A Finite-Interval Uniqueness Theorem for Bilateral Laplace Transforms

Abstract

Two or more bilateral Laplace transforms with a complex argument “s” may be equal in a finite vertical interval when, in fact, the transforms correspond to different functions. In this article, we prove that the existence of a bilateral Laplace transform in any finite horizontal interval uniquely determines the corresponding function. The result appears to be new as we could not find it in the literature. The novelty of the result is that the interval need not contain zero, the function need not be nonnegative and need not be integrable. The result has a potential to be useful in the context of fitting probability distributions to data using Laplace transforms or moment generating functions.

Random permutations without macroscopic cycles

“…Note that Theorem 2.2 in [12] shows that it is sufficient to compute the Laplace transform in Proposition 4.3 for s j ≥ 0 to prove Theorem 2.4.…”

Random permutations with logarithmic cycle weights

“…Although the expressions in (3.7) holds for general s ∈ C, we will calculate the asymptotic behaviour of the moment generating function of K 0n only on the positive half-line s ≥ 0. Theorem 2.2 in [7] shows that this is enough to prove statement of the theorem. Let r be the solution of e s P (r) = ne −r .…”

“…Proof of Theorem 4.1. Theorem 2.2 in [7] shows that it is sufficient to compute the Laplace transform for s ≥ 0 to establish the CLT. Therefore Lemma 4.2 immediately implies the second point of Theorem 4.1.…”

Efficient Estimation for Semiparametric Structural Equation Models With Censored Data

“…Note that ∫-∞∞e-st0.16667emfalse(f∗gfalse)0.16667emfalse(tfalse)0.16667emnormaldt=∫-∞∞e-stf(t)dt∫-∞∞e-stg(t)dt for any s such that the integrals are defined, where (f∗g)(t)≡∫-∞∞f(t-s)g(s)ds is the convolution of f and g . Therefore, ( f * g )(·) = 0 implies that ∫-∞∞e-stf0.16667emfalse(tfalse)0.16667emnormaldt∫-∞∞e-stg0.16667emfalse(tfalse)0.16667emnormaldt=0, which, if g is positive, implies that f (·) = 0 by the uniqueness of the bilateral Laplace 27 transform (Chareka 2007). Because ḡ...…”

“…∫-∞∞e-stf0.16667emfalse(tfalse)0.16667emnormaldt∫-∞∞e-stg0.16667emfalse(tfalse)0.16667emnormaldt=0, which, if g is positive, implies that f (·) = 0 by the uniqueness of the bilateral Laplace 27 transform (Chareka 2007). Because ḡ k (·) e (·) is positive, (A.4) implies that f η 1 | Y ( η 1 | Z , Y; ψ , ν ) = f η 1 | Y ( η̃ 1 | Z , Y; ψ̃ , ν̃ ), where η̃ 1 is defined in condition (C4′).…”