As we consider departures δa 1 then the pdf of δa 1

Info icon This preview shows pages 33–35. Sign up to view the full content.

View Full Document Right Arrow Icon
as we consider departures δa 1 , then the pdf of δa 1 has dispersion σ a 1 . Alternatively, we can say that in a large number of experiments, the pdf of δa 1 follows a chi-square pdf with one degree of freedom if we don’t care what happens to δa 2 . If, however, we are concerned about the pair, then we must look not at the projection down one axis or the other, but rather at the two-dimensional distribution. This is characterized by the tilted ellipses. Here, for a large number of experiments, the pair ( a 1 , a 2 ) follows a chi-square distribution with 2 degrees of freedom (if we don’t care about a 0 ; if we do, it’s 3 degrees of freedom and the ellipse becomes an ellipsoid, but this is very hard to plot!). For ν = 2, 68 . 3% of the points lie within Δ χ 2 = 2 . 3, where we have drawn the outer contour in Figure 9.1. The points inside this ellipse are darker; 68 . 3% of the points lie within that ellipse.
Image of page 33

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
– 34 – The best description of the specifics of calculating these ellipsoids is in BR § 11.5 (Confidence Intervals, Confidence Levels for Multiparameter Fits). To describe it, we’ll talk specifically about our numerical example, which has M = 4 measurements and N = 3 unknowns. The unknowns are a = [ a 0 , a 1 , a 2 ]. We’ll first begin by discussing the case of a single parameter; then we’ll generalize. 9.2. Calculating the uncertainties of a single parameter—gedankenexperiment First, suppose we want to know the value σ a 0 without regard to the values of a 1 and a 2 . Having already done the solution, we know the chi-square value of a 0 so we consider variations δa 0 around this best value. Pick a nonzero value of δa 0 and redo the least-squares solution for [ a 1 , a 2 ]; because of the covariance, these adopt values different from those when δa 0 = 0. This gives a new value for χ 2 which is, of course, larger than the minimum value that was obtained with δa 0 = 0. Call this difference Δ χ 2 δa 0 . Determine the dependence of Δ χ 2 δa 0 upon δa 0 and find the value of δa 0 such that Δ χ 2 δa 0 = 1. This is the desired result, namely the value σ a 0 without regard to the values of a 1 and a 2 . This value is σ 2 a 0 = [ α χ ] - 1 00 , the same result quoted in equation 8.10. Consider now what you’ve done in this process. For each least-squares fit you used a trial value of δa 0 . In specifying δa 0 you had exactly one degree of freedom because you are fixing one and only one parameter. Having done this, you could do a large number of experiments (or Monte Carlo trials) to determine the resultant distribution of Δ χ 2 δa 0 . It should be clear that this distribution follows a chi-square distribution with one degree of freedom ( ν = 1). So the uncertainty σ a 0 is that value for which Δ χ 2 δa 0 = 1. (The chi-square fit for the other two parameters has M 2 degrees of freedom, but this is irrelevant because—by hypothesis—you don’t care what happens to those variables.) 9.3. Calculating the uncertainties of two parameters—gedankenexperiment Suppose we want to know the value ( σ a 0 , σ a 2 ) without regard to the value of a 1 . Now we consider variations ( δa 0 , δa 2
Image of page 34
Image of page 35
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern