2 On Normalization Functions and ϕ-Families of Probability Distributions
29
Proof Observing that the normalizing function ψ is convex with ψ(0) = 0, we may
conclude that ψ u (α) = ψ(αu) is non-decreasing and continuous in [0, 1). Moreover,
(ψ u )
+ (α) is non-decreasing in [0, 1). Fix any function u in the boundary of B
ϕ
c such
that
T ϕ(c + u) dμ < ∞. Assume that ψ(αu) tends to ∞ as α ↑ 1. In this case,
one can note that
ϕ(c + αu − ψ(αu)u 0 ) ≤ ϕ(c + u1 {u>0} − ψ(αu)u 0 ) → 0,
as α ↑ 1.
Since ϕ(c + αu − ψ(αu)u 0 ) ≤ ϕ(c + u1 {u>0} ), we can use the Dominated Convergence Theorem to write
T
ϕ(c + αu − ψ(αu)u 0 ) dμ → 0,
as α ↑ 1,
which is a contradiction to
T ϕ(c + αu − ψ(αu)u 0 )dμ = 1. Thus ψ(αu) is bounded
in [0, 1), and ψ(αu) converges to some β ∈ (0, ∞) as α ↑ 1.
Now, consider any function u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞, then suppose
that, for some λ > 0, the function u satisfies ψ(αu) ≤ λ for all α ∈ [0, 1). Denote
A = {u ≥ 0}. Observing that
A
ϕ(c + αu − λu 0 )dμ ≤
T
ϕ(c + αu − λu 0 )dμ ≤
T
ϕ(c + αu − ψ(αu)u 0 )dμ = 1,
we obtain that
A ϕ(c + u − λu 0 )dμ < ∞. In addition, it is clear that
T \A
ϕ(c + u − λu 0 )dμ ≤
T \A
ϕ(c)dμ ≤ 1.
As a result, we have
T ϕ(c + u − λu 0 )dμ < ∞. From the condition (2.1), it follows
that
T ϕ(c + u)dμ < ∞, which is a contradiction.
Thus, we conclude that for any u ∈ ∂B
ϕ
c , if
T ϕ(c + u)dμ < ∞, then ψ(αu) →
β, and if
T ϕ(c + u)dμ = ∞, then ψ(αu) → ∞, as α ↑ 1.
In the next section, we will discuss the behavior of the normalizing function ψ
assuming that the deformed exponential function does not satisfy Condition 2.2.
2.3.2 Condition 2.2 is Not Satisfied
We know there are deformed exponential functions that do not satisfy Condition 2.2.
Throughout this section it will be clear that this assumption is sufficient to ensure that
the boundary of B
ϕ
c is not empty and, as a consequence, we can analyze the behavior
of the normalizing function near the boundary of the parametrization domain.
29
Proof Observing that the normalizing function ψ is convex with ψ(0) = 0, we may
conclude that ψ u (α) = ψ(αu) is non-decreasing and continuous in [0, 1). Moreover,
(ψ u )
+ (α) is non-decreasing in [0, 1). Fix any function u in the boundary of B
ϕ
c such
that
T ϕ(c + u) dμ < ∞. Assume that ψ(αu) tends to ∞ as α ↑ 1. In this case,
one can note that
ϕ(c + αu − ψ(αu)u 0 ) ≤ ϕ(c + u1 {u>0} − ψ(αu)u 0 ) → 0,
as α ↑ 1.
Since ϕ(c + αu − ψ(αu)u 0 ) ≤ ϕ(c + u1 {u>0} ), we can use the Dominated Convergence Theorem to write
T
ϕ(c + αu − ψ(αu)u 0 ) dμ → 0,
as α ↑ 1,
which is a contradiction to
T ϕ(c + αu − ψ(αu)u 0 )dμ = 1. Thus ψ(αu) is bounded
in [0, 1), and ψ(αu) converges to some β ∈ (0, ∞) as α ↑ 1.
Now, consider any function u ∈ ∂B
ϕ
c such that
T ϕ(c + u)dμ = ∞, then suppose
that, for some λ > 0, the function u satisfies ψ(αu) ≤ λ for all α ∈ [0, 1). Denote
A = {u ≥ 0}. Observing that
A
ϕ(c + αu − λu 0 )dμ ≤
T
ϕ(c + αu − λu 0 )dμ ≤
T
ϕ(c + αu − ψ(αu)u 0 )dμ = 1,
we obtain that
A ϕ(c + u − λu 0 )dμ < ∞. In addition, it is clear that
T \A
ϕ(c + u − λu 0 )dμ ≤
T \A
ϕ(c)dμ ≤ 1.
As a result, we have
T ϕ(c + u − λu 0 )dμ < ∞. From the condition (2.1), it follows
that
T ϕ(c + u)dμ < ∞, which is a contradiction.
Thus, we conclude that for any u ∈ ∂B
ϕ
c , if
T ϕ(c + u)dμ < ∞, then ψ(αu) →
β, and if
T ϕ(c + u)dμ = ∞, then ψ(αu) → ∞, as α ↑ 1.
In the next section, we will discuss the behavior of the normalizing function ψ
assuming that the deformed exponential function does not satisfy Condition 2.2.
2.3.2 Condition 2.2 is Not Satisfied
We know there are deformed exponential functions that do not satisfy Condition 2.2.
Throughout this section it will be clear that this assumption is sufficient to ensure that
the boundary of B
ϕ
c is not empty and, as a consequence, we can analyze the behavior
of the normalizing function near the boundary of the parametrization domain.