What is the difference between $E\{e^{-sx}\}$ and $E\{e^{sx}\}$ for MFG

Question

I am working in wireless communication.

When I cheek the books about MFG I found the MGF of random variable $X$ is given by the following formula

$$ M_X(s)=E\{e^{sx}\} $$

However when I read papers, I found the following definition

$$ M_X(s)=E\{e^{-sx}\} $$

What is the difference between this two?.

Answer 1

The moment generating function, or MGF, has the usual definition of:

$$MGF(s) = E\{ e^{sX} \} = \int_{-\infty}^{\infty}f(x)e^{sx}dx$$

The bilateral Laplace transform has the usual definition of:

$$ L(s) = \int_{-\infty}^{\infty}f(x)e^{-sx}dx$$

Comparing the two we can derive the following:

$$MGF(s) = L(-s) = \int_{-\infty}^{\infty}f(x)e^{sx}dx$$

Answer 2

If $f$ is the PDF of a random variable $X$ then, conventionally, the MGF of $X$ is defined as $M_X(s):=\mathbb{E}(e^{sX})$ and the Laplace transform of $f$ (often reworded as the Laplace transform of $X$, though technically incorrect) can be written as $\mathscr{L}[f](s)=M_X(-s)=\mathbb{E}(e^{-sX})$.

Both satisfy the lovely "independence implies multiplication" property and you can further recover the moments of $X$ from either form but you must account for the signs when using $M_X(-s)$, e.g. $$-\frac{d}{ds} \mathbb{E}(e^{-sX})=-\mathbb{E}(-Xe^{-sX}) |_{s=0}=\mathbb{E}(X),$$ etc.