According to Wikipedia, the moment-generating function $M_X (t)$ of a probability distribution $f_X (x)$ is given by $M_X (t) = \int_{-\infty}^\infty e^{tx}f_X (x)\ dx$.

  1. Is $t$ time? If so, why does it appear in the output of this transform rather than the input? In Differential Equations, the Laplace Transform transforms the time domain into the frequency domain. In this scenario, is the $x$ domain "one hop in the direction opposite the frequency domain" relative to the time domain, so that the Laplace Transform transforms the $x$ domain to the time domain, and a composition of two Laplace Transforms would transform the $x$ domain to the frequency domain?

  2. Why is $t$ negated relative to the Laplace Transform? My understanding from the derivation of a Laplace Transform from a power series is that making the exponent negative is a substitution choice. Doesn't choosing to make the exponent positive make the integral much less likely to converge?


1 Answer 1


In an MGF, $t$ isn't time or any other physical quantity, but a "dummy variable" whose dimension is inverse to that of $X$ (so if $X$ is dimensionless $t$ is too, but we certainly wouldn't pair a spatial length with time). The symbol $t$ is arbitrary; $k$ is often used instead, as is $s$.

Similarly, a Laplace transform doesn't require a physical interpretation such as the time/frequency one you gave, but note again this relates two physical quantities whose product is dimensionless. (This requirement exists because we exponentiate such a product; $e^z=1+z+z^2/2+\cdots$ only makes sense for dimensionless $z$.) Any physical examples, such as time and frequency or length and wavenumber, results from some physical context in which a Laplace transform naturally arises. Actually, this is even more commonplace with Fourier transforms.

The Laplace transform of a function $f(x)$ is defined as $\int_0^\infty e^{-sx}f(x)dx$, thus a function of $s$ (and if you prefer, change $x$ and $s$ to whichever letters you like). Note the integral is only over $x\ge0$, so the $e^{-sx}$ factor helps convergence if $\Re s$ is large enough. For example, the choice $f(x)=e^{ax}$ leads to a convergence Laplace transform provided $\Re s>\Re a$. The resemblance to MGFs is superficial.


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