Formal Error Propagation for the FWHM Suppose I have some pulse on an oscilloscope. I measure the amplitude in the vertical dimension of this pulse as $A \pm \delta A$ V.  I then want to measure the full-width-half-maximum. So I set a horizontal cursor at $A/2$, then two vertical cursors at the intersection of this horizontal cursor and my pulse, and make two measurements in the "horizontal dimension." Let us call these $a \pm \delta a$ and $b \pm \delta b,$ with $b>a.$
The FWHM is then equal to $b-a,$ so the uncertainty would be $\sqrt{ (\delta a)^2 + (\delta b)^2}.$ However, strictly speaking, my estimates of $a$ and $b$ depended on $A$, so I cannot just take $\delta a$ and $\delta b$ to be the error associated with my difficulties in using the oscilloscope precisely at those two points. In practice, of course, this does not really matter, and is probably not worth worrying about too much. But I am curious.
Let the pulse be the function $f(x),$ with $x$ being the horizontal dimension. Then $a=\text{min} \\{f^{-1}(A/2) \\},$ and $b= \text{max} \\{f^{-1}(A/2) \\}$ where $f^{-1}$ is the inverse image of $f$. Therefore,
$$\delta a = \left(\frac{\partial f}{\partial x}_a \right)^{-1} \cdot \delta A$$
and likewise for $b.$ This make intuitive sense. The steepness of the pulse affects how well you can estimate the FWHM. But I don't know $f(x)$ exactly at all, so I would have to make another estimate for this (at the two points in question). To do this, I would need to take two points close to my best guess for where $a$ is. But this seems circular, because my estimate of the derivative would again in turn rely on my original estimate for $A.$
I think the more accurate approach would be to measure the following four points:

*

*A position for what I have called $a$ assuming the amplitude is $A+ \delta A.$ I will call this $a_1.$

*A position for what I have called $a$ assuming the amplitude is $A - \delta A.$ I will call this $a_2.$

*A position for what I have called $b$ assuming the amplitude is $A+ \delta A.$ I will call this $b_1.$

*A position for what I have called $b$ assuming the amplitude is $A - \delta A.$ I will call this $b_2.$
Then I would take $\delta a_i=a_1 - a_2.$ However, there is some error associated $a_1$ and $a_2$ in not being able to use the oscilloscope perfectly precisely. It seems like the best thing to do would be to add in quadrature, so finally I am left up with $$\delta a_f = \sqrt{ \delta a_i ^2 + \delta a_1^2 + \delta a_2^2}.$$
Is this "rigorous?" (A term which I am not sure is even relevant to begin with here).
 A: As you said, you're asking a question that begs an answer with a level of precision that is well beyond what's needed for reading a value off an oscilloscope. Having said that, as a fun exercise, let's overanalyze this situation.
To formalize an experiment, we need probabilistic model for the measurement process. This boils down to specifying a likelihood function, which tells us the probability of observing the data, given some parameters.
The simplest model for a single oscilloscope reading is that there is some true value of the voltage, $V_{\rm true}$, and a probability distribution $p(V_{\rm obs}|V_{\rm true})$ which will tell us how likely we are to observe a voltage $V_{\rm obs}$ from the oscilloscope given that the true value is $V_{\rm true}$. In general, $V_{\rm obs}$ will be different because of effects like some of the components not being properly calibrated or fuzziness in the line drawn on the screen -- you know, experimental noise. The simplest and most common model of the data generation process is that it is Gaussian
\begin{equation}
p(V_{\rm obs}|V_{\rm true}) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp\left(-\frac{\left(V_{\rm obs}-V_{\rm true}\right)^2}{2\sigma^2}\right)
\end{equation}
where $\sigma$ is a new parameter that describes the uncertainty of the experiment.
The standard propagation of uncertainty formulas come when we combine measurements, assuming they are independent. In your example, we could have $V_{\rm obs}^{(a)}$ and $V_{\rm obs}^{(b)}$, corresponding to the observed values of the left and right peaks. Assuming that those measurements are independent and Gaussian (we'll get back to your question about non-independence in a minute), then we can model the process of observing the width, or difference in those values, $\Delta V = V^{(b)} - V^{(a)}$, with the likelihood function
\begin{eqnarray}
p(\Delta V_{\rm obs}|\Delta V_{\rm true}) = \frac{1}{\sqrt{2\pi \Sigma^2}} \exp\left(-\frac{\left(V_{\rm obs}-\Delta V_{\rm true}\right)^2}{2\Sigma^2}\right)
\end{eqnarray}
where
\begin{eqnarray}
\Delta V_{\rm true} &=& V^{(b)}_{\rm true} - V^{(a)}_{\rm true} \\
\Sigma &=& \sqrt{\sigma_a^2 + \sigma_b^2}
\end{eqnarray}
A careful derivation can be found on wikipedia. Also, as an aside, the more general error propagation rules you have learned all boil down to assuming Gaussian distribution for the residuals, linearizing the dependence of the likelihood on the "true" parameter, and combining the Gaussians. Again, see wikipedia if you are interested in the details.
What you are getting at in your question is that assuming that measurements of $V^{(a)}$ and $V^{(b)}$ are independent is problematic. A more satisfying likelihood would have a form like
\begin{eqnarray}
p(\Delta V_{\rm obs} | V^{(a)}, V^{(b)}, A) = p(\Delta V_{\rm obs} | V^{(a)}, V^{(b)}) p(V^{(a)}, V^{(b)} | A)
\end{eqnarray}
where you could specify some model for the likelihood of observing $V^{(a)}$ and $V^{(b)}$ given $A$ in the function $p(V^{(a)}, V^{(b)} | A)$, and where $p(\Delta V_{\rm obs} | V^{(a)}, V^{(b)})$ could be given by (for example) the Gaussian form defined above.
There are various techniques for handling statistical errors when the likelihood becomes more complicated. One way would be to explicitly assume a form for $p(\Delta V_{\rm obs} | V^{(a)}, V^{(b)})$ that can be handled analytically. If you assume everything is Gaussian, you could combine the factors above and arrive at something like an error propagation formula.
A more general approach is to use Bayesian statistics. There, you would specify a prior distribution for the parameters $V^{(a)}$, $V^{(b)}$ and $A$; call this function $\pi(V^{(a)}, V^{(b)}, A)$. There's no objectively correct choice of prior, which makes some people uncomfortable with Bayesian, but there are standard "uninformative" choices you can make such as assuming that any value for $V^{(a)}$, $V^{(b)}$, and $A$ in some range is equally likely, before you look at the data.  Then the posterior probability that the parameters take on a certain value is given by the product
\begin{equation}
p(V^{(a)}, V^{(b)}, A|\Delta V_{\rm obs}) = p(\Delta V_{\rm obs} | V^{(a)}, V^{(b)}, A) \pi(V^{(a)}, V^{(b)}, A)
\end{equation}
Your data analysis would then amount to calculating various properties of this distribution, such as its mean value and standard deviation. In a simple case like this with only a few parameters, you could probably evaluate the posterior on a grid of points on a computer and get all the information you want that way. In more complicated, research-level data analysis, you would need to use a sampling algorithm like Markov Chain Monte Carlo (MCMC) or nested sampling. Applying an MCMC to this problem would be extraordinary overkill, of course :-).
