TITLE: Reconciling two intuitions about convolution
QUESTION [12 upvotes]: There are two intuitive things convolution does. In the time domain, it represents the distribution of the sum of two independent random variables. In the frequency domain, it's just multiplication.
These seem sort of related but I can't make the connection. Is there a way to see one of these properties from the other?

REPLY [2 votes]: It's a way of gathering things that sum to a constant. For example,
\begin{align}
   \sum_{n=0}^{\infty}a_{n}z^{n}\sum_{n=0}^{\infty}b_{n}z^{n}
     & = \sum_{n=0}^{\infty}\left(\sum_{j+k=n}a_{j}b_{k}\right) z^{n} \\
     & = \sum_{n=0}^{\infty}\left(\sum_{j=0}^{n}a_{j}b_{n-j}\right)z^{n}.
\end{align}
A Fourier transform can be viewed as an integral "sum" of exponentials. So, when you multiply two of them together, you also collect like terms
\begin{align}
        &\int_{-\infty}^{\infty}e^{-ist}a(t)dt\int_{-\infty}^{\infty}e^{-ist}b(t)dt \\
  & = \int_{-\infty}^{\infty}\left(\int_{u+v=s}a(u)b(v)\right)e^{-ist}ds \\
  & = \int_{-\infty}^{\infty}\left(\int_{-\infty}^{\infty}a(u)b(s-u)du\right)e^{-ist}ds .
\end{align}
And, when you look at the sum of random variables, you also look at where the sum is a constant.