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Error propagation problem

  1. Feb 25, 2016 #1
    1. The problem statement, all variables and given/known data
    Hello,

    I have the following operation that I want to perform:

    [tex] f=\frac{\bar{X}}{100-\sum \bar{Y}_j}*K[/tex]
    [itex] \bar{X} [/itex] and [itex] \bar{Y} [/itex] are averages with variances [itex] S_{X}^2 [/itex] and [itex] S_{Y_j}^2 [/itex] and [itex] K [/itex] is a constant.

    How will the error propagate?

    2. Relevant equations
    According to Wikipedia:
    (1) [itex] f=a\bar{A} \Rightarrow S_f^2=a^2S_f^2[/itex] where [itex] a [/itex] is a constant.
    (2) [itex] f=\bar{A}\bar{B} \Rightarrow S_f^2=S_A^2+S_B^2[/itex]
    (3) [itex] f=\frac{\bar{A}}{\bar{B}} \Rightarrow S_f^2=f^2\left(\frac{S_A^2}{A^2}+\frac{S_B^2}{B^2}\right) [/itex]

    3. The attempt at a solution
    So then the error of the nominator will be [itex]S_{X}^2[/itex]
    Only looking at the denominator i will have: [itex]100-\sum S_{Y_j}^2[/itex]
    Using the third and first equation will then yield:

    [tex] S_f^2=f^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{100-\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2[/tex]

    Where [itex] K^2 [/itex] comes from the first equation.

    I am a little bit confused though. Is this correct?

    Thanks.

    Edit: Covariance=0
     
  2. jcsd
  3. Feb 25, 2016 #2

    Ray Vickson

    User Avatar
    Science Advisor
    Homework Helper

    No, it is not correct: the squared error in ##100 - \sum Y_j## is not ##100 - \sum S_{Y_j}^2##. For one thing, the '100' is a constant that has no error; for another thing, the ##Y_i## squared errors should not be subtracted from anything.
     
  4. Feb 26, 2016 #3
    Thank you for your answer. Would this be correct?

    [tex] S_f^2=f^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2[/tex]

    Thanks.
     
  5. Feb 26, 2016 #4
    To answer my own question:

    [tex] S_f^2=\left(\frac{\bar{X}}{100-\sum \bar{Y}_j}\right)^2 \left(\frac{S_{X}^2}{\bar{X}^2} + \frac{\sum S_{Y_j}^2}{(100-\sum \bar{Y}_j)^2} \right)K^2[/tex]

    It should be clarified that K is an exakt number.
     
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