Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

11.2. More Bayesian Terminology and Ideas 667

does not exist. However, such priors are used if, when combined with the likelihood,

we obtain a posterior pdf which is a proper pdf. Byproper,wemeanthatit

integrates to a positive constant. In this example, we obtain the posterior pdf of

f(θ|y)∝θy−^1 (1−θ)n−y−^1 ,

which is proper providedy>0andn−y>0. Of course, the posterior mean is
y/n.
Definition 11.2.2.LetX′=(X 1 ,X 2 ,...,Xn)be a random sample from the dis-
tribution with pdff(x|θ).Apriorh(θ)≥ 0 for this family is said to beimproper
if it is not a pdf, but the functionk(θ|x)∝L(x|θ)h(θ)can be made proper.

Anoninformative prioris a prior that treats all values ofθthe same, that is,
uniformly. Continuous noninformative priors are often improper. As an example,
suppose we have a normal distributionN(θ 1 ,θ 2 )inwhichbothθ 1 andθ 2 >0are
unknown. A noninformative prior forθ 1 ish 1 (θ 1 )=1,−∞<θ 1 <∞. Clearly,
this is not a pdf. An improper prior forθ 2 ish 2 (θ 2 )=c 2 /θ 2 , 0 <θ 2 <∞, zero
elsewhere. Note that logθ 2 is uniformly distributed between−∞<logθ 2 <∞.
Hence, in this way, it is a noninformative prior. In addition, assume the parameters
are independent. Then the joint prior, which is improper, is

h 1 (θ 1 )h 2 (θ 2 )∝ 1 /θ 2 , −∞<θ 1 <∞, 0 <θ 2 <∞. (11.2.1)

Using this prior, we present the Bayes solution forθ 1 in the next example.

Example 11.2.1.LetX 1 ,X 2 ,...,Xnbe a random sample from aN(θ 1 ,θ 2 ) distri-
bution. Recall thatXandS^2 =(n−1)−^1

∑n

i=1(Xi−X)

(^2) are sufficient statistics.
Suppose we use the improper prior given by (11.2.1). Then the posterior distribution
is given by
k 12 (θ 1 ,θ 2 |x, s^2 )∝
(
1
θ 2
)(
1
√
2 πθ 2
)n
exp
[
−
1
2
{
(n−1)s^2 +n(x−θ 1 )^2
}
/θ 2
]
∝
(
1
θ 2
)n 2 +1
exp
[
−
1
2
{
(n−1)s^2 +n(x−θ 1 )^2
}
/θ 2
]
.
To get the conditional pdf ofθ 1 ,givenxands^2 ,weintegrateoutθ 2
k 1 (θ 1 |x, s^2 )=
∫∞
0
k 12 (θ 1 ,θ 2 |x, s^2 )dθ 2.
To carry this out, let us change variablesz=1/θ 2 andθ 2 =1/z, with Jacobian
− 1 /z^2 .Thus
k 1 (θ 1 |x, s^2 )∝
∫∞
0
z
n 2 +1
z^2
exp
[
−
{
(n−1)s^2 +n(x−θ 1 )^2
2
}
z
]
dz.
Referring to a gamma distribution withα=n/2andβ=2/{(n−1)s^2 +n(x−θ 1 )^2 },
this result is proportional to
k 1 (θ 1 |x, s^2 )∝{(n−1)s^2 +n(x−θ 1 )^2 }−n/^2.