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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?.

Monir
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    Have you ever heard of the Laplace transform? See: https://en.wikipedia.org/wiki/Laplace_transform#Probability_theory – Nap D. Lover Jan 20 '19 at 23:07
  • I know the Laplace transform, But in paper they talk about MGF, since for $Y=X+Z$, where $X$ and $Z$ are independent we have $E{e^{-sY}}=E{e^{-sX}}E{e^{-sZ}}$. So I am confuse? Dose this properties hold in Laplace transform and what relation between MGF and Laplace transform? – Monir Jan 20 '19 at 23:14
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    Both of these questions are answered in the link I provided, which apparently you did not bother to even take a cursory reading of—if $f$ is the PDF of a random variable $X$, then conventionally, the MGF of $X$ is defined as $M_X(s):=\mathbb{E}(e^{s X})$ and the Laplace transform can be written as $\mathscr{L}(f)=M_X(-s)=\mathbb{E}(e^{-sX})$. – Nap D. Lover Jan 20 '19 at 23:18
  • I have read the reference, thanks, So the authors In paper should say the Laplace transform not the MGF – Monir Jan 20 '19 at 23:21
  • Technically, yes. But I think probabilists will just often call it the MGF anyway because we are unpredictable :). With either version you can still recover the moments of $X$ but you have to account for the signs. – Nap D. Lover Jan 20 '19 at 23:43

2 Answers2

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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:

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

Gustav Streicher
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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.

Nap D. Lover
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