# How to perform nonlinear regression with regressors affected by gaussian error?

I am trying to calibrate a sensor and I have a data set consisting of several observations of a 3-dimensional vector $X_i$, with

$X_i=w_i + \epsilon_i$

where $w_i$ is the value that the sensor should measure (if uncalibrated but unaffected by noise) which is fixed for each $i$ and unobservable, and $\epsilon_i$ is gaussian error vector with zero mean and a known diagonal covariance matrix, which is the same for all $i$. The error vectors $\epsilon_i$ are uncorrelated for different $i$.

The response variable is

$Y_i=\|C(w_i-B)\|=1$ for all $i$.

$C$ is a 3x3 matrix that can be eigendecomposed by $C=R(\psi,\theta,\phi)\,E\,R(\psi,\theta,\phi)^T$, where $R$ is a rotation matrix parametrized by the three angles $\psi,\theta,\phi$. $E$ is a diagonal matrix where each entry in the diagonal is positive and bounded from above by a known value. $B$ is a 3x1 vector.

This is a typical nonlinear regression with additive measurement error, but without additional additive error in $Y$ (as opposed to the traditional setting of this problem that I have seen so far). Is it possible to obtain an unbiased estimate of the model parameters $p$ (the three attitude angles that parametrize $R$, the three diagonal entries of $E$ and the vector $B$) from the observations $X_i$?

Alternatively, is there a way to find out if solving the nonlinear least squares problem $\hat{p}= \min_p \sum_{i=1}^n (\|C(X_i-B)\|-1)^2$ yields an unbiased estimate of $p$?

• Would it be enough for your application to get an unbiased estimator of the entries of $C$, rather than of its eigendecomposition? The nonlinearity in the mapping from $C$ to its eigendecomposition seems like it will present difficulties for you. – Ian Jun 25 '15 at 14:22
• It would be fine to get an estimate of the entries of $C$ as long as we can force $C$ to be "decomposable" as $R\,E\,R^T$. I guess this would mean $C$ would have to be a real, symmetric matrix. By the way, the only reason I parametrized $C$ in such a way was to be able to easily specify upper and lower bounds on each of the parameters, so that I could run nonlinear least squares. – Pedro Caldeira Jun 25 '15 at 14:25
• Yes, an orthogonally diagonalizable matrix over $\mathbb{R}$ is necessarily symmetric. As expected from before that means that $C$ is specified by six numbers (the diagonal and the superdiagonals) just as it was before (the eigenvalues and the rotational angles). – Ian Jun 25 '15 at 14:26
• Unfortunately, solving the nonlinear least squares problem will generally not give an unbiased estimator for $C$ and $B$. My (limited) understanding is actually that unbiased estimators are "holy grails" that we usually can't get except in very nice situations. – Ian Jun 25 '15 at 14:31