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Moment-Generating Function

Given a Random Variable $x \in R$, if there exists an $h > 0$ such that

M(t) \equiv \left\langle{e^{tx}}\right\rangle{} = \cases{ \s...
...fty}^\infty e^{tx}P(x)\,dx & for a continuous distribution\cr}
\end{displaymath} (1)

for $\vert t\vert < h$, then
M(t) \equiv \langle e^{tx}\rangle
\end{displaymath} (2)

is the moment-generating function.
$\displaystyle M(t)$ $\textstyle =$ $\displaystyle \int_{-\infty}^\infty (1+tx+{\textstyle{1\over 2!}}t^2x^2+\ldots)P(x)\,dx$  
  $\textstyle =$ $\displaystyle 1+tm_1+{\textstyle{1\over 2!}}t^2 m_2+\ldots,$ (3)

where $m_r$ is the $r$th Moment about zero. The moment-generating function satisfies
$\displaystyle M_{x+y}(t)$ $\textstyle =$ $\displaystyle \langle e^{t(x+y)}\rangle = \langle e^{tx}e^{ty}\rangle$  
  $\textstyle =$ $\displaystyle \langle e^{tx}\rangle \langle e^{ty}\rangle = M_x(t)M_y(t).$ (4)

If $M(t)$ is differentiable at zero, then the $n$th Moments about the Origin are given by $M^{n}(0)$
M(t) = \langle e^{tx}\rangle \qquad M(0) = 1
\end{displaymath} (5)

M'(t) = \langle xe^{tx}\rangle \qquad M'(0) = \langle x\rangle
\end{displaymath} (6)

M''(t) = \langle x^2e^{tx}\rangle \qquad M''(0) = \langle x^2\rangle
\end{displaymath} (7)

M^{(n)}(t) = \langle x^ne^{tx}\rangle \qquad M^{(n)}(0)=\langle x^n\rangle.
\end{displaymath} (8)

The Mean and Variance are therefore
$\displaystyle \mu$ $\textstyle \equiv$ $\displaystyle \langle x\rangle = M'(0)$ (9)
$\displaystyle \sigma^2$ $\textstyle \equiv$ $\displaystyle \langle x^2\rangle -\langle x\rangle^2 = M''(0)-[M'(0)]^2.$ (10)

It is also true that
\mu_n=\sum_{j=0}^n{n\choose j}(-1)^{n-j}\mu_j'(\mu'_1)^{n-j},
\end{displaymath} (11)

where $\mu'_0=1$ and $\mu_j'$ is the $j$th moment about the origin.

It is sometimes simpler to work with the Logarithm of the moment-generating function, which is also called the Cumulant-Generating Function, and is defined by

$\displaystyle R(t)$ $\textstyle \equiv$ $\displaystyle \ln[M(t)]$ (12)
$\displaystyle R'(t)$ $\textstyle =$ $\displaystyle {M'(t)\over M(t)}$ (13)
$\displaystyle R''(t)$ $\textstyle =$ $\displaystyle {M(t)M''(t)-[M'(t)]^2\over[M(t)]^2}.$ (14)

But $M(0)=\left\langle{1}\right\rangle{}=1$, so
$\displaystyle \mu$ $\textstyle =$ $\displaystyle M'(0) = R'(0)$ (15)
$\displaystyle \sigma^2$ $\textstyle =$ $\displaystyle M''(0)-[M'(0)]^2 = R''(0).$ (16)

See also Characteristic Function, Cumulant, Cumulant-Generating Function, Moment


Kenney, J. F. and Keeping, E. S. ``Moment-Generating and Characteristic Functions,'' ``Some Examples of Moment-Generating Functions,'' and ``Uniqueness Theorem for Characteristic Functions.'' §4.6-4.8 in Mathematics of Statistics, Pt. 2, 2nd ed. Princeton, NJ: Van Nostrand, pp. 72-77, 1951.

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© 1996-9 Eric W. Weisstein