6.2. Rao–Cram ́er Lower Bound and Efficiency 367Definition 6.2.2(Efficiency).In cases in which we can differentiate with respect
to a parameter under an integral or summation symbol, the ratio of the Rao–Cram ́er
lower bound to the actual variance of any unbiased estimator of a parameter is called
theefficiencyof that estimator.Example 6.2.3(Poisson(θ) Distribution).LetX 1 ,X 2 ,...,Xndenote a random
sample from a Poisson distribution that has the meanθ>0. It is known thatXis
an mle ofθ; we shall show that it is also an efficient estimator ofθ.Wehave
∂logf(x;θ)
∂θ=∂
∂θ(xlogθ−θ−logx!)=
x
θ−1=
x−θ
θ.Accordingly,E[(
∂logf(X;θ)
∂θ) 2 ]
=E(X−θ)^2
θ^2=σ^2
θ^2=θ
θ^2=1
θ.The Rao–Cram ́er lower bound in this case is 1/[n(1/θ)] =θ/n.Butθ/nis the
variance ofX. HenceXis an efficient estimator ofθ.
Example 6.2.4(Beta(θ,1) Distribution). LetX 1 ,X 2 ,...,Xndenote a random
sample of sizen>2 from a distribution with pdff(x;θ)={
θxθ−^1 for 0<x< 1
0elsewhere,
(6.2.14)where the parameter space is Ω = (0,∞). This is the beta distribution, (3.3.9),
with parametersθand 1, which we denote by beta(θ,1). The derivative of the log
offis
∂logf
∂θ
=logx+1
θ. (6.2.15)
From this we have∂^2 logf/∂θ^2 =−θ−^2. Hence the information isI(θ)=θ−^2.
Next, we find the mle ofθand investigate its efficiency. The log of the likelihood
function is
l(θ)=θ∑ni=1logxi−∑ni=1logxi+nlogθ.The first partial ofl(θ)is∂l(θ)
∂θ=∑ni=1logxi+n
θ. (6.2.16)
Setting this to 0 and solving forθ,themleisθ̂=−n/
∑n
i=1logXi.Toobtain
the distribution of̂θ,letYi=−logXi. A straight transformation argument shows