Probability proof of inversion formula for Laplace transform

Question

Let $f:[0, \infty[\longrightarrow \mathbb{R}$ be bounded and continuous and define $L(\lambda)=\int_0^\infty e^{-\lambda x}f(x)dx$.

Let $X_n$ be a sequence of independent random variables with exponential distribution of rate $\lambda$. Using the fact that the sum $S_n=X_1+...+X_n$ has Gamma distribution we can see that $$ (-1)^{n-1}\frac{\lambda^nL^{(n-1)}(\lambda)}{(n-1)!} =Ef(S_n)$$ where $L^{(n-1)}$ is the $n-1$ derivative of $L$. How can we use this to prove $$ f(y)=\lim_{n}(-1)^{n-1}\frac{(\frac{n}{y})^nL^{(n-1)}(\frac{n}{y})}{(n-1)!} $$

We know that by the strong law of large numbers $S_n/n$ converges to $E(X_1)$ almost everywhere, so I could consider the parameter $\lambda=f(y)$ but that doesn't seem useful..

Answer 1

Try the Weierstrass Approximation Theorem.

Define $B_n(y) := E[f(S_n)](n/y)$, $Y_n := |f(S_n) - f(y)|$ and $Z_n := |S_n - y|$

Since f is bounded and continuous on $[0,\infty)$, it is bounded and continuous on $[0,M] \ \forall M > 0$ and hence uniformly continuous in $[0,M]$.

$\to \forall \epsilon > 0, \exists \delta > 0 \forall x, y \in [0,M]$ s.t.

$$Z_n \le \delta \to Y_n < \epsilon/2$$

Let us try to show that $\forall \epsilon > 0, \exists N > 0$ s.t. $n > N \ \to \ |B_n(y) - f(y)| < \epsilon$:

$$|B_n(y) - f(y)|$$

$$ = |E[f(S_n)](n/y) - f(y)|$$

$$ = |E[f(S_n)](\lambda)|_{\lambda = n/y} - f(y)|$$

$$ = |(E[f(S_n)] - f(y))(\lambda)|_{\lambda = n/y}|$$

$$ = |(E[f(S_n) - f(y)])(\lambda)|_{\lambda = n/y}|$$

$$ = (|E[f(S_n) - f(y)]|)(\lambda)|_{\lambda = n/y}$$

By Jensen's Inequality,

$$ \le (E[|f(S_n) - f(y)|])(\lambda)|_{\lambda = n/y}$$

Omitting $\lambda$ for now

$$ = E[|f(S_n) - f(y)|]$$

$$ = E[Y_n]$$

$$ = E[Y_n 1_{Z_n \le \delta} + Y_n 1_{Z_n > \delta}]$$

$$ = E[Y_n 1_{Z_n \le \delta}] + E[Y_n 1_{Z_n > \delta}]$$

$$ < E[\epsilon/2 1_{Z_n \le \delta}] + E[Y_n 1_{Z_n > \delta}]$$

$$ = \epsilon/2 E[1_{Z_n \le \delta}] + E[Y_n 1_{Z_n > \delta}]$$

$$ = \epsilon/2 P(Z_n \le \delta) + E[Y_n 1_{Z_n > \delta}]$$

$$ = \epsilon/2 (1) + E[Y_n 1_{Z_n > \delta}]$$

$$ = \epsilon/2 + E[Y_n 1_{Z_n > \delta}]$$

$$ \le \epsilon/2 + E[2K 1_{Z_n > \delta}]$$

Note: $\exists K > 0$ s.t. $|f(x)| \le K \forall x \in [0, M]$

$\to \ Y_n = |f(S_n) - f(y)| \le |f(S_n)| + |f(y)| \le K + K = 2K$

$$ = \epsilon/2 + 2K E[ 1_{Z_n > \delta}]$$

$$ = \epsilon/2 + 2K P(Z_n > \delta)$$

By Chebyshev's Inequality, $\forall \delta > 0, P(Z_n > \delta) = P(|S_n - \frac{n}{\lambda}| > \delta) = P(|S_n - \frac{n}{\lambda}| > n\delta)$

$= P(|\frac{S_n}{n} - \frac{1}{\lambda}| > \delta) \le \frac{1}{n \lambda^2 \delta^2}$

$$ \le \epsilon/2 + 2K \frac{1}{n \lambda^2 \delta^2}$$

$$\le \epsilon/2 + \epsilon/2 = \epsilon$$

if we choose $\color{red}{N}$ as such:

$$2K \frac{1}{n \lambda^2 \delta^2} < \epsilon/2$$

$$\frac{1}{n \lambda^2 \delta^2} < \frac{\epsilon}{4K}$$

$$\frac{1}{n} < \frac{\lambda^2 \delta^2 \epsilon}{4K}$$

$$n > \frac{4K}{\lambda^2 \delta^2 \epsilon} \color{red}{= N}$$

Not sure what to do about the $\lambda$, but it seems that $\forall \epsilon > 0$,

$$n > \frac{4K}{\lambda^2 \delta^2 \epsilon} \to Y_n < \epsilon/2$$

$$\to |B_n(y) - f(y)| < \epsilon \ QED$$

$$\therefore \ \lim_{n \to \infty} B_n(y) = f(y)$$

Answer 2

Old question, but I was working on the same problem recently. Here's an idea:

The functions $g_n$ defined by $g(x) = f(x/n)$ are bounded and continuous. Let $L_h$ denote the Laplace transform of a function $h$, and let the independent exponential random variables $X_1,X_2,\dots$ have rate $\frac{1}{y}$ (using $y\neq 0$). By the first part of the problem $$ \mathbb{E}\left(f\left(\frac{S_n}{n}\right)\right) = \mathbb{E}(g_n(S_n)) = (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!} $$

Yet the strong law from Chapter 7 gives

$$\lim_{n\to\infty}\frac{S_n}{n} = \mathbb{E}(X_1) = y \quad \text{almost surely}$$

and since $f$ is bounded the expected values converge: $$ \lim_{n\to\infty} (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!} = \lim_{n\to\infty} \mathbb{E}\left(f\left(\frac{S_n}{n}\right)\right) = f(y) $$ The rest is just a matter of rewriting the terms $L_{g_n}^{(n-1)}$ in terms of $L_f^{(n-1)}$. For any bounded continuous (say) function $h$ one can compute for each $\lambda > 0$ that $$ \quad L_h^{(n-1)}(\lambda) = (-1)^{n-1}\int_0^\infty e^{-\lambda x} x^{n-1} h(x)\,dx $$ In our case, then, $$ L_{g_n}^{(n-1)}(\lambda) = (-1)^{n-1}\int_0^\infty e^{-\lambda x} x^{n-1} f(x/n) \, \mathrel{\underset{u:=x/n}{=}} (-1)^{n-1}\int_0^\infty e^{-n\lambda u} n^{n-1}u f(u) n \, du = n^n L_f(n\lambda) $$ Setting $\lambda = \frac{1}{y}$ yields $$ f(y) = \lim_{n\to\infty} (-1)^{n-1}\frac{\frac{1}{y^n}L^{(n-1)}_{g_n}\left(\frac{1}{y}\right)}{(n-1)!} = \lim_{n\to\infty} (-1)^{n-1}\frac{\frac{n^n}{y^n}L^{(n-1)}_{f}\left(\frac{n}{y}\right)}{(n-1)!} $$