Confidence interval for a constrained fit to Gaussian-like data

I'm working with data from an instrument which is expected a priori to produce Gaussian (normally) distributed data:

$$G = A\exp\left(-\dfrac{(x - \mu)^2}{\sigma} \right)$$

The data are normally sparse, with only about 2-3 measurements representing each $G$. In this question, I'm focussing on a single $G$, but in reality, we often have analyte signals causing overlapping $G$ that are then fitted simultaneously as described below.

To fit $G$ to our measurements, we calibrate for $\mu$ and $\sigma$ a priori using reference signals, then use these calibrations to constrain all parameters except $A$. So the fit reduces to

$$G = AG_0$$

which is then fitted by least-squares minimization to determine $A$.

My main question is: what is the uncertainty ($\sigma_A$) in the fitted $A$?

My initial approach was to estimate $\sigma_A$ as the RMSE of the fit. But since the RMSE is essentially the standard deviation of the residuals, this seems like an overestimate: I want the confidence interval of the fitted parameter.

Can I safely use the following textbook equation for the confidence interval of a linear-regression slope in this context? (the $x_i$ are the predictor variables, $\bar{x}$ is the mean of the predictor variables, the $e_i$ are fit residuals, and 2 degrees of freedom were consumed by the 1 fitted variable and the fact that the residuals sum to zero)

$$\sigma_A = \dfrac{s_e}{\Sigma _i (x_i - \bar{x})^2 } = \dfrac{\frac{1}{n- 2}\Sigma_i e_i^2}{\Sigma _i (x_i - \bar{x})^2 }$$

I think "yes": all I have done is transform my $x$ before fitting a linear parameter. I also think "no" because I'm not sure of the meaning of $x_i - \bar{x}$ in this equation -- is it specific to linear regression?

To give you a visualization, my data are almost as bad as these simulated data (but often the signal ontains one or two data points more):

Note: I am aware that Bayesian analysis would be a better method for passing information about $\mu$ and $\sigma$ to my fit, but I am not at liberty to change the analysis software right now. I need to limit myself to an estimate of $\sigma_A$.

edit: Another note that might exclude some solutions: I am analyzing thousands of measurements in bulk with no known true value.

Jeff,
1) The steps you described correspond to fitting the model: $$G_i = AG_0(x_i) + \epsilon_i$$
with {epsilon_i} iid, normally distributed, with mean 0 and variance sigma, which is a standard linear model (linear in the way the observations are assumed to depend on the un-known parameter A), without intercept.
The least-square estimator is $$\hat A = \dfrac{\Sigma _i G_i G_0(x_i)}{\Sigma _i G_0(x_i)^2 }$$
whose variance can be estimated by
$$se^2(\hat A) = \dfrac{\frac{1}{n- 1}\Sigma_i e_i^2}{\Sigma _i G_0(x_i)^2 }.$$
There would be an \bar{x} if the model had an intercept.

2) It seems, from what you say, that to reflect the uncertainty in A, a model of the form:
$$ln(G_i) = a + ln(G_0(x_i)) + \epsilon_i$$
would be better. One could fit the constant a, taking the ln(G_0(x_i)) term as a fixed offset, and translate the uncertainty in the estimator of a into uncertainty in A.

• Thank you! So (if I understand correctly) it was true that any transformation of the $x$ variable, followed by a linear fit, can be afterwards analyzed by standard least-squares approach. However, I don't understand why you suggest switching from my physically-based Gaussian approach, to your log-transformation approach? Why have you chosen a logarithmic function? To represent a more general form of the exponential Gaussian function? Aug 6, 2014 at 8:34
• The log transformation is just a suggestion. Ordinary least squares in the original scale would assume that {epsilon_i} have the same variance, but G0 and G_i decrease very fast; the assumption that errors for observations in the tails are of the same size as around \mu would not hold. You can still work in the scale of the G's; the estimator would be the BLUE estimator; use of the least-squares machinery (p-value computations, confidence intervals, etc) would not be justified since a key assumption underlying those results (equal variance of the remainders) would not be true. Aug 14, 2014 at 18:28
• Note that the log transformation also effectively changes the weighting of the data points in the least-squares minimization. Equal weights in linear scale correspond to equal absolute uncertainty, while equal weights in log scale correspond to equal relative uncertainty in linear scale. Oct 18, 2017 at 17:24