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I have a ball on a table located in position $x', y'$.

I am using many different rulers to measure the coordinates $x_i, y_i$ of the ball. I do this with $N$ different rulers, so $i = 1\ldots N$. Each measurement comes with an uncertainty $\epsilon_{x,i}, \epsilon_{y,i}$ drawn with mean 0 and uncorrelated but known variances $\sigma_{x,i}^2, \sigma_{y,i}^2$.

The ball hasn't changed in its location, but I have many noisy measurements for where it is located. How can I combine all of these measurements to give my best guess as to the true coordinates $x',y'$?

rhombidodecahedron
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    Why not simply take the average for x and for y and use that as the coordinate estimates. Don't you mean the noise terms are normal with mean 0 and the two known variances? – Michael R. Chernick Apr 24 '17 at 01:35
  • @rhomb When you say that the uncertainties are normal, do you mean they are jointly normal? – Glen_b Apr 24 '17 at 01:56
  • @Glen_b Yes I suppose – rhombidodecahedron Apr 24 '17 at 03:16
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    @MichaelChernick Just the mean? What if $\sigma_{x,1} \gg \sigma_{x,2}$? The variance of each measurement tool is known - can't I use that information to get a better estimate? – rhombidodecahedron Apr 24 '17 at 03:18
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    @MichalChernick "Don't you mean the noise terms are normal with mean 0 and the two known variances?" That is what I mean - Isn't that what I wrote? If not, can you please inform me how I should write it? – rhombidodecahedron Apr 24 '17 at 03:19
  • Yes that's what you wrote, that part's already clear. – Glen_b Apr 24 '17 at 03:20
  • It is not a big deal. I was just saying that the measurements are different from the noise. – Michael R. Chernick Apr 24 '17 at 03:27
  • @Glen_b I guess the normal distribution part isn't important, I edited that out – rhombidodecahedron Apr 24 '17 at 05:03
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    Can you say what is critically different than your last question on this? https://stats.stackexchange.com/questions/202931/combining-measurements-with-known-uncertainties?rq=1 Once again, the MLE is probably appropriate, though I think it requires the assumption of normal errors. – AJK Apr 24 '17 at 05:35
  • Do you mean you are given the values and uncertainties for each measurement, i.e. all the $x_i$ and $\epsilon_{x,i}$? If that is the case, this is a very standard experimental physics problem and I can give you an immediate answer. Or do you mean you are given only the $x_i$ and the fact that the unknown $\epsilon_{x,i}$ are drawn from $N(0, \sigma_x)$ for which you know $\sigma$. In which case, why do you put indexes on $\sigma_x$? – David Wright Apr 24 '17 at 05:42
  • @DavidWright we are given the measurements $x_i$ and the stated $1\sigma$ uncertainty on the measurement as well. If the exact $\epsilon_{x,i}$ were actually known, I would simply correct $x_i$ to obtain the exact value. I put indexes on each $\sigma_x$ because each ruler has a different $\sigma$. – rhombidodecahedron Apr 24 '17 at 14:41
  • @rhombidodecahedron: Now I understand. This is in fact the simple, well-known case. Jeremy's answer gave the formula for the combined best value, and I have added an answer with the formula for the combined uncertainty. – David Wright Apr 24 '17 at 17:35
  • See Page 10 of https://www.nature.com/articles/s41598-018-28130-5.pdf This might hold the key to your question. – Deep Mukherjee Jun 23 '19 at 21:14

2 Answers2

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If I'm understanding your question properly, this sounds like you need Inverse variance weighting.

https://en.wikipedia.org/wiki/Inverse-variance_weighting

The estimate of your $x'$ that would minimize the variance (so giving you the "best guess") will be given by

\begin{equation} \hat{x} = \frac{\Sigma_ix_i/\sigma^2_{x,i}}{\Sigma_i1/\sigma^2_{x,i}} \end{equation}

You stated that the uncertainties in your measurements were "iid". If they have different variances, then they are not identical, just independent.

For Inverse Variance Weighting to work, they only need to be independent.

Jeremy Voisey
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  • Sorry you're right about the iid, I will edit that part out – rhombidodecahedron Apr 24 '17 at 14:41
  • Thanks. I think that might have been the source of some confusion :-) – Jeremy Voisey Apr 24 '17 at 16:08
  • I tried this out and compared the standard deviation that I get from this method vs. doing some kind of a simulation where I do 1000 random perturbations of all the x's with a normal distribution having variance $\sigma^2$. I find a complete mismatch between them: the inverse-variance weighted standard deviation is orders of magnitude smaller than the "simulation." Can you comment on this please? – rhombidodecahedron Apr 24 '17 at 17:24
  • I don't quite follow your method sorry. Can you give an example? – Jeremy Voisey Apr 24 '17 at 17:48
  • The inverse-variance weighted standard deviation should be smaller than any individual measure's standard deviation. That's the point of it. – Jeremy Voisey Apr 24 '17 at 17:59
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And the inverse square of the error on the combined value is the sum of the inverse squares of the individual errors:

$$ \frac{1}{\sigma^2} = \sum_i \frac{1}{\sigma_{x,i}^2}$$

For a derivation, see the section on statistical methods of any experimental physics handbook.

(The fact that you have each measurement has an x and y value doesn't add any complexity; only the x values contribute to the combined x value and only the y values contribute to the combined y value.)

David Wright
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