Averaging measurement with stat +sys errors

[ChrisVer]

Homework Statement

You make a measurement of two variables with 100% correlated systematic uncertainty:
x_1 \pm \Delta x_1^{stat} \pm \Delta x_1^{sys} = 1.0 \pm 0.1 \pm 0.1
x_2 \pm \Delta x_2^{stat} \pm \Delta x_2^{sys} = 1.2 \pm 0.1 \pm 0.2

The average is taken by:

\bar{x} = \sum_{i=1}^2 w_i x_i

where w_i = \frac{\sum_j (C^{-1})_{ij}}{ \sum_{kj} (C^{-1})_{kj}} and C=C^{stat}+ C^{sys} the covariance matrix of the measurement.

Homework Equations

All given above

The Attempt at a Solution

I calculate C to get its inverse and find the weights.
For that I deduced that:
C^{stat} = \begin{pmatrix} (\sigma^{stat}_1)^2 & 0 \\ 0 & (\sigma_2^{stat})^2 \end{pmatrix}
and
C^{sys} =\begin{pmatrix} (\sigma^{sys}_1)^2 & \sigma^{sys}_1 \sigma^{sys}_2 \\ \sigma^{sys}_1 \sigma^{sys}_2 & (\sigma_2^{sys})^2 \end{pmatrix}
due to the 100% correlated systematic uncertainties \sigma_{12}^{sys} = \rho \sigma_1^{sys} \sigma_2^{sys}= \sigma_1^{sys} \sigma_2^{sys}.

When I go to get C then:

C=C^{stat} +C^{sys}= \begin{pmatrix} 0.01 & 0 \\ 0 & 0.01 \end{pmatrix} +\begin{pmatrix} 0.01 & 0.02 \\ 0.02 & 0.04 \end{pmatrix} =\frac{1}{100} \begin{pmatrix}2 & 2 \\ 2 & 5 \end{pmatrix}

The inverse of this matrix is C^{-1} = \frac{50}{3} \begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix}.

My problem is that with such a matrix I am getting for the weights:
w_1 =\frac{\sum_j (C^{-1})_{1j}}{ \sum_{kj} (C^{-1})_{kj}}= \dfrac{\frac{50}{3} (5-2)}{ \frac{50}{3}(5+2-2-2)}= 1

And
w_2 = 0 (since C_{21}^{-1}= - C_{22}^{-1}).

I don't know why this is happening... Any idea?
Obviously this doesn't seem to make sense because in the averaging I won't get any contribution from x_2...

 

[mfb]

Including the second measurement blows up the systematic error without reducing the statistical error much. To check this, you can give the second measurement the weight $w_2 = \epsilon \ll 1$ and see what the combined uncertainty is (compared to w₂=0).
I can imagine that not averaging at all is the best you can do in this special case where the systematics are weird (100% correlated, but much larger in the second case).

 

[ChrisVer]

The thing is that this makes it a bit more strange... Because I tried before doing the same for x_1= 0.1 \pm 0.0 \pm 0.1 and x_2= 1.0 \pm 0.0 \pm 0.2 (no statistical error). The covariance matrix was:
C= \frac{1}{100} \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix} \Rightarrow C^{-1} = \begin{pmatrix} 4 & 8 \\ 8 & 16 \end{pmatrix}
And the weigths were found to be w_1= \frac{1}{3} and w_2= \frac{2}{3} which make sense...

I will try to work out with w_2= \epsilon \ll 1 then... do you think w_1 = 1 - \epsilon as well?

 

[ChrisVer]

Also that's a weird inverse, since [C^{-1} C ]_{11}= \frac{1}{100} (4+16) \ne 1...

*edit and just realized that the determinant is zero and wolfram was giving me a pseudoinverse matrix*

 

[mfb]

Also that's a weird inverse, since [C^{-1} C ]_{11}= \frac{1}{100} (4+16) \ne 1...

*edit and just realized that the determinant is zero and wolfram was giving me a pseudoinverse matrix*

Ah, that could be the problem.

Without statistical errors the weights should certainly be 1 and 0, as using the value with the larger (but 100% correlated) systematics is pointless.

$1-\epsilon$ for the other weight, sure.