Reconciling two intuitions about convolution 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?
 A: Another "intuitive" thing convolution does is smoothing.  If you convolve a function $f$ with Dirac's delta function $\delta$ (which isn't a function is the sense of a mapping), you get $f$, and if you convolve $f$ with a smooth approximation to $\delta$, you get a smooth approximation to $f$.  The convolution will be at least as smooth as the approximation is.  (More generally, if you convolve $f$ with a well-behaved function, you usually get a function that is at least as well-behaved in the same respects.  In particular, if you convolve a function with a polynomial, you get a polynomial of the same degree.)
The following will be an incomplete answer, giving one relationship between multiplication and convolution.  Remember how you were taught in childhood to multiply numbers:
$$
\begin{array}{cccccccccc}
& & & 9 & 6 & 2 \\
& & \times & 9 & 2 & 4 \\
\hline
& & 3 & 8 & 4 & 8 \\
& 1 & 9 & 2 & 4 \\
8 & 6 & 5 & 8 \\
\hline
8 & 8 & 8 & 8 & 8 & 8
\end{array}
$$
Here you are multiplying by convolving.  You have two sequences: $9,6,2$ and $9,2,4$.  And what you're doing above is convolving: you multiply each term in the sequence by a suitable shift of the other sequence. Then you add the resulting sequences together, each one shifted to the right or left by the same number of places by which the earlier shift was done.  That's convolution.
There is a difference between this and the Fourier transform.  In this case, multiplication of numbers corresponds to convolution of sequences, but multiplication of sequences does not correspond to convolution of numbers (if that latter idea made sense).  With Fourier transforms, it goes both ways: $\widehat{(fg)} = \hat f * \hat g$ and $\widehat{(f*g)} = \hat f\cdot\hat g$.
With Fourier transforms you have a sum or integral of products of some factor with $(e^{ix})^t$.  With numbers above you do the same thing except you have $10$ instead of $e^{ix}$.  One difference between $10$ and $e^{ix}$ is that the sequence of powers of $10$ never returns to its starting point, whereas a high enough power of $e^{ix}$ is the same as $(e^{ix})^0$.
A: 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.
