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