updated definition of convolution

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Josia Pietsch 2023-07-19 23:54:35 +02:00
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@ -86,17 +86,23 @@ characteristic functions of Fourier transforms.
\subsection{Convolutions${}^\dagger$} \subsection{Convolutions${}^\dagger$}
\begin{definition}+[Convolution] \begin{definition}+[Convolution]
Let $\mu$ and $\nu$ be probability measures on $\R^d$ Let $\mu$ and $\nu$ be probability measures on $\R^d$.
with Lebesgue densities $f_\mu$ and $f_\nu$.
Then the \vocab{convolution} of $\mu$ and $\nu$, Then the \vocab{convolution} of $\mu$ and $\nu$,
$\mu \ast \nu$, $\mu \ast \nu$,
is the probability measure on $\R^d$ is the probability measure on $\R^d$
with Lebesgue density defined by
\[ \[
f_{\mu \ast \nu}(x) \coloneqq (\mu \ast \nu)(A) = \int_{\R^d} \int_{\R^d} \One_A(x + y) \mu(\dif x) \nu(\dif y).
\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
\] \]
\end{definition} \end{definition}
\begin{fact}
If $\mu$ and $\nu$ have Lebesgue densities $f_\mu$ and $f_\nu$,
then the convolution has Lebesgue density
\[
f_{\mu \ast \nu}(x) \coloneqq
\int_{\R^d} f_\mu(x - t) f_\nu(t) \lambda^d(\dif t).
\]
\end{fact}
\begin{fact}+[Exercise 6.1] \begin{fact}+[Exercise 6.1]
If $X_1,X_2,\ldots$ are independent with If $X_1,X_2,\ldots$ are independent with