Let $X$ be a real valued random variable. For every $t\in\mathbb{R}$, let us denote by $L(t)$ the quantity $\mathbb{E}(\exp(tX))$. Of course $L$ may not be defined everywhere.
This is well known ( for example by standard complex analysis) that if $L$ is defined on an interval of the form $]-t_0,t_0[$ then $L$ characterizes the law of $X$.
Now let us assume that $X$ is non negative and that for every $t>0$, $L(t)$ is infinite.
Then the proposition above can not be applied anymore. I am pretty sure that in this situation the knowledge of $L$ on $\mathbb{R}_{-}$ is not characterizing the law of $X$ anymore but I do not see a concrete counter example. Does someone know this kind of counter example?
To be more clear:
I want to find $X$ and $Y$ that are non negative and that have $\underline{\text{different}}$ distributions such that for every $t>0$, $\mathbb{E}(\exp(tX))=\mathbb{E}(\exp(tY))=+\infty$ and such that for every $t\leq 0$, $\mathbb{E}(\exp(tX))=\mathbb{E}(\exp(tY))$.
Best Answer
It does in fact characterize the law of $X$.
Suppose $X$ is nonnegative. You have $L(-k)=\mathbb E[\exp(-kX)]=\mathbb E[Y^k]$ for every integer $k\in\mathbb N$, where $Y:=\exp(-X)$. Note that $Y$ has support in the compact $[0,1]$, so the distribution of $Y$ is characterized by its moments (by Weierstrass' theorem). Hence the family $(L(-k))_{k\in\mathbb N}$ characterizes the distribution of $Y$, and thus of $X=-{\log Y}$.