Enlace entre la función generadora de momento y la función característica

Estoy tratando de entender el vínculo entre la función generadora de momento y la función característica. La función de generación de momento se define como:

M_{X} (t) = E (\exp (t X)) = 1 + \frac{t E (X)}{1} + \frac{t^{2} E (X^{2})}{2!} + \dots + \frac{t^{n} E (X^{n})}{n!}

$M_X(t) = E(\exp(tX)) = 1 + \frac{t E(X)}{1} + \frac{t^2 E(X^2)}{2!} + \dots + \frac{t^n E(X^n)}{n!}$

Using the series expansion of $\exp(tX) = \sum_0^{\infty} \frac{(t)^n \cdot X^n}{n!}$ , I can find all the moments of the distribution for the random variable X.

The characteristic function is defined as:

φ_{X} (t) = E (\exp (i t X)) = 1 + \frac{i t E (X)}{1} - \frac{t^{2} E (X^{2})}{2!} + \dots + \frac{(i t)^{n} E (X^{n})}{n!}

$\varphi_X(t) = E(\exp(itX)) = 1 + \frac{it E(X)}{1} - \frac{t^2 E(X^2)}{2!} + \ldots + \frac{(it)^n E(X^n)}{n!}$

I don't fully understand what information the imaginary number $i$ gives me more. I see that $i^2 = -1$ and thus we don't have only $+$ in the characteristic function, but why do we need to subtract moments in the characteristic function? What's the mathematical idea?

— Giuseppe
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One important point is that the moment-generating function is not always finite! (See this question, for example.) If you want to build a general theory, say, about convergence in distribution, you'd like to be able to have it work with as many objects as possible. The characteristic function is, of course, finite for any random variable since

| \exp (i t X) | \leq 1

$|\exp(itX)| \leq 1$ .

— cardinal

The similarities in the Taylor expansions still allow one to read off the moments, when they exist, but note that not all distributions have moments, so the interest in these functions goes far beyond this! :)

— cardinal

Another point to note is that the MGF is the Laplace transformation of a random variable and the CF is the Fourier transform. There are fundamental relationships between these integral transforms, see here.

— tchakravarty

I thought CF is the inverse fourier transform (and not the fourier transform) of a propability distribution?

— Giuseppe

The distinction is only a matter of sign in the exponent, and possibly a multiplicative constant.

— Glen_b -Reinstate Monica

As mentioned in the comments, characteristic functions always exist, because they require integration of a function of modulus $1$ . However, the moment generating function doesn't need to exist because in particular it requires the existence of moments of any order.

When we know that $E[e^{tX}]$ is integrable for all $t$ , we can define $g(z):=E[e^{zX}]$ for each complex number $z$ . Then we notice that $M_X(t)=g(t)$ and $\varphi_X(t)=g(it)$ .

— Davide Giraudo
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