Long tail property of Laplace transforms

Question

A function $F: \mathbb R_+ \rightarrow \mathbb R_+$ is said to be long tailed if $F(\infty)=0$ and for all $y \geq 0$ $$\frac{F(x+y)}{F(x)} \rightarrow 1, \quad x\rightarrow \infty.$$

Let $\mu$ be a measure on $[0, \infty)$ with full support and denote $f(x)=\mathcal L(\mu,x)$ its Laplace transform. I am trying to figure out whether or not it is true that $f$ is long tailed however given $\mu$.

• If $\mu (x) \sim x^\gamma g(x) dx$, $\gamma>-1$ and $g$ analytic in 0, $g(0)\neq 0$ then there should be no problem since we can essentially invoke Watson's lemma, and then $f(x) \sim g(0)\Gamma(\gamma+1)x^{-\gamma-1}$.
• There are also things like $\mu(x)=e^{-1/x}dx$. Then $f(x) =2 K_1(2 \sqrt{x})/\sqrt{x}$ with $K_\nu$ the modified Bessel function, and $f(x) \sim \sqrt{\pi}e^{-2 \sqrt{x}}x^{-3/4}$. Even though the decay is faster than every polynomial order the long tail property is verified.

Purely discrete measures are ruled out by definition, I would initially be happy with the absolutely continuous case only (or a counterexample).

Answer 1

Yes, this is true, assuming that \begin{equation*} f(s):=\int_{[0,\infty)} e^{-sx}\mu(dx)<\infty \end{equation*} for some real $s_0$ and all real $s\ge s_0$. As in the OP, we are also assuming that $\mu$ is a measure on $[0, \infty)$ with full support. We want to show that then for each real $y\ge0$ we have \begin{equation*} r_y(s):=\frac{f(s-y)}{f(s)}\to1 \tag{10}\label{10} \end{equation*} (as $s\to\infty$).

To do that, take any real $h>0$ and write \begin{equation*} f(s)=I_1(s)+I_2(s), \end{equation*} where \begin{equation*} I_1(s):=I_{h,1}(s):=\int_{[0,h)} e^{-sx}\mu(dx),\quad I_2(s):=I_{h,2}(s):=\int_{[h,\infty)} e^{-sx}\mu(dx). \end{equation*} So, for real $s\ge s_0$, \begin{equation*} r_y(s)\le\frac{I_1(s-y)}{I_1(s)}\Big(1+\frac{I_2(s-y)}{I_1(s-y)}\Big). \tag{20}\label{20} \end{equation*} Next, \begin{equation*} I_1(s-y)=\int_{[0,h)} e^{yx}e^{-sx}\mu(dx)\le e^{yh}I_1(s) \tag{30}\label{30} \end{equation*} and, for real $s\ge s_0$, \begin{equation*} I_2(s)\le e^{-(s-s_0)h}f(s_0)=Ce^{-sh}, \end{equation*} where $C:=e^{s_0h}f(s_0)\in[0,\infty)$. On the other hand, \begin{equation*} I_1(s)\ge \int_{[0,h/2)} e^{-sx}\mu(dx)\ge ce^{-sh/2}, \end{equation*} where $c:=\int_{[0,h/2)} \mu(dx)>0$, since $\mu$ is a measure on $[0,\infty)$ with full support. So, for real $s\ge s_0$ we have $I_2(s)\le \frac Cc\, I_1(s)\,e^{-sh/2}$. So, $\dfrac{I_2(s-y)}{I_1(s-y)}\to0$ (as $s\to\infty$).

It now follows from \eqref{20} and \eqref{30} that $L_y:=\limsup_{s\to\infty}r_y(s)\le e^{yh}$ for any real $h>0$, and hence $L_y\le1$ for any real $y\ge0$. On the other hand, the function $f$ is nonincreasing and hence $r_y(s)\ge1$ for any real $y\ge0$ and $s\ge s_0$. Thus, \eqref{10} follows. $\quad\Box$

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It is easily seen from the above proof that the condition that $\mu$ is a measure on $[0, \infty)$ with full support can be relaxed to the condition that $0$ is in the support of $\mu$.