigor behind it. (I've removed the ${}^x$ from the integrals because I don't really know what I'd do with it. Jaynes credits this proof to Cox (1961), which I looked up - it didn't include any intermediate steps to help me, but it also omitted the ${}^x$, so I feel somewhat comfortable doing so myself.)

I think there are a few specific things confusing me, where the notation just doesn't mean what I expect it to:

• I expect expressions like $\int^x f(s) \mathrm d s$ to be a function of $x$, where $s$ is a bound variable, and we can rewrite $s$ as $p$ or $\alpha$ or anything, and rewriting $s$ as $x$ is just about the most confusing choice we can possibly make. In this case, it seems that the symbol used inside the integration is relevant?

• $w$ looks like a function $\mathbb R \rightarrow \mathbb R $, but $w(v)$ depends upon $\mathrm d v$ as well as $v$?

Can anyone clear this up for me?

Answer 1

So you have a scalar field $F$ such that

$$\nabla F = [e_1/H(y) + e_2/H(z)] [H \circ F](y,z)$$

You can, of course, rearrange this to read

$$\frac{\nabla F|_{y,z}}{(H \circ F)(y,z)} = \frac{e_1}{H(y)} + \frac{e_2}{H(z)}$$

Now integrate on some curve $\ell(t) = \ell^y(t) e_1 + \ell^z(t) e_2$:

$$\int \left( \frac{e_1}{H \circ \ell^y} + \frac{e_2}{H \circ \ell^z} \right) \cdot \ell' \, dt = \int \frac{\nabla F|_\ell}{H \circ F \circ \ell} \cdot \ell' dt$$

Now, define

$$w(x;\lambda)=\exp \int^x \frac{\lambda'(t)}{(H \circ \lambda)(t)} \, dt$$

where $\lambda$ is some function. It should be clear that exponentiating the LHS just gives

$$\exp \left[\int^x \frac{(\ell^y)'}{H \circ \ell^y} \, dt + \int^x \frac{(\ell^z)'}{H \circ \ell^z} \, dt \right] = w(x; \ell^y)w(x; \ell^z)$$

The RHS is a little trickier. You need to realize that

$$\frac{d}{dt} (F \circ \ell)(t) = \ell' \cdot \nabla F|_{\ell(t)}$$

So the RHS can be written in terms of the function $G(t) = (F \circ \ell)(t)$ as

$$\int^x \frac{\ell'(t) \cdot \nabla F|_{\ell(t)}}{H \circ F \circ \ell} \, dt= \int^x \frac{G'(t)}{(H \circ G)(t)} \, dt$$

So exponentiating this, we get

$$w(x; F \circ \ell) = w(x; \ell^y) w(x; \ell^z)$$

Now, most people would call $y = \ell^y$ and $z = \ell^z$; this is a trivial choice of parameterization. Most people wouldn't bother writing $F \circ \ell$ and would just write $v$ as the authors did. So the letter of the proof holds: that $w(x;v) = w(x;y) w(x;z)$. I think writing it with one argument instead of two (as I did) is considerably less clear. Using the same variable for a bound of integration as the dummy variable is a big no-no, but some people do that when renaming would be less clear.

I think at the very least, $w$ ought to be a function of two arguments: a number $x$ and a function $\lambda: \mathbb R \mapsto \mathbb R$.