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

where
1 / *h* is the rms spread
of about
*,

Since as defined in Eq. (3) is
1 / *h* , we have

(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 * in Eq. 6. We differentiate Eq. 5 with respect to . The answer is

Using this in Eq. 7 gives

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"
_{i}.

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

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

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

Note that use of Eq. 7 for
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* (; *x*). Under
these circumstances we wish to determine
averaged over many repeated experiments
consisting of *N* events each. For one event we have

for *N* events

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

The last integral vanishes if one integrates before the differentiation because

Thus

and Eq. (7) leads to

(8) |

__Example 1__

Assume in the *µ*-e decay distribution function,
*f* (; *x*) =
(1 + *x*) / 2 ,
that
_{0} = - 1/3. How
many *µ*-e decays are needed to establish a
to a 1% accuracy (i.e.,
/
= 100)?

Note that

For

For this problem