Weighted average that takes error into account?

My data consists of values with an error (from non-linear curve fitting), for example:

value#1=2.3322e-10 error#1=8.7707e-13

value#2=2.3257e-10 error#2=1.2317e-12

What would be the most appropriate way to produce an average of the values that would take the error of each value into account?

For example, value#2 has a larger error so it would have less "weight" in the total average.

Thanks.

[Walter Roberson]

Can the error ever be 0? If so then are all of the other entries to be ignored, or would they contribute anyhow? Can the error ever be negative? If one entry has error 1E-13 and another has error 1E-12 then how much relative weight should the two have ? Is the weight to be derived somehow from the overall errors,or should it be absolute ?

[guyklx]

Hi Walter, no the error cannot be zero just a very small number.

[Walter Roberson]

If one entry has error 1E-13 and another has error 1E-12 then how much relative weight should the two have ? Is the weight to be derived somehow from the overall errors,or should it be absolute ?

[guyklx]

"If one entry has error 1E-13 and another has error 1E-12 then how much relative weight should the two have ?" <= that is exactly what I'm asking.

No, the weight should not be derived from the overall errors. Every error is value specific.

[Walter Roberson]

Then it is really up to you. There are circumstances where the weighting should be inverse linear. There are circumstances where it should be . There are circumstances where the relationship should be exponential.

[guyklx]

Thanks Walter.

Answer 1

[Star Strider]

It depends on what you mean by ‘error’. In regressions, data are commonly weighted by the inverse of the variance, so the larger the variance, the smaller the weight.

[guyklx]

Hi,

The errors are derived from non-linear curve fitting.

[Star Strider]

Is it appropriate to assume the ‘errors’ are residuals?

Inverse-variance weighting characteristically assumes the variances are from a group of dependent observations at a common independent variable under essentially identical conditions.

I would assume that weighting individual observbations by the inverse of their residuals would not result in any significant improvement in the estimated parameters.

[guyklx]

Thanks.

[Star Strider]

My pleasure.