This question may be related to [this one](https://math.stackexchange.com/q/4310801/498394).


Now I try to make some statistical estimator using Laplace transform, but I face the following serious problem.


Let $f$ be some one-sided probability distribution defined on $[0,\infty)$, and $\hat{f}$ be its Laplace transform.


Now assume that we only have information of $\hat{f}(s)$ near $s=0$ (for example, $\hat{f}(s) = \hat{g}(s) / (2-\hat{g}(s))$ with known probability distribution $g(x)$), and using this, we want to find the convergence speed of the tail:


\begin{align\*}
 \int\_x^{\infty} f(x)\, dx = O(?).
\end{align\*}


According to Tauberian remainder theory in [J. Korevaar's book](https://books.google.co.kr/books?hl=en&lr=&id=aWfsCAAAQBAJ&oi=fnd&pg=PA1&dq=Tauberian+theory&ots=Z5Nl1wlXq5&sig=JyZpb6E6H8gCrPRIIIekyRYwFu0&redir_esc=y#v=onepage&q=Tauberian%20theory&f=false) (Examples 2.3., page 348), he said


* If $|\hat{f}(s) - \hat{f}(0)| \le Cs^{\alpha}$ with some $\alpha > 0$, then the tail rate is $O(1/\log x)$;
* If $|\hat{f}(s) - \hat{f}(0)| \le Ce^{-\alpha/s}$ with some $\alpha > 0$, then the tail rate is $O(1/\sqrt{x})$.


In fact, this is a disaster for statisticians because any higher-order Taylor approximation of $\hat{f}(s)$ near $s=0$ cannot guess whether the tail is light or super super heavy ($1/\log x$).


So the question is the following.



> 
> **Question.** What other condition on $\hat{f}$ near $0$ is required to guarantee the tail has at least a power-tail $O(x^{-\beta})$? If the power tail cannot be guaranteed by any of information about $\hat{f}$ near $s=0$, then what condition is required for $g$?
> 
> 
> 


Any help would be appreciated. (suggesting books, papers, or anything!)


Thanks for the reading,



Convergence speed of the tail of distribution using Tauberian remainder theorem
Let $f$ be a pdf on $[0,\infty)$. Let $\hat f$ be the Laplace transform of $f$, so that
\begin{equation}
 \hat f(s)=\int\_0^\infty f(x)e^{-sx}\,dx
\end{equation}
for real $s\ge0$.


Suppose that
\begin{equation}
 |\hat f(s)-\hat f(0)|\le Cs^a
\end{equation}
for some real $a,C>0$ and all real $s\ge0$.


Then for all real $s>0$
\begin{equation}
 Cs^a\ge|\hat f(s)-\hat f(0)|=\int\_0^\infty f(u)(1-e^{-su})\,du
 \ge(1-e^{-1})\int\_{1/s}^\infty f(u)\,du,
\end{equation}
whence for all real $x>0$
\begin{equation}
 \int\_x^\infty f(u)\,du\le C\_1/x^a,
\end{equation}
where $C\_1:=C/(1-e^{-1})\in(0,\infty)$, as desired.


(Your difficulty was due to the fact that you tried to use general Tauberian theorems, valid without the nonnegativity condition. On the other hand, all pdf's are of course nonnegative. Once this nonnegativity condition is taken into account, everything becomes much simpler and better.)


