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ChrisVer
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Homework Statement
You make a measurement of two variables with 100% correlated systematic uncertainty:
[itex] x_1 \pm \Delta x_1^{stat} \pm \Delta x_1^{sys} = 1.0 \pm 0.1 \pm 0.1 [/itex]
[itex]x_2 \pm \Delta x_2^{stat} \pm \Delta x_2^{sys} = 1.2 \pm 0.1 \pm 0.2 [/itex]
The average is taken by:
[itex] \bar{x} = \sum_{i=1}^2 w_i x_i[/itex]
where [itex]w_i = \frac{\sum_j (C^{-1})_{ij}}{ \sum_{kj} (C^{-1})_{kj}}[/itex] and [itex]C=C^{stat}+ C^{sys}[/itex] the covariance matrix of the measurement.
Homework Equations
All given above
The Attempt at a Solution
I calculate [itex]C[/itex] to get its inverse and find the weights.
For that I deduced that:
[itex]C^{stat} = \begin{pmatrix} (\sigma^{stat}_1)^2 & 0 \\ 0 & (\sigma_2^{stat})^2 \end{pmatrix}[/itex]
and
[itex]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}[/itex]
due to the 100% correlated systematic uncertainties [itex]\sigma_{12}^{sys} = \rho \sigma_1^{sys} \sigma_2^{sys}= \sigma_1^{sys} \sigma_2^{sys}[/itex].
When I go to get [itex]C[/itex] then:
[itex]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} [/itex]
The inverse of this matrix is [itex]C^{-1} = \frac{50}{3} \begin{pmatrix} 5 & -2 \\ -2 & 2 \end{pmatrix} [/itex].
My problem is that with such a matrix I am getting for the weights:
[itex]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[/itex]
And
[itex] w_2 = 0[/itex] (since [itex]C_{21}^{-1}= - C_{22}^{-1}[/itex]).
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 [itex]x_2[/itex]...
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