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It is true that the Laplace transform of a (positive) random variable characterises that random variable, just like its density?

($L_X(z) = E(exp(-Xz))$)

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Up to a change of sign in the exponent, the Laplace transform is the moment generating function.

The MGF only characterizes a random variable if the MGF converges in an open interval around 0.

This isn't automatically the case, even when all moments exist (e.g. see the lognormal).

By contrast the characteristic function (which up to sign in the exponent is the same as the Fourier transform) does characterize a random variable.

For more information, see this question, and the several threads linked to from there.

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  • $\begingroup$ If I understand correctly, the Laplace transform characterises the random variable if it converges in an open interval around 0. Is it correct. Could you expand a little bit on that condition. What does it mean? $\endgroup$
    – user7064
    Jul 9, 2014 at 6:20
  • $\begingroup$ Yes, for the same reason that the MGF does. To be specific, the claim applies to the two sided Laplace transform. Details are here. Were you after a definition of 'converges' or 'open interval'? $\endgroup$
    – Glen_b
    Jul 9, 2014 at 8:35
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    $\begingroup$ The result is basically dealt with here $\endgroup$
    – Glen_b
    Jul 9, 2014 at 8:37
  • $\begingroup$ "If $L_X(\cdot) = L_Y(\cdot)$, then $X$ and $Y$ have the same distribution." This statement is wrong in general, right? $\endgroup$
    – user7064
    Jul 9, 2014 at 10:19
  • $\begingroup$ The thing is, if the integrals for $E(e^{-Xz})$ converge (so that it's meaningful to compare them for equality) then the equality of the two in an open interval around 0 would imply they do have the same distribution. $\endgroup$
    – Glen_b
    Jul 9, 2014 at 10:32

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