**4.3 Estimator for the Poisson Distribution**

Suppose we have *n* measurements of samples, *x*_{1},
*x*_{2}, *x*_{3}, . . ., *x*_{n}, from a
Poisson distribution with mean *µ*. The likelihood function for this
case is then

To eliminate the product sign, we take the logarithm

Differentiating and setting the result to zero, we when find

which yields the solution

Equation (39), of course, is just the sample mean. This is of no great surprise, but it does confirm the often unconscious use of (39).

The variance of can be found by using (33); however, in this particular case, we will use a different way. From (9) we have the definition

Applying this to the sample mean and rearranging the terms, we thus have

Expanding the square of the sum, we find

If now the expectation value is taken, the cross term vanishes, so that

As the reader may have noticed, (43) was derived without reference to the Poisson distribution, so that (43) is, in fact, a general result: the variance of the sample mean is given by the variance of the parent distribution, whatever it may be, divided by the sample size.

For a Poisson distribution, ^{2} = *µ*, so that the error on the estimated
Poisson mean is

where have substituted the estimated value
for the theoretical
*µ*.