Error propagation of Gaussian measures I am a physicist and I’ve never taken any statistics or probability classes. Nevertheless, I'm trying to build a set of lecture notes for my students of experimental physics regarding error propagation. Sadly, I haven’t found much literature on the subject from a rigorous point of view (it probably has a name and I just don’t know it).
In any case, in physics we usually denote our measurements by $x\pm\Delta x$ where $x$ is our “best guess” and $\Delta x$ is an interval where we are reasonably certain that our measurement lies. If we have multiple measurements $x_1\pm\Delta x_1,\dots,x_n\pm\Delta x_n$ and we want to calculate a function of these, we would report it as $$f(x_1,\dots,x_n)\pm\sqrt{\sum_{i=1}^n\big(\partial_i f(x_1,\dots,x_n)\Delta x_i\big)^2}.$$
In my set of notes, I wanted to take a more rigorous approach. Instead of using the $x\pm \Delta x$ notation, I insisted on measurements being probability measures on an outcome space $X$. Then error propagation in a random variable $f:X\rightarrow Y$ is just the probability measure induced on $Y$ by $f$ $$P_f(F)=P(f^{-1}(F)).$$ Nevertheless, for my students to make calculations, I wanted to relate back to the old notation. I have two options:


*

*I can define $x\pm\Delta x$ as the probability measure on $\mathbb{R}$ whose probability density function is a gaussian centered at $x$ with standard deviation $\Delta x$. If I take this option, I would like to prove that for all random variables $f:\mathbb{R}^n\rightarrow \mathbb{R}$, the induced measure is reasonably approximated by a Gaussian measure described by the physicists’ formula. Moreover, I find the approach very appealing because of the central limit theorem. Am I right in that that theorem can be interpreted as saying that the average of a bunch of probability measures (the process of averaging understood as a random variable) tends to induce a Gaussian probability measure?

*I can define the expected value of a measure and its variance. I introduce the notation $x\pm \Delta x$ as stating that it is just some measure with expected value $x$ and variance $\Delta x$. Can I justify the physicists’ formula?
I would really appreciate if the reader can pick its favorite choice and help me build arguments for the questions asked.
 A: I would wrap up the mathematical background of error propagation as follows:


*

*All basic error-propagation rules are based on the assumption of independent, normal distributions.
The assumption of normal distributions is often justified by the central limit theorem as a typical measurement error is the accumulation of several error sources.
The latter may be distributed in several ways, but when accumulated you get a normal distribution due to the central limit theorem.
 (Also see this answer of mine on Cross Validated).

*Under these assumptions, it’s easy to show that $Δ(λx) = λ·Δx$.

*It’s also easy to show that $Δ(x+y) = \sqrt{Δx^2 + Δy^2}$ (convolute the distributions).

*If you assume that the errors are small with respect to the second derivatives of $f$, you can linearise $f$ and apply points 2 and 3 to get the desired formula.
With this is mind, to your individual questions:

Moreover, I find the approach very appealing because of the central limit theorem. Am I right in that that theorem can be interpreted as saying that the average of a bunch of probability measures (the process of averaging understood as a random variable) tends to induce a Gaussian probability measure?

I don’t think that the central limit theorem is of relevance to this particular step (see above).
But then it is not very clear to me why you do.

I can define the expected value of a measure and its variance. I introduce the notation $x\pm \Delta x$ as stating that it is just some measure with expected value $x$ and variance $\Delta x$. Can I justify the physicists’ formula?

No, you additionally need to assume a normal distribution.
You can easily see this with a numerical experiment multiplying two exponentially distributed variables.
