What does Jaynes' continuous pdf notation "$g(x)~\mathrm dx$" actually mean?

Question

Something has been bugging me about E.T. Jaynes' treatment of continuous parameters.

In his book Probability Theory: The Logic of Science, uses notation that I am unfamiliar with when getting probabilities for continuous parameters from their densities. For example, in chapter 4, page 420, equation 4-49, Jaynes "discretizes" the continuous parameter $f \in [0,1]$ by taking the interval $(f, f+df)$ and putting a probability on it via $g$, the density of $f$:

.. so we will go back to the original probability from (4-3):

$$P(A|DX)=\frac{P(D|AX)P(A|X)}{P(D|X)}$$

Letting $A$ now stand for the proposition "The fraction of bad ones is in the range $(f, f+df)$", there is a prior pdf

$$P(A|X)=g(f|X)~df, $$

which gives the probability that the fraction of bad ones is in the range $df;$

Why isn't the above written as $P(A|X)=\displaystyle\int_f^{f+dx} g(t|X)~dt$ (with dummy variable $t$)? $g(f|X)~df$ for a finite $df$ is the left-rectangle approximation (a la Riemann sums), not $P(A|X).$ If you were to partition the $[0,1]$ into many separate proposition and assigned probability as above, it may not sum to $1.$

Or does $df$ denote some sort of infinitesimal notation I'm not familiar with? I've never seem that notation used 'alone'-- as far as I know it's only meaningful when used in context of differentiation/integration, like $\frac{d}{dx}$ or $\int \textrm{foo}()~dx$. If it is some infinitesimal notation, this seems to run contrary to Jaynes' philosophy of sticking to finite number of propositions until the end, where the limit may be taken explicitly.

As another example, in chapter 15, page 1514, equation (15-38), he writes the bivariate normal probability with correlation $\rho$ as:

$$p(dx~dy|I)=\frac{\sqrt{1-\rho^2}}{2\pi}\,\exp\left[-\frac{1}{2}(x^2 + y^2 -2 \rho x y)\right]\,dx\,dy$$

which is a little different again. Presumably $p(dx~dy|I)$ means "Probability that the true value of $(x,y)$ falls in $(x+dx,y+dy)$", but here again he writes the Riemann-sum-like approximation instead of the integral of the density on $[(x,x+dx) \times (y,y+dy)].$

What am I missing?

Answer 1

The $df$ should be taken as approaching zero. See differential (infinitesimal).

Answer 2

About the first quote: Your first interpretation is correct: Jaynes replaces the integral with a rectangle approximation. Since it is an approximation to the integral, you are also correct in remarking that, for a given $df$, the finite sum over the rectangles does not give $1$.

About the second quote: I am afraid Jaynes made a notational mistake in this definition. Using the Lebesgue measure on $\mathbb{R}^2$ as $\text{d}x\,\text{d}y$, we should have $$ \text{d}p(x,y|I)=\sqrt{1−ρ^2}/\sqrt{2π}\exp[−1/2(x^2+y^2−2ρxy)]\text{d}x\text{d} y $$ for the normal measure. However, since Jaynes is overall very much opposed to measure theory, I think your interpretation as a Riemann approximation is completely right!

Answer 3

The notation $\Pr(X \in dx)=f(x)dx$ is a standard notation to say that the law of the random variable $X$ has density $f$. Note that this notation is coherent with $\frac{\Pr(X\in dx)}{dx}=f(x)$ which actually says that $f$ is the Radon-Nikodym derivative of the law of $X$ with respect to the Lebesgue measure - in other words, the density. Thus the notation $p(dx)=f(x)dx$ is correct when denoting by $p$ the distribution of $X$. This notation is also coherent with the fact that $\Pr(X \in [x, x+h]) \approx hf(x)$ when $h$ is small.