Let $X,Y$ be be random variables whose moment generating functions $s\mapsto \mathbb{E}(e^{sX})$ exist and agree on either the interval $(-\delta,0]$ or on the interval $[0,\delta)$ for some $\delta > 0$. Do $X$ and $Y$ have the same distribution?

In particular, is the following argument outline valid: The Laplace transforms (with $s$ now in $\mathbb{C}$), $s\mapsto \mathbb{E}(e^{sX})$ exist on some strip $\text{Re}(s)\in (-\delta,0)$ or $\text{Re}(s)\in (0,\delta)$ and are analytic there. Therefore, they agree on that strip, and so they agree on the boundary $\text{Re}(s)=0$, so the characteristic functions are the same. That implies the distributions are the same.

  • $\begingroup$ why "analytic there"? $\endgroup$ – mathworker21 Jun 24 '20 at 0:27
  • $\begingroup$ @mathworker21 I think you can get it from Morera's theorem. As a reference: en.wikipedia.org/wiki/Two-sided_Laplace_transform $\endgroup$ – user3281410 Jun 24 '20 at 10:48
  • $\begingroup$ @mathworker21 I am confused about your objection. From the article: "Analogously, the two-sided transform converges absolutely in a strip of the form a < Re(s) < b, and possibly including the lines Re(s) = a or Re(s) = b.[2] The subset of values of s for which the Laplace transform converges absolutely is called the region of absolute convergence or the domain of absolute convergence. In the two-sided case, it is sometimes called the strip of absolute convergence. The Laplace transform is analytic in the region of absolute convergence." $\endgroup$ – user3281410 Jun 24 '20 at 16:49
  • $\begingroup$ Also, for $s = a+ib$, we have $|e^{-sx}f_{X}(x)| = e^{-ax}f_{X}(x)$, so saying that the transform is defined on a small interval on the real line guarantees absolute convergence on the strip in $\mathbb{C}$ corresponding to that interval? $\endgroup$ – user3281410 Jun 24 '20 at 16:55
  • $\begingroup$ in your question, u just said the laplace transform exists in some strip. how do you know it converges absolutely in that strip? $\endgroup$ – mathworker21 Jun 24 '20 at 19:52

Yes it is valid, see Theorem 2 in A note on moment generating functions by Mukherjea, Rao, and Suen.

For convenience I will quote the theorem here:

Let $0<a<b$, $M_n(t)=Ee^{tX_n}$ and $M(t)=E(e^{tX})$ such that $\lim_{n\to\infty} M_n(t)=M(t)$ whenever $a<t<b$. Then $F_n$, the cumulative distribution function of $X_n$, converges weakly to $F$, the cumulative distribution function of $X$.

This theorem includes your case as well, since you can just take $X_n$ to be a constant sequence.

Link to paper: https://www.sciencedirect.com/science/article/abs/pii/S016771520500475X


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