276 Chapter 7: Parameter Estimation
that it will be betweenμandμ+a? If the answer is positive, then we accept, as a working
hypothesis, that our prior feelings aboutθcan be expressed in terms of a prior distribution
that is normal with meanμ. To determineσ, the standard deviation of the normal prior,
think of an interval centered aboutμthat youa priorifeel is 90 percent certain to contain
θ. For instance, suppose you feel 90 percent (no more and no less) certain thatθwill lie
betweenμ−aandμ+a. Then, since a normal random variableθwith meanμand
varianceσ^2 is such that
P{
−1.645<θ−μ
σ<1.645}
=.90or
P{μ−1.645σ<θ<μ+1.645σ}=.90it seems reasonable to take
1.645σ=a or σ=a
1.645Thus, if your prior feelings can indeed be reasonably described by a normal distribution,
then that distribution would have meanμand standard deviationσ=a/1.645. As a test
of whether this distribution indeed fits your prior feelings you might ask yourself such
questions as whether you are 95 percent certain thatθwill fall betweenμ−1.96σand
μ+1.96σ, or whether you are 99 percent certain thatθwill fall betweenμ−2.58σand
μ+2.58σ, where these intervals are determined by the equalities
P{
−1.96<θ−μ
σ<1.96}
=.95P{
−2.58<θ−μ
σ<2.58}
=.99which hold whenθis normal with meanμand varianceσ^2.
EXAMPLE 7.8c Consider the likelihood functionf(x 1 ,...,xn|θ) and suppose thatθis
uniformly distributed over some interval (a,b). The posterior density ofθgivenX 1 ,...,Xn
equals
f(θ|x 1 ,...,xn)=f(x 1 ,...,xn|θ)p(θ)
∫b
af(x^1 ,...,xn|θ)p(θ)dθ=f(x 1 ,...,xn|θ)
∫b
af(x^1 ,...,xn|θ)dθa<θ <bNow themodeof a densityf(θ) was defined to be that value ofθthat maximizesf(θ).
By the foregoing, it follows that the mode of the densityf(θ|x 1 ,...,xn) is that value ofθ