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

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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.
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– 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
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