4.9. Bootstrap Procedures 30740 60 80 100 120 140 1600100200300400500600Frequencyx*Figure 4.9.1:Histogram of the 3000 bootstrapx∗s. The 90% bootstrap confidence
interval is (61. 655 , 120 .48).One way of improving the bootstrap is to use a pivot random variable, a variable
whose distribution is free of other parameters. For instance, in the last example,
instead of usingX,useX/ˆσX,whereˆσX=S/√
nandS=[∑
(Xi−X)^2 /(n−1)]^1 /^2 ;
that is, adjustXby its standard error. This is discussed in Exercise 4.9.6. Other
improvements are discussed in the two books cited earlier.
Remark 4.9.1.∗Briefly, we show that the normal assumption on the distribution
of̂θ, (4.9.1), is transparent to the argument around expression (4.9.3); see Efron
and Tibshirani (1993) for further discussion. SupposeHis the cdf of̂θand thatH
depends onθ. Then, using Theorem 4.8.1, we can find an increasing transformation
φ=m(θ) such that the distribution ofφ̂=m(̂θ)isN(φ, σc^2 ), whereφ=m(θ)
andσ^2 c is some variance. For example, take the transformation to bem(θ)=
Fc−^1 (H(θ)), whereFc(x)isthecdfofaN(φ, σc^2 ) distribution. Then, as above,
(φ̂−z(1−α/2)σc,φ̂−z(α/2)σc)isa(1−α)100% confidence interval forφ. But note
that
1 −α = P[
φ̂−z(1−α/2)σc<φ<φ̂−z(α/2)σc)]= P[
m−^1 (φ̂−z(1−α/2)σc)<θ<m−^1 (φ̂−z(α/2)σc)]. (4.9.6)
Hence, (m−^1 (φ̂−z(1−α/2)σc),m−^1 (φ̂−z(α/2)σc)) is a (1−α)100% confidence interval
forθ. Now supposeĤis the cdfHwith a realization̂θsubstituted in forθ, i.e.,
analogous to theN(θ, σ̂ θb^2 ) distribution above. Supposeθ̂∗is a random variable with