Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

4.1. Sampling and Statistics 229

The two partial derivatives simplify to

∂l(μ, σ)

∂μ

= −

∑n

i=1

(

xi−μ

σ

)(

−

1

σ

)

(4.1.5)

∂l(μ, σ)

∂σ

= −

n

σ

+

1

σ^3

∑n

i=1

(xi−μ)^2. (4.1.6)

Setting these to 0 and solving simultaneously, we see that the mles are

̂μ = X (4.1.7)

̂σ^2 = n−^1

∑n

i=1

(Xi−X)^2. (4.1.8)

Notice that we have used the property that the mle ofσ̂^2 is the mle ofσsquared. As
we have shown in Chapter 2, (2.8.6), the estimatorXis an unbiased estimator for
μ. Further, from Example 2.8.7 of Section 2.8 we know that the following statistic

S^2 =

1

n− 1

∑n

i=1

(Xi−X)^2 (4.1.9)

is an unbiased estimator ofσ^2. Thus for the mle ofσ^2 ,E(̂σ^2 )=[n/(n−1)]σ^2.
Hence, the mle is a biased estimator ofσ^2. Note, though, that the bias of̂σ^2 is
E(̂σ^2 −σ^2 )=−σ^2 /n, which converges to 0 asn→∞. In practice, however,S^2 is
the preferred estimator ofσ^2.
Rasmussen (1991), page 65, discusses a study to measure the concentration of
sulfur dioxide in a damaged Bavarian forest. The following data set is the realization
of a random sample of sizen= 24 measurements (micro grams per cubic meter) of
this sulfur dioxide concentration:
33.4 38.6 41.7 43.9 44.4 45.3 46.1 47.6 50.0 52.4 52.7 53.9
54.3 55.1 56.4 56.5 60.7 61.8 62.2 63.4 65.5 66.6 70.0 71.5.
These data are also in the R data filesulfurdio.rdaat the site listed in the Preface.
Assuming these data are in the R vectorsulfurdioxide, the following R segment
obtains the estimates of the true mean and variance (boths^2 and̂σ^2 are computed):
mean(sulfurdioxide);var(sulfurdioxide);(23/24)*var(sulfurdioxide)
53.91667 101.4797 97.25139.
Hence, we estimate the true mean concentration of sulfur dioxide in this damaged
Bavarian forest to be 53.92 micro grams per cubic meter. The realization of the
statisticS^2 iss^2 = 101.48, while the biased estimate ofσ^2 is 97.25. Rasmussen notes
that the average concentration of sulfur dioxide in undamaged areas of Bavaria is
20 micro grams per cubic meter. This value appears to be quite distant from the
sample values. This will be discussed statistically in later sections.

In all three of these examples, standard differential calculus methods led us to
the solution. For the next example, the support of the random variable involvesθ
and, hence, it is not surprising that for this case differential calculus is not useful.