Moment Generating Function of Pareto Distribution

Theorem
Let $X$ be a continuous random variable with a Pareto distribution with parameters a and b for $a, b \in \R_{> 0}$.

Then the moment generating function $M_X$ of $X$ is given by:


 * $\map {M_X} t = \begin {cases} a \paren {-b t}^a \map \Gamma {-a, -b t} & t < 0 \\ 1 & t = 0 \\ \text {does not exist} & t > 0 \end {cases}$

Proof
From the definition of the Pareto distribution, $X$ has probability density function:


 * $\map {f_X} x = \dfrac {a b^a} {x^{a + 1} }$

From the definition of a moment generating function:


 * $\ds \map {M_X} t = \expect { e^{t X} } = \int_b^\infty e^{t x} \map {f_X} x \rd x$

First take $t < 0$.

Then:


 * $\ds \map {M_X} t = a b^a \int_b^\infty x^{-\paren {a + 1} } e^{t x} \rd x $

let:

Then:

Now take $t = 0$.

Our integral becomes:

Finally take $t > 0$.

Then:

As a consequence of Exponential Dominates Polynomial, we have:
 * $x^a < e^{t x}$

for sufficiently large $x$.

Therefore, in this case, the integrand increases without bound.

We conclude that the integral is divergent.

Hence $\expect {e^{t X} }$ does not exist for $t > 0$.