# Homework Help: Averaging measurement with stat +sys errors

1. Jun 18, 2015

### ChrisVer

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

All given above

3. 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 $x_2$...

Last edited: Jun 18, 2015
2. Jun 18, 2015

### Staff: Mentor

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).

3. Jun 18, 2015

### 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?

4. Jun 18, 2015

### 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*

5. Jun 18, 2015

### Staff: Mentor

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.