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Averaging measurement with stat +sys errors

  1. Jun 18, 2015 #1

    ChrisVer

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    Gold Member

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

    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.

    2. Relevant equations

    All given above

    3. 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]...
     
    Last edited: Jun 18, 2015
  2. jcsd
  3. Jun 18, 2015 #2

    mfb

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    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).
     
  4. Jun 18, 2015 #3

    ChrisVer

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

    I will try to work out with [itex]w_2= \epsilon \ll 1[/itex] then... do you think [itex]w_1 = 1 - \epsilon[/itex] as well?
     
  5. Jun 18, 2015 #4

    ChrisVer

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

    *edit and just realized that the determinant is zero and wolfram was giving me a pseudoinverse matrix*
     
  6. Jun 18, 2015 #5

    mfb

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    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.
     
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