Is finding the CDF from the Laplace transform well-posed?

Question

In my study of Dynamic Light Scattering, I came across the following inverse problem. Let $F(s):[0,T]\rightarrow[0,T]$ be the Laplace transform of a probability distribution $f(t)$ on the real line with finite support. This means that $$F(s)=\int_0^{L}e^{-st}f(t)dt$$ for all $s$, while $$\int_0^{L}f(t)dt=1$$ and $f(t)\geq0$ for all $t\in[0,L]$. The numbers $T$ and $L$ are fixed positive real constants. I found that the inverse problem: obtain $f(t)$ from $F(s)$, is ill-posed. Specifically, we can make a sequence of Laplace transforms $F_n(s)$ converge under the $L_2$ norm to some limiting $F(s)$, while the sequence of PDFs $f_n(t)$ does not converge to $f(t)$. I'll give an example at the end of the question.

However, it seems to be different for the cumulative distribution function. Let the CDF be defined by $$C(t):= \int_0^t f(t')dt'.$$ The question: If the Laplace transforms $F_n(s)$ of $f_n(t)$ converge under the $L_2$ norm to the Laplace transform $F(s)$ of $f(t)$, does it follow that the CDFs $C_n(t)$ converge under the $L_2$ norm to $C(t)$?

Own attempts Both the Laplace transforms $F_n(s)$ and the CDFs $C_n(t)$ are monotone functions. So when we assume smoothness, convergence under the $L_2$ norm is equivalent to pointwise convergence. So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise?

Finding the PDF is ill-posed: Let $$f_n(t):=\frac{1}{t\sigma\sqrt{2\pi}} e^{-\frac{-(\ln(t)-\mu)^2}{2\sigma^2}}(1+\sin(2\pi n(\ln t-\mu)/\sigma)).$$ This is a lognormal distribution multiplied by a sinusoid. Here we take $\mu$ such that $e^{\mu}$ lies inside $[0,L]$, and we take $\sigma$ very small. Now $f_n(t)$ is a probability distribution for all $n$ (up to a very small normalizing factor). The Laplace transforms $F_n(s)$ of $f_n(t)$ converge uniformly on $[0,T]$ to the Laplace transform $F(s)$ of the lognormal distribution: $$f(t):=\frac{1}{t\sigma\sqrt{2\pi}} e^{-\frac{-(\ln(t)-\mu)^2}{2\sigma^2}}.$$ However, the PDFs $f_n(t)$ do not converge to $f(t)$ under the $L_2$ norm. Here is an illustration of $f(t)$ and one instance of $f_n(t)$:

Answer 1

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So the question can be restated as follows: if the Laplace transforms converge pointwise, do the CDFs also converge pointwise?

The answer is yes.

Indeed, let $P_n$ ($n=1,2,\ldots$) and $P$ be any probability distributions supported on a finite interval $[a,b]$ such that $LP_n\to LP$ pointwise on an interval $[0,T]$ for some real $T>0$, where $LQ$ is the Laplace transform of a probability distribution $Q$, so that $$(LQ)(t):=\int e^{-tx}Q(dx)$$ for real $t$.

Take any subsequence $(P_{n_k})$ of the sequence $(P_n)$ that converges weakly to a probability distribution $D$. Since $e^{-tx}$ is continuous in $x$, it follows that $(LP_{n_k})(t)\to (LD)(t)$ for all real $t$. So, $$LP=LD \text{ on } [0,T]. \tag{10}\label{10}$$

Since the distribution $P$ is supported on a finite interval, it is determined by its moments $m_k(P)$ ($k=1,2,\ldots$) -- see e.g. Theorem 30.1 in Billingsley. But $$m_k(P)=(-1)^k (LP)^{(k)}(0)$$ for $k=0,1,\dots$, so that the $m_k(P)$'s are in turn determined by the values of $LP$ on $[0,T]$. So, $P$ is determined by the values of $LP$ on $[0,T]$. So, by \eqref{10}, $D=P$. So, any weakly convergent subsequence of the sequence $(P_n)$ weakly converges to $P$.

Also, since all the $P_n$'s are supported on a finite interval, the sequence $(P_n)$ is tight.

So, by the Corollary to Theorem 29.3 on p. 381 in Billingsley's book, $(P_n)$ weakly converges to $P$.

That is, the c.d.f. $C_n$ of $P_n$ converges to the c.d.f. $C$ of $P$ at every point of continuity of $C$.

Finally, if $C$ is absolutely continuous (or just continuous), then $C_n\to C$ pointwise. $\quad\Box$