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

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

kjetil b halvorsen
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user7064
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1 Answers1

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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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Glen_b
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  • 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? – user7064 Jul 09 '14 at 06:20
  • Yes, for the same reason that the MGF does. To be specific, the claim applies to the two sided Laplace transform. Details are [here](http://stats.stackexchange.com/a/46068/805). Were you after a definition of 'converges' or 'open interval'? – Glen_b Jul 09 '14 at 08:35
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    The result is basically dealt with [here](http://stats.stackexchange.com/questions/34956/a-proof-involving-properties-of-moment-generating-functions) – Glen_b Jul 09 '14 at 08:37
  • "If $L_X(\cdot) = L_Y(\cdot)$, then $X$ and $Y$ have the same distribution." This statement is wrong in general, right? – user7064 Jul 09 '14 at 10:19
  • 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. – Glen_b Jul 09 '14 at 10:32
  • Ok, I understand. So, in fact, it suffices to have the equality in an open interval around 0 rather than on the whole positive half line. – user7064 Jul 09 '14 at 11:03