On injectivity of two-sided Laplace transform

In summary, the article discusses the conditions under which the two-sided Laplace transform is injective, meaning that it uniquely determines a function from its transform. It explores the implications of injectivity in various contexts, such as signal processing and differential equations, and provides mathematical proofs and examples to illustrate the criteria for injectivity. The findings emphasize the importance of the transform in analyzing systems and signals, highlighting its role in both theoretical and practical applications.
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I am studying a theorem in probability that the Laplace transform of a (nonnegative) random variable determines the law of that random variable, which is equivalent to its injectivity. The book only treats Laplace transforms of nonnegative random variables, but since the moment generating function (mgf) is in fact the two sided Laplace transform, and since mgfs exist for arbitrary sign random variables, I wonder how to extend the result to say the two-sided Laplace transform is also injective.
I will omit the theorem and its proof here, since it would mean a lot of typing. But the relevant part of the proof of the theorem is that we are considering the set ##H## of functions consisting of ##\psi_\lambda(x)=e^{-\lambda x}## for ##x\geq 0## and ##\lambda\geq 0##. We extend the functions by continuity so they are defined on all of ##[0,\infty]##, namely we put ##\psi_\lambda(\infty)=0## if ##\lambda>0## and ##\psi_0(\infty)=1##. Having a compact set ##[0,\infty]##, we can apply the Stone-Weierstrass theorem to say that ##H## is dense in ##C([0,\infty])##.

I have seen another proof of the injectivity of the Laplace transform (not related to probability), but that also uses the Stone-Weierstrass theorem. So how would one go about showing the two-sided Laplace transform is injective? I guess one would like to consider ##[-\infty,\infty]## and apply the Stone-Weierstrass theorem also, but I'm unsure how one would modify ##H##.
 
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I hope this can help.
 

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Gavran said:
I hope this can help.
Thank you. This was helpful. In proving the main theorem, Theorem 2.2, they rely on "Curtiss' theorem". I have looked at Curtiss paper but I am unsure which theorem Chareka means. Do you know which theorem Chareka is referring to?
 
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psie said:
Thank you. This was helpful. In proving the main theorem, Theorem 2.2, they rely on "Curtiss' theorem". I have looked at Curtiss paper but I am unsure which theorem Chareka means. Do you know which theorem Chareka is referring to?
If ## \{M_n(t)\} ## is a sequence of moment generating functions corresponding to a sequence of distribution functions ## \{F_n(x)\} ##, then convergence of ## \{M_n(t)\} ## to a moment generating function ## M(t) ## on ## (-\delta,\delta) ## implies that ## \{F_n(x)\} ## converges weakly to ## F(x) ##, where ## F(x) ## is the distribution function with moment generating function ## M(t) ##.
 
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Gavran said:
If ## \{M_n(t)\} ## is a sequence of moment generating functions corresponding to a sequence of distribution functions ## \{F_n(x)\} ##, then convergence of ## \{M_n(t)\} ## to a moment generating function ## M(t) ## on ## (-\delta,\delta) ## implies that ## \{F_n(x)\} ## converges weakly to ## F(x) ##, where ## F(x) ## is the distribution function with moment generating function ## M(t) ##.
Ok, but how does it follow from this that $$L_{u^\ast}(s')=L_{v^\ast}(s')\implies u^\ast=v^\ast,$$where ##u^\ast,v^\ast## are probability density functions (and ##L_u## the bilateral Laplace transform of ##u##)? This is what Chareka concludes in Theorem 2.2 due to Curtiss theorem. This is the part I have a hard time understanding.
 
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psie said:
Ok, but how does it follow from this that $$L_{u^\ast}(s')=L_{v^\ast}(s')\implies u^\ast=v^\ast,$$where ##u^\ast,v^\ast## are probability density functions (and ##L_u## the bilateral Laplace transform of ##u##)? This is what Chareka concludes in Theorem 2.2 due to Curtiss theorem. This is the part I have a hard time understanding.
I do not know because I have not seen it.
Probably they used the equation $$ M(s)=E(e^{sX})=E(e^{-(-s)X})=B(-s) $$ where ## B(s) ## is a bilateral Laplace transform.
 
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