Find the distribution of an absolutely continueous random variable given its moment-generating function

Question

Let $K$ be a random variable such that $\mathbb E\left[e^{tK}\right]$ exists for $t$ in some neighborhood of $0$, and that $$\varphi(t):= \mathbb E\left[e^{tK}\right] = \frac3{(3−t)}.$$

Find the distribution of $K$.

Answer 1

The mgf of an exponential $Exp(\lambda)$ random variable is $\frac{\lambda}{\lambda-t}$. We also have an important fact about the uniqueness of mgfs; that if two random variables have the same mgf then they have the same distribution. Therefore $K\sim Exp(3)$ and therefore $\mathbb{P}(K\leq 5)=\int_0^5 3e^{-3s}ds = 1-e^{-15}$.

I appreciate this isn't a very intuitive answer: "spot that it's m.g.f. is of a certain form" - and I would imagine there may be a better more general answer, but it is indeed helpful to know the standard forms of common mgfs, like geometric, exponential, gaussian, poisson etc.

Answer 2

As shown by Post (1930), given a continuous map $f:[0,\infty)\to\mathbb R$ for which there exists some $b\in\mathbb R$ such that $$ \sup_{t>0} |f(t)|e^{bt}<\infty, $$ for all $s>b$, the Laplace transform $F(s):= \lim_{\varepsilon\to 0^+} \int_{-\varepsilon}^\infty f(t)e^{-st}$ for $f(t)$ exists and is infinitely differentiable with respect to $s$, and we may obtain $f$ from $F$ by the equality $$ f(t) = \lim_{k\to\infty}\frac{(-1)^k}{k!} F^{(k)}\left( \frac kt \right). $$ Since the map $t\mapsto \varphi(t)\cdot e^{-2t}$ is decreasing on $(0,3)$ (as one may readily see by its derivative being negative on this interval, indeed $\sup_{t>0}|\varphi(t)|e^{-2t}=1<\infty$.

It follows that $\varphi$ is the moment-generating function corresponding to an absolutely random variable, say $X$, with density $f(x) = 3\cdot e^{-3\cdot x}\cdot\mathsf 1_{[0,\infty)(x)}$, and indeed: $$ \mathbb E[tX] = \int_0^\infty 3e^{-3x}e^{tx}\cdot\mathsf dx = \frac3{3-t},\ \ t<3. $$