Questions on Inverting Laplace transforms and Probability

Question

From Williams' Probability w/ Martingales:

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1. Are we allowed to switch derivative and integral as follows:

$$\frac{\partial}{\partial \lambda} \int_{0}^{\infty} e^{-\lambda x} f(x) = \int_{0}^{\infty} \frac{\partial}{\partial \lambda} e^{-\lambda x} f(x) $$

?

Why/Why not?

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2. Assuming the $E[f(S_n)]$ equation is true, how does one prove the $f(y)$ equation?

That is, consider $E[f(S_n)]$ as a function of $\lambda$:

$$E[f(S_n)] = E[f(S_n)](\lambda) = \frac{(-1)^n (\lambda)^n L^{n-1}(\lambda)}{(n-1)!}$$

If $\lambda = \frac{n}{y}$, then

$$E[f(S_n)](\frac{n}{y}) = \frac{(-1)^n (\frac{n}{y})^n L^{n-1}(\frac{n}{y})}{(n-1)!}$$

How does one prove that

$$\lim_{n \to \infty} E[f(S_n)](\frac{n}{y}) \left(= \lim_{n \to \infty} \frac{(-1)^n (\frac{n}{y})^n L^{n-1}(\frac{n}{y})}{(n-1)!} \right) = f(y)?$$

Answer 1

I also worked on this problem several days before. My point of view:

(1) Yes we are allowed to. Because $\partial_{\lambda}^{n-1}(e^{-\lambda x}f(x))$ exists and it is continuous.

(2) By weak law $$\frac{S_n}{n}\to \mathbb{E}(X_1)=\frac{1}{\lambda}~~\text{in probability.}$$ In other words, $$S_n \to \frac{n}{\lambda}=:y~~\text{in probability.}$$ Composition with a continuous function $f$ preserves convergence, so $$f(S_n)\to f(y)~~\text{in probability.}$$ Now apply bounded convergence theorem, and replace $\lambda$ with $\frac{n}{y}$, $$f(y)=\mathbb{E}(f(y))=\lim\limits_{n\rightarrow\infty}\mathbb{E}\left[f(S_n)\right]$$

Answer 2

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 $\forall \epsilon > 0$,

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

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

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