# Moment generating functions of continous random variables at 0

I'm studying mgfs of various distributions and I have a doubt about a property of mgfs:
My book says that a common feature of discrete function's mgf is that it always exists, and takes value 1 at t=0 It gives this proof for the discrete case: $$E(e^0) = E(1)= \sum 1 \space p(x) = 1$$ How do I do the proof for the continuous case? $$E(e^0) = E(1)= \int_{-\infty}^{\infty} 1 \space p(x) = ???$$

• For a continuous RV, you probably do not mean $p(x)$ but rather $f(x)$, the probability density function. For this, $$E[e^0] = \int_{-\infty}^{\infty}f(x)dx = 1.$$ Commented Mar 3, 2018 at 17:00
• You're right, I just copied the formula and forgot to change that, but... How to compute that integral? How does it give 1? Commented Mar 3, 2018 at 17:12
• The integral of the PDF over the entire range is 1 almost by definition. Commented Mar 3, 2018 at 17:13
• For the statement "an mgf exists" to be meaningful, the mgf ought to be defined and finite in some open neighborhood of $0,$ not just at $0$ itself. But then an mgf does not always exist. The Pareto distribution is a standard example. By "discrete" your book might mean "of finite support," which implies bounded support (all finite sets of real numbers are bounded). Unfortunately that rules out many important distributions like Poisson, Negative Binomial, Pareto, etc.
– whuber
Commented Mar 12, 2023 at 15:14

For example, a Normal random variable domain is $(-\infty, \infty)$,
a gamma random variable domain is $(0,\infty)$