Laplace Transform as a function of s - Intuitive Understanding

Question

I have worked with Laplace and Fourier transforms on numerous occasions, but I still struggle with understanding these transforms intuitively. As I prepare for qualifying exams in probability theory, they appear once again. The textbook I am working with presents this idea:

For a random variable $X \geq 0$, the Laplace transform of $X$ as a function of $s$ is defined as $L(s) = \text{E}\left[ e^{-sX} \right]$. This function $L \colon [0,\infty) \to \mathbb{R}$ is both well-defined and finite for all $s \in [0,\infty)$.

I know that if $X$ has PDF $f_X(x)$ (for either continuous or discrete random variables, in the latter case we use Dirac-delta functions for integrability), we may write $$ L(s) = \text{E}\left[ e^{-sX} \right] = \int_{0}^{\infty} e^{-sx} f_X(x) \text{d}x $$ Further, I understand that $e^{-sX}$ is decreasing in $y = sX$ since both $s \geq 0$ and $X \geq 0$. I can believe that this function $L(s)$ is well-defined since the integral expression clearly shows that if $s_1 = s_2$ then $L(s_1) = L(s_2)$.

What I am struggling to understand is why this expression is finite for all $s \in [0,\infty)$. Is there an intuitive explanation as to why this is true, perhaps related to the decreasing nature of $e^{-sX}$ or the Laplace transform's relation to frequency-space? Any help is greatly appreciated!

Answer 1

For all $s, x \ge 0$, $f_X(x)e^{-s x} \le f_X(x)$. Since $f_x(x)$ is a probability distribution, $\int_0^\infty f_X(x)dx = 1 < \infty$. Thus, $f_X(x)e^{-sx}$ is bounded by an integrable function, so it too must be integrable.

Answer 2

For heuristic explaination You could look first at what correspond Fourier transform for a given function, and them looking at the correspondance or, precisely, the similarities between the two. Laplace's transform is a bridge from temporal/spatial to frequential. Note that $s$ is complex in original Laplace transform.

It is finite because

$$ L : s \to \int_0^\infty e^{-sx}f_X(x)dx$$

And so because $f_X$ is of integral of one.

$$ |L(s)| \leq \int_0^\infty e^{-2sx}dx\int_0^\infty f_X(x)^2dx\leq \int_0^\infty e^{-2sx}dx=\dfrac{1}{2s}$$

Answer 3

For a great intuitive explanation of the Laplace transform I highly recommend watching the lecture on Laplace transforms by Prof. Arthur Mattuck of MIT from the course on Differential Equations.

You can find a video recording of the lecture here.

I can't find enough words to explain how great this lecture is and how well the intuition behind Laplace transforms is explained.
In essence Mattuck presents the Laplace transform as a continuous version of power series. He doesn't enter the topic of probability theory, but I still believe that the video is worth watching.
In the end, the Laplace transform of a random variable $X$ is just the Laplace transform of the PDF of $X$, so you find the parallel in computing the power series $A(x)=\sum p_Y(n)x^n$ of a discrete random variable $Y$ with probability mass function $p_Y(n) = \mathbb{P}(X=n)$.