Laplace transform of r.v. to a CDF on a bounded interval

Question

Let $X$ be a non-negative random variable with a cumulative distribution function (cdf) $F$. Define the Laplace transform of $F$ by \begin{align} L_X(t)=E[e^{-tX}], t \ge 0 \end{align} Some fact about laplace transform

1. $L_X(t)$ uniquely defines the distribution $F$
2. Inversion \begin{align} \sum_{ n \le t x} \frac{(-t)^n}{n!} L_X^{(n)}(t) \to F(x) \text{ as } t \to \infty \tag{$*$} \end{align} where $x$ is a point of continuity of $F$ and $L_X^{(n)}$ is $n$-th derivative.
3. Laplace tansform is unique on any open interval.

My questions: Suppose we only know the Laplace transform on the interval $(t_0,t_2)$ where $t_0< t_1$

1) Can we recover the cdf $F $? If so, what is the inversion formula?

2) If so, can this inversion be done with formula in $(*)\text{?}$

Note: The question about the cdf, not the pdf. We are not assuming that pdf even exists.

My thoughts: Because Laplace transform is unique at any interval, we can, of course, recover $F$. The question then, is what is the inversion formula?

Answer 1

Allowing $t$ to be complex with $\text{Re}(t) > 0$, the Laplace transform is analytic in the right half plane. An analytic function is uniquely determined by its values on any set that has a limit point. So yes, in principle the restriction of the Laplace transform to the interval $(t_0, t_1)$ does uniquely determine the Laplace transform on the right half plane, and thus the CDF.

Answer 2

This will be only a partial answer. It appears to me that the Laplace transforms of two different probability distributions on $[0,+\infty)$ cannot be identical on a bounded open interval. That means there should be an inversion formula, but at this point I'm not going to try to work that out. $$ L_X(t) = \operatorname E(e^{-tX}). $$ The exponential function is defined for $t\in\mathbb C.$ Let $\gamma$ be a smooth simple closed curve in the right half-plane in $\mathbb C.$ Then \begin{align*} & \int_\gamma L_X(t) \, dt \\[8pt] = {} & \int_\gamma \int_{[0,+\infty]} e^{-tx}\, dF_X(x) \, dt \\[8pt] = {} & \int_{[0,+\infty)} \int_\gamma e^{-tx} dt\,dF_X(x) & & \text{Is this step valid? See below.} \\[8pt] = {} & \int_{[0,+\infty)} 0 \, dF_X(x) & & \text{Why? See below.} \\[8pt] = {} & 0. \end{align*} Why is the interchange of integrals valid? Funbini's theorem says it's valid when the integral of the absolute value of the function is finite. Since $x$ is real and non-negative, we have $$ |e^{-tx}| = e^{-x\operatorname{Re}(t)} \le 1. $$ If $\operatorname{Re}(t)<0,$ that last inequality might not hold. Hence $$ \int_\gamma |e^{-tx}| \, |dt| < +\infty. $$ (Here we need $|dt|,$ not just $dt,$ since $dt$ has non-zero imaginary parts as $t$ moves along $\gamma.$)

In the theory of complex variables, we learn that if we integrate a function that is holomorphic in an open set includes a smooth simple closed curve $\gamma$ and includes that region that $\gamma$ bounds, we get $0.$ So the integral along $\gamma$ of the exponential function is $0.$

Morera's theorem is a converse: If a function integrated along every smooth simple closed curve in a simply connected region in the plane is always $0,$ then the function is holomorphic in that region.

Finally, two different holomorphic functions cannot agree everywhere in an open interval.

Unfortunately this answer cannot be understood without some basics of the theory of complex variables. Probably there is some way to do this using only real variables, but I don't know what it is.