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6. MAXIMUM-LIKELIHOOD ERROR, ONE PARAMETER

It can be shown that for large N, curlyL(alpha) approaches a Gaussian distribution. To this approximation (actually the above example is always Gaussian in alpha), we have

Equation

where 1 / sqrth is the rms spread of alpha about alpha*,

Equation

Since Deltaalpha as defined in Eq. (3) is 1 / sqrth , we have

Equation 7     (7)

It is also proven in Cramer [4] that no method of estimation can give an error smaller than that of Eq. 7 (or its alternate form Eq. 8). Eq. 7 is indeed very powerful and important. It should be at the fingertips of all physicists. Let us now apply this formula to determine the error associated with alpha* in Eq. 6. We differentiate Eq. 5 with respect to alpha. The answer is

Equation

Using this in Eq. 7 gives

Equation

This formula is commonly known as the law of combination of errors and refers to repeated measurements of the same quantity which are Gaussian-distributed with "errors" sigmai.

In many actual problems, neither alpha* nor Deltaalpha may be found analytically. In such cases the curve curlyL(alpha) can be found numerically by trying several values of alpha and using Eq. (2) to get the corresponding values of curlyL(alpha). The complete function is then obtained by drawing a smooth curve through the points. If curlyL(alpha) is Gaussian-like, ð2w / ðalpha2 is the same everywhere. If not, it is best to use the average

Equation

A plausibility argument for using the above average goes as follows: If the tails of curlyL(alpha) drop off more slowly than Gaussian tails, img 24 is smaller than

Equation

Thus, use of the average second derivative gives the required larger error.

Note that use of Eq. 7 for Deltaalpha depends on having a particular experimental result before the error can be determined. However, it is often important in the design of experiments to be able to estimate in advance how many data will be needed in order to obtain a given accuracy. We shall now develop an alternate formula for the maximum-likelihood error, which depends only on knowledge of f (alpha; x). Under these circumstances we wish to determine img 24 averaged over many repeated experiments consisting of N events each. For one event we have

Equation

for N events

Equation

This can be put in the form of a first derivative as follows:

Equation

The last integral vanishes if one integrates before the differentiation because

Equation

Thus

Equation

and Eq. (7) leads to

Equation 8     (8)

Example 1

Assume in the µ-e decay distribution function, f (alpha; x) = (1 + alpha x) / 2 , that alpha0 = - 1/3. How many µ-e decays are needed to establish a to a 1% accuracy (i.e., alpha / Deltaalpha = 100)?

Equation

Note that

Equation

For

Equation

For this problem

Equation

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