Nonlinear regression: What's the best estimator for this problem?

Context and goals:

Consider $$x$$ a scalar (deterministic) independent variable and $$y = f(x;\beta)+\eta$$ be a dependent random variable obtained trough $$f$$ with some parameters $$\beta$$ and $$\eta$$ is some noise modeled as a Gaussian with zero mean and unknown variance. Concretely, chose $$f(x;\beta) = \beta_1 + \frac{\beta_2}{1+e^{\beta_3x}}$$ Now, assume that we made $$M$$ experiments (with $$M$$ presumably low) and obtained $$M$$ pairs of measurements $$(x_1,y_1),\dots,(x_M,y_M)$$.

• Question: Given the data, How to obtain a way to estimate of $$y$$ for any $$x$$ as well as its "confidence"?

Let me explain my attempt to solve this problem:

My idea was to fit the parameters to the data using nonlinear regression by means of optimization (gradient descent) for example. Thus obtain an approximate model $$f(\bullet;\hat{\beta})$$. Then, for certain value of $$x$$ we can model that its corresponding $$y$$ is a Gaussian centered at $$f(\bullet;\hat{\beta})$$ (right?). However, since I don't know the variance of the noise $$\eta$$, the most we can do is to compute the sample variance of $$y_1,\dots,y_M$$ and use a student's-t distribution instead.

The problem is that, I'm not totally convinced that $$f(\bullet;\hat{\beta})$$ can be used as the mean of $$y$$. I guess one has to check if $$f(\bullet;\hat{\beta})$$ is an unbiased estimate of $$y$$. However, I don't know how to check that. Moreover, by using the sample variance I only get a number which represents the "confidence" for all $$y$$ which seems unsatisfactory to me since I would expect to be more confident of values of $$y$$ for $$x$$ near the samples $$x_1,\dots,x_M$$. But maybe my intuition is not correct.

Given the nonlinear regression model $$Y_{i} = f( x_{i}, \underline{\beta} ) + \eta_{i} \quad \textrm{where} \quad \eta_i \overset{\text{iid}}{\sim} \mathcal{N}(0, \sigma^{2})$$

and given the data $$(x_{i}, y_{i}), i = 1, ..., M$$ a standard approach to estimate the unknown parameters $$\underline{\beta} = (\beta_{1}, ..., \beta_{p})$$, is to apply the method of least squares by minimizing the sum of squared residuals $$\sum_{i=1}^{M} \left( y_{i} - f(x_{i}, \underline{\beta}) \right)^{2}$$ Since this problem is not linear in the unknowns $$\beta_{1}, ..., \beta_{p}$$, the estimation can not be derived explicitly as in linear regression, but we can solve it with the help of an iterative numeric procedure such as Gauss-Newton method, see: http://en.wikipedia.org/wiki/Gauss-Newton_method.

The least squares estimator $$\hat{\beta}$$ is asymptotically multivariate normally distributed with mean $$\beta$$ and therefore unbiased.

For a given $$x_{0}$$, the function $$f$$ can be approximated with: $$f(x_{0}, \underline{\hat{\beta}}) \approx f(x_{0}, \underline{\beta}) + \frac{\partial f(x_0, \underline{\beta}) }{ \partial \underline{\beta}} (\underline{\hat{\beta}} - \underline{\beta})$$ and the approximate $$(1 - \alpha)$$ confidence interval for the function value $$f(x_{0}, \underline{\beta})$$ is then $$f(x_{0}, \underline{\hat{\beta}}) \pm q_{1-\alpha/2}^{t_{m - p}} \cdot \textrm{se}( f(x_{0}, \underline{\hat{\beta}}) )$$

We can estimate the expected value $$E(Y_{0}) = f(x_{0}, \underline{\beta})$$ with $$f(x_{0}, \underline{\hat{\beta}})$$. And to get the interval where the observations will lie with high probability we also have to take into account the variance of the error $$\eta_{0}$$ $$f(x_{0}, \underline{\hat{\beta}}) \pm q_{1-\alpha/2}^{t_{m - p}} \cdot \sqrt{ \hat{\sigma}^{2} + \textrm{se}( f(x_{0}, \underline{\hat{\beta}}) )^{2} }$$ This is called prediction interval or forecast interval.

• This is potentially what I was looking for. Just wanted to know if you could provide a reference were I can look at this in more detail. Thanks sir Nov 16 '20 at 17:19
• Sure, great introductory notes are available here: stat.ethz.ch/~stahel/courses/cheming/nlreg10E.pdf They are based mostly on the book: Nonlinear Regression Analysis & Its Applications by Bates et. al. Nov 16 '20 at 17:46
• awesome reference, this is precisely what I was looking for. Thanks! Nov 16 '20 at 17:51