4.3 Estimator for the Poisson Distribution

Suppose we have n measurements of samples, x1, x2, x3, . . ., xn, from a Poisson distribution with mean µ. The likelihood function for this case is then

(36)

To eliminate the product sign, we take the logarithm

(37)

Differentiating and setting the result to zero, we when find

(38)

which yields the solution

(39)

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

(40)

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

(41)

Expanding the square of the sum, we find

(42)

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

(43)

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

(44)

where have substituted the estimated value for the theoretical µ.