Let x = (x1,...,xn), and let g(x, ) be a family of probability densities depending on
Question:
Let x = (x1,...,xn), and let gθ(x, ξ) be a family of probability densities depending on θ = (θ1,...,θr) and the real parameter ξ, and jointly measurable in x and ξ. For each θ, let hθ(ξ) be a probability density with respect to a σ-finite measure ν such that pθ(x) = gθ(x, ξ)hθ(ξ) dν(ξ) exists. We shall say that a function f of two arguments u = (u1,...,ur), v = (v1,...,vs) is nondecreasing in (u, v) if f(u
, v)/f(u, v) ≤ f(u
, v
)/f(u, v
) for all (u, v) satisfying ui ≤ u
i, vj ≤ v
j (i = 1,...,r; j = 1,...,s). Then pθ(x) is nondecreasing in (x, θ)
provided the product gθ(x, ξ)hθ(ξ) is
(a) nondecreasing in (x, θ) for each fixed ξ;
(b) nondecreasing in (θ, ξ) for each fixed x;
(c) nondecreasing in (x, ξ) for each fixed θ.
[Interpreting gθ(x, ξ) as the conditional density of x given ξ, and hθ(ξ) as the a priori density of ξ, let ρ(ξ) denote the a posteriori density of ξ given x, and let ρ
(ξ) be defined analogously with θ in place of θ. That pθ(x) is nondecreasing in its two arguments is equivalent to gθ(x
, ξ)
gθ(x, ξ) ρ(ξ) dν(ξ) ≤
gθ (x
, ξ)
gθ (x, ξ) ρ
(ξ) dν(ξ).
By
(a) it is enough to prove that D = gθ(x
, ξ)
gθ(x, ξ) [ρ
(ξ) − ρ(ξ)] dν(ξ) ≥ 0.
Let S− = {ξ : ρ
(ξ)/ρ(ξ) < 1} and S+ = {ξ : ρ(ξ)/ρ(ξ) ≥ 1}. By
(b) the set S−
lies entirely to the left of S+. It follows from
(c) that there exists a ≤ b such that D = a
S−
[ρ
(ξ) − ρ(ξ)] dν(ξ) + b
S+
[ρ
(ξ) − ρ(ξ)] dν(ξ), and hence that D = (b − a)
S+ [ρ
(ξ) − ρ(ξ)] dν(ξ) ≥ 0.]
Step by Step Answer:
Testing Statistical Hypotheses
ISBN: 9781441931788
3rd Edition
Authors: Erich L. Lehmann, Joseph P. Romano