Optimality

142 D. M. Dabrowska
where λ =−1,
ηθ(t) =
� τ
0
Kθ(t, u)ρ − ˙ Γ (u, θ)EN(du),
ρ−Γ ˙ (u, θ) = v(u, θ) ˙ Γθ(u)+ρ(u, θ) and Bθ is given by (2.4). For fixed θ, the kernel Kθ
is symmetric, positive definite and square integrable with respect to Bθ. Therefore it
can have only positive eigenvalues. For λ =−1, the equation has a unique solution
given by
(2.9) ψθ(t) = ηθ(u)−
� τ
0
∆θ(t, u,−1)ηθ(u)Bθ(du),
where ∆θ(t, u, λ) is the resolvent corresponding to the kernel Kθ. By definition, the
resolvent satisfies a pair of integral equations
Kθ(t, u) = ∆θ(t, u, λ)−λ
= ∆θ(t, u, λ)−λ
� τ
0
� τ
0
∆θ(t, w, λ)Bθ(dw)Kθ(w, u)
Kθ(t, w)Bθ(dw)∆θ(w, u, λ),
where integration is with respect to different variables in the two equations. For
λ =−1 the solution to the equation is given by
ψθ(t) =
� τ
0
� τ
−
Kθ(t, u)ρ − ˙ Γ (u, θ)EN(du)
0
∆θ(t, w,−1)Bθ(dw)
� τ
0
Kθ(w, u)ρ − ˙ Γ (u, θ)EN(du)
and the resolvent equations imply that the right-hand side is equal to
(2.10) ψθ(t) =
� τ
0
∆θ(t, u,−1)ρ − ˙ Γ (u, θ)EN(du).
For θ = θ0, substitution of this expression into the formula for the matrices
Σ1,ϕ(θ0, τ) and Σ2,ϕ(θ0, τ) and application of the resolvent equations yields also
Σ1,ϕ(θ0, τ) = Σ2,ϕ(θ0, τ)
=
� τ
v − ˙ Γ (u, θ0)EN(du)
0
� τ � τ
−
0
0
∆θ0(t, u,−1)ρ − ˙ Γ (u, θ0)ρ − ˙ Γ (t, θ0) T EN(du)EN(dt).
It remains to find the resolvent ∆θ. We shall consider first the case of θ = θ0.
To simplify algebra, we multiply both sides of the equation (2.8) byPθ0(0, t) −1 =
exp � t
0 s′ (θ0,Γθ0(u), u)Cθ0(du). For this purpose set
˜ψ(t) =Pθ0(0, t) −1 ψ(t),
˙ G(t) =Pθ0(0, t) −1 ˙ Γθ0(t),
˜v(t, θ0) = v(t, θ0)Pθ0(0, t) 2 , ˜ρ − ˙ G (t, θ0) =Pθ0(0, t)ρ − ˙ Γ (t, θ0),
b(t) =
� t
0
˜v(u, θ0)dEN(u), c(t) =
Multiplication of (2.8) byPθ0(0, t) −1 yields
(2.11)
˜ ψ(t) +
� τ
0
k(t, u) ˜ ψ(u)b(du) =
� τ
0
� t
0
Pθ0(0, u) −2 dCθ0(u).
k(t, u)˜ρ − ˙ G (u, θ0)EN(du),