If two PDFs have the same moment generating function that converges in an open set around 0, then the PDFs are same.

This is a well known fact, but I can't find its proof. If the PDFs are defined for only non-negative values, the MGF is essentially Laplace transform and the uniqueness problem is just that of Laplace transform. However, if the PDFs are defined on the whole real line, the MGF is not Laplace transform. I guess the condition that the MGF converges around zero may take care of something in this case. Could someone provide the proof or reference for it?


1 Answer 1


If the random variable $X$ has pdf $f(x)$, $x\in \mathbb{R}$, then the moment generating function $M_X(t)$, provided it exists, is the two-sided Laplace transform of $X$. If $\mathcal{L}\{f(x)\}(t)$ is the (one-sided) Laplace Transform of $f(x)$, then

$$ M_X(-t)=\mathcal{L}\{f(x)\}(t)+\mathcal{L}\{f(-x)\}(-t) $$

and $M_X(t)$ is unique iff both $\mathcal{L}\{f(x)\}(t)$ and $\mathcal{L}\{f(-x)\}(-t)$ are unique. But we already know that the Laplace transform of a function is unique, provided it exists.

  • $\begingroup$ This answer is complete. I'd also like to add that there are references provided in Casella and Berger (2002, page 66) to Widder (1946) and Feller (1971) that may also be of interest. $\endgroup$ Commented Feb 26, 2015 at 15:13
  • 1
    $\begingroup$ @VivekSubramanian I don’t see at all how the answer is complete, it just pushes the question from “why do mgfs determine the distribution?” to “why does the Laplace transform determine the distribution?” This would never fly, for example, in a measure theory class. $\endgroup$
    – guy
    Commented Oct 28, 2018 at 14:37
  • $\begingroup$ I don't believe that, although leading to the right conclusion, the logic used here is correct. A counterexample of the argument is: stats.stackexchange.com/a/651648 $\endgroup$ Commented 13 hours ago

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.