7.2 Linear Fits. The Straight Line

In the case of functions linear in their parameters a_j, i.e., there are no terms which are products or ratios of different a_j, (71) can be solved analytically. Let us illustrate this for the case of a straight line

(74)

where a and b are the parameters to be determined. Forming S, we find

(75)

Taking the partial derivatives with respect to a and b, we then have the equations

(76)

To simplify the notation, let us define the terms

(77)

Using these definitions, (76) becomes

(78)

This then leads to the solution,

(79)

Our work is not complete, however, as the errors on a and b must also be determined. Forming the inverse error matrix, we then have

where (80)

Inverting (80), we find

(81)

so that

(82)

To complete the process, now, it is necessary to also have an idea of the quality of the fit. Do the data, in fact, correspond to the function f(x) we have assumed? This can be tested by means of the chi-square. This is just the value of S at the minimum. Recalling Section 2.4, we saw that if the data correspond to the function and the deviations are Gaussian, S should be expected to follow a chi-square distribution with mean value equal to the degrees of freedom, . In the above problem, there are n independent data points from which m parameters are extracted. The degrees of freedom is thus = n - m. In the case of a linear fit, m = 2, so that = n - 2. We thus expect S to be close to = n - 2 if the fit is good. A quick and easy test is to form the reduced chi-square

(83)

which should be close to 1 for a good fit.

A more rigorous test is to look at the probability of obtaining a ² value greater than S, i.e., P(² S). This requires integrating the chi-square distribution or using cumulative distribution tables. In general, if P(² S) is greater than 5%, the fit can be accepted. Beyond this point, some questions must be asked.

An equally important point to consider is when S is very small. This implies that the points are not fluctuating enough. Barring falsified data, the most likely cause is an overestimation of the errors on the data points, if the reader will recall, the error bars represent a 1 deviation, so that about 1/3 of the data points should, in fact, be expected to fall outside the fit!

Example 6. Find the best straight line through the following measured points

x	0	1	2	3	4	5
y	0.92	4.15	9.78	14.46	17.26	21.90
	0.5	1.0	0.75	1.25	1.0	1.5

Applying (75) to (82), we find

a = 4.227 b = 0.878

(a) = 0.044 (b) = 0.203 and
cov(a, b) = - 0.0629.

To test the goodness-of-fit, we must look at the chi-square

² = 2.078

for 4 degrees of freedom. Forming the reduced chi-square, ² / 0.5, we can see already that his is a good fit. If we calculate the probability P(² > 2.07) for 4 degrees of freedom, we find P 97.5% which is well within acceptable limits.

Example 7. For certain nonlinear functions, a linearization may be affected so that the method of linear least squares becomes applicable. One case is the example of the exponential, (69), which we gave at the beginning of this section. Consider a decaying radioactive source whose activity is measured at intervals of 15 seconds. The total counts during each period are given below.

t [s]	1	15	30	45	60	75	90	105	120	135
N [cts]	106	80	98	75	74	73	49	38	37	22

What is the lifetime for this source?

The obvious procedure is to fit (69) to these data in order to determine . Equation (69), of course, is nonlinear, however it can be linearized by taking the logarithm of both sides. This then yields

Setting y = ln N, a = -1/ and b = ln N₀, we see that this is just a straight line, so that our linear least-squares procedure can be used. One point which we must be careful about, however, is the errors. The statistical errors on N, of course, are Poissonian, so that (N) = N. In the fit, however, it is the logarithm of N which is being used. The errors must therefore be transformed using the propagation of errors formula; we then have

Using (75) to (82) now, we find

a = - 1/ = - 0.008999 (a) = 0.001

b = ln N₀ = 4.721 (b) = 0.064.

The lifetime is thus

= 111 ± 12 s.

The chi-square for this fit is ² = 15.6 with 8 degrees of freedom. The reduced chi-square is thus 15.6/8 1.96, which is somewhat high. If we calculate the probability P(² > 15) 0.05, however, we find that the fit is just acceptable. The data and the best straight line are sketched in Fig. 7 on a semi-log plot.

While the above fit is acceptable, the relatively large chi-square should, nevertheless, prompt some questions. For example, in the treatment above, background counts were ignored. An improvement in our fit might therefore be obtained if we took this into account. If we assume a constant background, then the equation to fit would be

N(t) = N₀ exp(-t / ) + C.

Fig. 7. Fit to data of Example 7. Note that the error bars of about 1/3 of the points do not touch the fitted line. This is consistent with the Gaussian nature of the measurements. Since the region defined by the errors bars (± 1) comprises 68% of the Gaussian distribution (see Fig. 5), there is a 32% chance that a measurement will exceed these limits!

Another hypothesis could be that the source has more than one decay component in which case the function to fit would be a sum of exponentials. These forms unfortunately cannot be linearized as above and recourse must be made to nonlinear methods. In the special case described above, a non-iterative procedure [Refs: 2, 3, 4, 5, 6] exists which may also be helpful.