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Josia Pietsch 2023-07-12 15:25:52 +02:00
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@ -113,9 +113,11 @@ We have
$X: (\Omega, \cF) \to (\R, \cB(\R))$ is a random variable.
Then we can define
\[
\phi_X(t) \coloneqq \bE[e^{\i t x}] = \int e^{\i t X(\omega)} \bP(d \omega) = \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t)
\phi_X(t) \coloneqq \bE[e^{\i t x}]
= \int e^{\i t X(\omega)} \bP(\dif \omega)
= \int_{\R} e^{\i t x} \mu(dx) = \phi_\mu(t),
\]
where $\mu = \bP x^{-1}$.
where $\mu = \bP \circ X^{-1}$.
\end{remark}
\begin{theorem}[Inversion formula] % thm1