Timeline for How do I calculate the variance-covariance matrix for a set of 2-D data points with errors: (x, y, dy)
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 30, 2016 at 14:21 | history | edited | Steven C. Howell | CC BY-SA 3.0 |
correctly describe how weight the fitting
|
| Dec 24, 2016 at 1:42 | comment | added | Steven C. Howell | @whuber I see your point but the error on the type of data i am considering is actually quite large. I am actually trying to fully propagate this large error, and the variance from a straight line, doing both of these in an accurate manner so I can make a case that the fit values are not as certain as people typically treat them. | |
| Dec 24, 2016 at 1:27 | history | edited | Steven C. Howell | CC BY-SA 3.0 |
made it clear this answer will not be accepted
|
| Dec 23, 2016 at 23:03 | comment | added | GeoMatt22 | You could perhaps introduce a "minimum uncertainty", $\sigma_0$, and then use $\sigma_i^2=dy_i^2+\sigma_0^2$ to compute the weights (a simple shrinkage estimate). | |
| Dec 23, 2016 at 22:48 | comment | added | whuber♦ | Although this appears to be a correct description of one way to carry out weighted regression (modulo the caveat by @GeoMatt), it doesn't seem likely to solve your problem. The issue is that the $dy$ are only part of the error variance: there is another component reflecting other random deviations of the data from the idealized linear function you are fitting. Weighting by $1/dy$ in this fashion can grossly inflate the influence of data with small $dy$. To see why, consider an extreme case where $y$ is measured perfectly precisely, so $dy=0$. Your solution will force the line through $(x,y)$! | |
| Dec 23, 2016 at 17:20 | comment | added | GeoMatt22 | The trick of "divide observations and predictors by error" (i.e. multiply by weight) works very generally as far as I know. For example the Wikipedia link by the OP shows the problem being solved is $\min_\beta\|W^{1/2}(y-X\beta)\|^2$, which shown the $\beta$ will be unchanged by multiplying the $W^{1/2}$ through. However note that in standard regression, not just $x$ but also "$1$" are predictors, i.e. $X=[x,1]$. I am not sure if your formulas account for the $1/dy_i$ "predictor". | |
| Dec 23, 2016 at 17:08 | history | answered | Steven C. Howell | CC BY-SA 3.0 |