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Let $X$ be a non-negative random variable with a CDF $F$. Let $L_X(t)$ denote the Laplace transform of $F$, i.e., \begin{align} L_X(t)=E[ e^{-tX}], \quad t \ge 0 \end{align}

It is known that $L_X(t)$ is unique for every open interval.

Therefore, if $L_X(t)$ is known on an interval $(t_0, t_1)$, there must exist an inversion formula from $L_X(t)$ to $F$.

My Question: Given that we know $L_X(t)$ only on $(t_0, t_1)$ what is the inversion formula back to $F$?

I want to point out the if $L_X(t)$ is known for all $t \ge 0$, then the inversion formula is given by \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.

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  • $\begingroup$ It is analytic in $\Re(t)$ thus once you know it on $(t_0,t_1)$ you know it on $\Re(t) > 0$ and the inversion formula is from the contour integral en.wikipedia.org/wiki/Inverse_Laplace_transform your inversion formula is wrong $\lim_{t \to \infty}\sum_{n \le tx} \frac{(-1)^n}{n!} L_X^{(n)}(t) =\lim_{t \to \infty} \sum_{n \le tx} \int_0^\infty \frac{y^n}{n!} e^{-ty}F(y)dy\le \lim_{t \to \infty}\int_0^\infty e^{(1-t)y} F(y)dy = 0 $ $\endgroup$
    – reuns
    Commented Oct 28, 2019 at 2:23
  • $\begingroup$ @reuns Note that I have $(-t)^n$ and not $(-1)^n$. I would have to think about the first part of your comment about the inversion. $\endgroup$
    – Boby
    Commented Oct 28, 2019 at 18:40
  • $\begingroup$ @reuns Question? Would the inversion formula you suggest produce pdf or cdf? I think it would produce pdf. $\endgroup$
    – Boby
    Commented Oct 28, 2019 at 21:04
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    $\begingroup$ If you know that $L_X(t)$ converges on $(t_0, t_1)$, then it converges on all of $(t_0, \infty)$. Moreover, if $t, \delta > 0$ are chosen so that $t_0 < t-\delta < t < t_1$, then for any integer $k\geq 1$ we have $$\sum_{n_1=0}^{\infty}\frac{\delta^{n_1}}{n_1!}\cdots\sum_{n_k=0}^{\infty} \frac{\delta^{n_k}}{n_k!}L_X^{(n_1+\cdots+n_k)}(t) = L_X(t+k\delta). $$ So, at least theoretically, we can recover values of $L_X$ only using those on $(t_0, t_1)$. (Remark. The above iterated summations cannot be reduced to multiple summation in general.) $\endgroup$
    – Sangchul Lee
    Commented Nov 1, 2019 at 9:59
  • $\begingroup$ @SangchulLee Thanks. Could you explain how you prove the formula? Maybe you can put it as an answer. $\endgroup$
    – Boby
    Commented Nov 1, 2019 at 11:17

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