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Error Propagation

  1. Nov 20, 2012 #1
    I am confused about calculating errors. I have learned if you take the variance covariance matrix [itex]\Sigma_{ij}[/itex] of a fit of function f(x,p) to data for parameters [itex]p_i[/itex] (for example by using Levenberg-Marquart) that the one sigma error interval for [itex]p_i[/itex] is [tex]\sigma_{p_i}=\sqrt{\Sigma_{ii}}[/tex] I only understand this, if there are no covariance terms. Why do we do this? I would have thought a better way to find the error would be to do diagonalize [itex]\Sigma[/itex], say the diagonal form is [itex]\Xi[/itex] with normalized eigenvectors [itex](\vec{v})_k[/itex]. Then we would have independent variables that have a Gaussian distribution and one can calculate the error on [itex]p_i[/itex] using error propagation, i.e. [tex]\sigma_{p_i} = \sqrt{\sum \Xi_{kk}\left\langle(\vec{v})_k\mid l_i \right\rangle}[/tex] where [itex]\left\langle(\vec{v})_k\mid l_i \right\rangle[/itex] is the [itex]i^\text{th}[/itex] component of [itex](\vec{v})_k[/itex]. If this is permissible, is there a name for it?
     
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  3. Nov 20, 2012 #2

    Stephen Tashi

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    It is rather confusing how any process can purport to calculate a standard deviation for the paramters of a fit y = f(x,p) in the case when the data is of the form [itex] (x_i,y_i) [/itex]. There is no random sample of the parameters. How can any variation be assigned to them? My best guess is in post #7 of the thread: https://www.physicsforums.com/showthread.php?t=645291&highlight=parameters

    I'm not sure what you mean by "the variance covariance matrix [itex] \Sigma_{i,j} [/itex] of a fit of the function f(x,p) to the data for parameters [itex] p_i [/itex]". What is the definition of that matrix?
     
  4. Nov 21, 2012 #3
    Well I guess that you know the theory better than I do, but the idea is somehow a correspondence between least squares and maximum likelihood.
    So you have the sum of the squares of a fit function [itex]f(x,p_1,p_2,\dots)[/itex] to data [itex]x_i,y_i[/itex]

    [tex]
    sq(p_1,p_2,\dots) = \sum_i (f(x_i,p_1,p_2,\dots)-y_i)^2
    [/tex]

    And the residuals
    [tex]r_i=f(x_i,p_1,p_2,\dots)-y_i[/tex]

    for some optimal set of parameters [itex]p_k[/itex] that minimizes sq. If the residuals are gaussian then the variance of the residuals times the reciprocal of the Hessian of [itex]sq(p_1,p_2,\dots)[/itex] is somehow a measure of how confident one can be in the fitted parameters and it is also a variance-covariance matrix. This is how I understand it, but if I would really understand the theory I wouldn't be asking questions. Anyhow my question was why one only uses the diagonal elements of that matrix.
     
  5. Nov 25, 2012 #4

    Stephen Tashi

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    Can your original question can be considered outside of the context of curve-fitting.

    Suppose the [itex] p_i [/itex] are simply a set of random variables, not necessarily having the meaning of parameters in a curve fit. If the covariance matrix is [itex] \Sigma [/itex], are you proposing a method to get a different estimate for each [itex] \sigma^2_{p_i} [/itex] than using the diagonal element [itex] \Sigma_{i,i} [/itex] ?
     
  6. Nov 26, 2012 #5
    Exactly. Maybe the thing I am looking for already has a name. If we have a covariance matrix like this

    [tex]\Sigma = \left( \begin{matrix} .1&100\\ 100&1000 \end{matrix} \right)[/tex]

    The first parameter is varying very little while the second one is varying a lot. But the second parameter also has a large influence on the first parameter, and it seems to me that this does not get captured if we use .1 as the variance for the first parameter. So I was suggesting to diagonalize the matrix to get independent parameters and then something like error propagation to determine the "real" uncertainty of the first parameter. I tried to make an example but I don't know how to make random numbers with a given covariance matrix.
     
  7. Nov 26, 2012 #6

    Stephen Tashi

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    You could use a bivariate normal distribution and try to get the desired covariance matrix.

    if you don't want to use the variance of a random variable to define its uncertainty, you'll have to state what definition for uncertainty that you want to use.

    The variance of one random variable in a joint distribution, doesn't define a joint confidence interval for several variables. Perhaps you are trying to find a joint confidence interval.
     
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