# Equality of Moment Generating Functions

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.

• why "analytic there"? – mathworker21 Jun 24 '20 at 0:27
• @mathworker21 I think you can get it from Morera's theorem. As a reference: en.wikipedia.org/wiki/Two-sided_Laplace_transform – user3281410 Jun 24 '20 at 10:48
• @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." – user3281410 Jun 24 '20 at 16:49
• 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? – user3281410 Jun 24 '20 at 16:55
• in your question, u just said the laplace transform exists in some strip. how do you know it converges absolutely in that strip? – mathworker21 Jun 24 '20 at 19:52

Let $$0, $$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. 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.