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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

Equation 36 (36)

To eliminate the product sign, we take the logarithm

Equation 37 (37)

Differentiating and setting the result to zero, we when find

Equation 38 (38)

which yields the solution

Equation 39 (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 xbar can be found by using (33); however, in this particular case, we will use a different way. From (9) we have the definition

Equation 40 (40)

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

Equation 41 (41)

Expanding the square of the sum, we find

Equation 42 (42)

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

Equation 43 (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, sigma2 = µ, so that the error on the estimated Poisson mean is

Equation 44 (44)

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

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