Moment Generating Function of Pareto Distribution

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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:

\(\ds u\) \(=\) \(\ds -t x\)
\(\ds \leadsto \ \ \) \(\ds \frac {\d u} {\d x}\) \(=\) \(\ds -t\) Derivative of Power


Then:

\(\ds \map {M_X} t\) \(=\) \(\ds -\dfrac {a b^a} t \int_{-bt}^\infty \paren {-\dfrac u t}^{-\paren {a + 1} } e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds -\dfrac {a b^a} t \int_{-bt}^\infty \paren {-\dfrac 1 t}^{-\paren {a + 1} } u^{-\paren {a + 1} } e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds a \paren {-b t}^a \int_{-bt}^\infty u^{-\paren {a + 1} } e^{-u} \rd u\)
\(\ds \) \(=\) \(\ds a \paren {-b t}^a \map \Gamma {-a, -b t}\) Definition of Gamma Distribution

$\Box$


Now take $t = 0$.

Our integral becomes:

\(\ds a b^a \int_b^\infty x^{-a - 1} \rd x\) \(=\) \(\ds a b^a \bigintlimits {-\dfrac 1 {a x^a} } b \infty\) Primitive of Power, Fundamental Theorem of Calculus
\(\ds \) \(=\) \(\ds - b^a \lim_{x \mathop \to \infty} \dfrac 1 {x^a } - a b^a \paren {-\dfrac 1 {a b^a} }\)
\(\ds \) \(\to\) \(\ds 1\) since $a > 0$, $\dfrac 1 {x^a} \to 0$ as $x \mathop \to \infty$

$\Box$


Finally take $t > 0$.

Then:

\(\ds \map {M_X} t\) \(=\) \(\ds a b^a \int_b^\infty \dfrac {e^{t x} \rd x } {x^{a + 1} }\)
\(\ds \) \(=\) \(\ds a b^a \paren {\bigintlimits {\dfrac {e^{t x} } {\paren {a + 1 - 1} x^{a + 1 - 1} } } b \infty + \frac {t} {a + 1 - 1} \int_b^\infty \frac {e^{t x} \rd x} {x^{a + 1 - 1} } }\) Primitive of Exponential of a x over Power of x
\(\ds \) \(\to\) \(\ds \infty\)

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 $t > 0$.

$\blacksquare$

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