The average of a continuous value: $\overline{O} = \int O(x) \rho(x) dx$, but coordinate invariant

Question

I am trying to solve a Lagrange multiplier problem for the following equation

$$ L= - \int_{-\infty}^\infty \rho(x) \ln \frac{\rho(x)}{q(x)} dx + \alpha \left( 1- \int_{-\infty}^\infty \rho(x) dx \right) +\beta \left( \overline{O} - \int_{-\infty}^\infty O(x) \rho(x) dx \right) \tag{1} $$

for $\frac{\partial L}{\partial \rho(w)} =0$.

Where the first term is the relative entropy, the second term is the constraint that it sums to one, and the last term is the constraint of an observable involving its average value. The problem solves for the probability measure that maximizes the relative entropy of a continuous parametrization x.

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An interesting property of the relative entropy is that its equations remains invariant with respect to a change in variable. Indeed,

$$ \int_{-\infty}^\infty \rho(x)\ln \frac{\rho(x)}{q(x)} dx \to \int_{-\infty}^\infty \rho(y(x)) \left|\frac{\partial y}{\partial x}\right|\ln \frac{\rho(y(x))\left|\frac{\partial y}{\partial x}\right|}{q(y(x)) \left|\frac{\partial y}{\partial x}\right|} dx = \int_{-\infty}^\infty \rho(y)\ln \frac{\rho(y)}{q(y)} dy \tag{2} $$

and

$$ \int_{-\infty}^\infty \rho(x)dx \to \int_{-\infty}^\infty \rho(y(x)) \left|\frac{\partial y}{\partial x}\right| dx = \int_{-\infty}^\infty \rho(y) dy \tag{3} $$

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However, the last term isn't. Indeed:

$$ \int_{-\infty}^\infty O(x) \rho (x) dx \to \int_{-\infty}^\infty O(y(x)) \left|\frac{\partial y}{\partial x}\right| \rho (y(x)) \left|\frac{\partial y}{\partial x}\right| dx = \int_{-\infty}^\infty O(y) \rho (y(x)) \left|\frac{\partial y}{\partial x}\right| dy \tag{4} $$

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How can I modify the integral that contains O so that it is coordinate invariant?

What is the expression for the average of a function O(x) over a continuous parametrization x, such that the average value is coordinate invariant?

Answer 1

lpz is correct that you are including an erroneous factor of $|dy/dx|$ in your transformation of $O(x)$. You may be getting misled by the transformation of the density, which uses the same symbol for the density of both $x$ and $y$ in your notation. Your conclusion in the comments that all functions must transform the same way under a change of variables is incorrect.

The transformation is more accurately written $$\rho_X(x) = \rho_Y(y) \left|\frac{dy}{dx}\right|$$ where $\rho_X(x)$ is the probability density of $x$ and $\rho_Y(y)$ is the probability density of $y$; these two densities are different functions. They are related here under the assumption that we may view $x$ as a function of $y$ or vice versa. The derivative here comes from the Jacobian of the integral under the change of variables, which is then combined with $\rho_X(x)$ to define $\rho_y(y)$.

The transformation of $\int dx~O(x) \rho_x(x)$ should therefore be $$\int dx~O(x) \rho_x(y) = \int dy \left|\frac{dx}{dy}\right| O(x(y)) \rho_X(x(y)) = \int dy \frac{1}{\left|\frac{dy}{dx}\right|} O(x(y)) \rho_X(x(y)) = \int dy~ O'(y) \rho_Y(y),$$ where I used the fact that $dx/dy = 1/(dy/dx)$ and defined $O'(y) = O(x(y))$, with $x(y)$ is the variable $x$ interpreted as a function of $y$. Note that $O(x)$ and $O'(y)$ will not have the same shape unless $x(y) = y$, and this is similar to the fact that $\rho_X(x)$ and $\rho_Y(y)$ are not the same function.

The values of the integrals $\int dx~O(x) \rho_X(x)$ and $\int dy~O'(y) \rho_Y(y)$ will be the same.