Why is the MGF the Laplace transform?

[Joan Fernandez]

TL;DR

The MGF of a probability distribution is its Laplace transform. However, LTs have domain and codomains in the complex plane, whereas MGFs are real. Why is this not an issue?

The Laplace transform gives information about the exponential components in a function, as well as oscillatory components. To do so there is a need for the complex plane (complex exponentials).

I get why the MGF of a distribution is very useful (moment extraction and classification of the distribution in terms of its tails (exponential, subexponential, fat tailed, etc). But since the MGF has as domain an interval in the real line in which E[exp{tX}] is defined, and maps to [0, infinity], all the analysis of oscillatory components in the function is left out, and within the realm of the characteristic function. All that the MGF achieves is the "scalar product" with an exponential function. How can therefore be called the Laplace transform?

 

[anuttarasammyak]

\frac{d}{dt} e^{tX}=X e^{tX}
for any complex variable t. So you don't have to limit t a real number. In the case t is pure imaginary number it is called Fourier transform. So in general complex t, the transformation would be called Laplace-Fourier transform. The final results $E^{(n)}(t=0)$ remain real.

 

[Joan Fernandez]

\frac{d}{dt} e^{tX}=X e^{tX}
for any complex variable t. So you don't have to limit t a real number. In the case t is pure imaginary number it is called Fourier transform. So in general complex t, the transformation would be called Laplace-Fourier transform. The final results $E^{(n)}(t=0)$ remain real.

In terms of recovering moments it makes no difference, but what about capturing any oscillating sinusoidal components?

 

[anuttarasammyak]

Even in usual case of real t, I am afraid you do not find a meaning of $e^{tX}$. Neither I do not think you have to worry about meaningless in $e^{tX}$ for pure imaginary or complex t.

 

[pasmith]

Summary:: The MGF of a probability distribution is its Laplace transform. However, LTs have domain and codomains in the complex plane, whereas MGFs are real. Why is this not an issue?

The Laplace transform gives information about the exponential components in a function, as well as oscillatory components. To do so there is a need for the complex plane (complex exponentials).

I get why the MGF of a distribution is very useful (moment extraction and classification of the distribution in terms of its tails (exponential, subexponential, fat tailed, etc). But since the MGF has as domain an interval in the real line in which E[exp{tX}] is defined, and maps to [0, infinity], all the analysis of oscillatory components in the function is left out, and within the realm of the characteristic function. All that the MGF achieves is the "scalar product" with an exponential function. How can therefore be called the Laplace transform?

If f is real-valued, then <br /> \int_0^\infty f(x)e^{-tx}\,dx is real-valued for real t. What set t is in depends on your application. If you want to invert the transform, then t must be complex. On the other hand, if you want to extract \int_0^\infty x^n f(x)\,dx = \left. (-1)^n \frac{d^n}{dt^n}\left(\int_0^\infty f(x)e^{-tx}\,dx\right)\right|_{t=0} then t only needs to be in an interval which includes 0.

 

[mathman]

$\int\limits_0^\infty e^{-tx}f(x)dx=\sum\limits_{n=0}^\infty \frac{(-t)^n}{n!}\int\limits_0^\infty x^nf(x)dx$. Which can give moments for non-negative random variables. The Fourier transform removes the non-negative restriction.

 

[Joan Fernandez]

If f is real-valued, then <br /> \int_0^\infty f(x)e^{-tx}\,dx is real-valued for real t. What set t is in depends on your application. If you want to invert the transform, then t must be complex. On the other hand, if you want to extract \int_0^\infty x^n f(x)\,dx = \left. (-1)^n \frac{d^n}{dt^n}\left(\int_0^\infty f(x)e^{-tx}\,dx\right)\right|_{t=0} then t only needs to be in an interval which includes 0.

Right, I follow... So my question could really be re-phrased as: when t is complex we have a bona fides LT, but what is it that we are doing when in your first integral t is real? Are we just "dotting" the function with an exponential to see how it decays?