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I'm trying to get the analysis of this (I hope) simple case correct in my head.

Lets say I measure the same thing two different ways, on the same set of data. Call these measurements $y_1$ and $y_2$. They have Gaussian errors $s_1$ and $s_2$. But since they are measured on common data, they have a correlation coefficient $\rho$. Say $\rho$ is measured with a toy monte carlo or similar, though I don't think that matters.

Just to make sure we are on the same page, this will generate a co-variance matrix that has $s_1^2$ and $\rho s_1s_2$ on the top row, and $\rho s_1 s_2$ and $s_2^2$ along the bottom row.

It is easy to calculate a weighted average, $ \bar y$. Under certain circumstances it is possible to find that $y$ is outside the range $y_1, y_2$. This happens, in particular, when $\rho >\frac {s_1}{s_2}$.

Since these measurements are on the same data, and are measuring the same thing, this is just not possible. In short, "opps." I must have messed up my measurements of $s_1, s_2$, or $\rho$. I feel comfortable with this. :-)

Now, lets say I do exactly as above, but now I take it to the next level and measure a number of systematic errors associated with each method. Say, for the sake of argument, that the systematic errors for method 1 are not correlated with any of the systematic errors for method 2.

After thinking about this for a while... well, I'm not entirely sure what to think. Do I have such a neat test to see if I've made a mistake anymore? I actually think I must, as you might imagine that I have systematic errors but they are vanishingly small. So, I can do something like decrease $\rho$ - because they now have uncorrelated error (the systematic error) and then re-run the test. This is pretty straight forward, I think: I use $\rho$ to split the statistical error into a correlated and uncorrelated part, add the uncorrelated parts in quad with the systematic errors, and then re-calculate $\rho$. But does the test still make sense? I fear I'm missing a fundamental concept here or I'd be able to determine what was going on.

Things in my current problem domain get arbitrarily complex. I have to combine measurements of the same thing in uncorrelated data sets, but with substantial correlated systematic errors. The procedure I describe just above for adjusting $\rho$ applies here just as well (the statistical error is uncorrelated, but some of the systematic errors will be), but here it seems incorrect. Another clue to my missing fundamental concept. :-)

I'd appreciate some help getting my thinking straight.

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  • $\begingroup$ Ranabir - thanks for editing and tex-ifying the post. For me, in my browser, this just shows up as $y_1$ - is there something I should be doing to get it properly formatted? $\endgroup$
    – Gordon
    Commented Aug 3, 2012 at 16:20
  • $\begingroup$ Seems to me from the context s1 and s2 are not the errors but rather the standard deviation of the errors. $\endgroup$ Commented Aug 3, 2012 at 21:44
  • $\begingroup$ The statistical errors are Gaussian, and they have widths $s_1$ and $s_2$. That was what I was trying to say. I can update the post. $\endgroup$
    – Gordon
    Commented Aug 3, 2012 at 22:16
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    $\begingroup$ Possible duplicate of Can an optimal weighted average ever have negative weights? $\endgroup$ Commented Nov 15, 2018 at 9:43
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    $\begingroup$ I voted to close this question because it is too difficult to distill a clear question out of it. Maybe you could reduce it to something of under 300 words and/or add some equations (Stephen Hawking was wrong that more equations make a book less interesting, "Someone told me that each equation I included in the book would halve the sales", it is the words that make a text not interesting) $\endgroup$ Commented Nov 19, 2018 at 15:14

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It turns out that the estimate lying outside the interval [y1,y2] actually makes sense. See the article: http://cds.cern.ch/record/183996/files/OUNP-88-05.pdf

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  • $\begingroup$ I like their explanation - since both y1 and y2 have an error, there is nothing that says they can't be both on the high side of the actual value. So perhaps the words I used in the question "wrong" are too strong there. So the basis here is I should not worry about this (which is the direction I took in this work anyway). Also. Good to see BLUE showing up. $\endgroup$
    – Gordon
    Commented Nov 22, 2018 at 21:30

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