Why is one graph reversed under convolution?

Question

In this Wikipedia article, it is said that a convolution function "is defined as the integral of the product of the two functions after one is reversed and shifted." The bold is added by me, and shows where my question lies. Why is it reversed?

Answer 1

Consider a similar problem with sequences. If $$ A = \sum_{n=0}^\infty a_n,\qquad B = \sum_{n=0}^\infty b_n, $$ then $$ AB = \sum_{n=0}^\infty c_n $$ where $c_n$ is the convolution $$ c_n= \sum_{k=0}^n a_{k}b_{n-k} $$ where you multiply the sequences, but one is reversed and shifted.

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For convolution of functions:
For example, of $\mathcal F$ is the Fourier transform, then $$ \mathcal F(fg) = \mathcal F{f} * \mathcal F(g) $$ where $*$ is the convolution, defined as in that wikipedia article. Same for Laplace transform.

It is true that the pointwise product $fg$ is interesting and useful. But these uses show that the convolution $f * g$ is also useful.

Answer 2

Another way to think of convolution is that you are integrating $f(s) f(t)$ over all $s,t$ with $s+t = x.$ So, you are looking at all ways of decomposing $x$ as $s+t$.

This is useful eg in probability, where if $S$ and $T$ are independent random variables then $$P(S+T = x) = \sum_{s+t = x} P(S=s) P(T=t),$$ ie a convolution. Similarly, in the continuous case, the probability density function of $S+T$ is $f * g$ (convolution), where $f,g$ are the pdfs of $S,T$.

For example, if you want the chance that you roll two weighted dice $D_1, D_2$ and you want to know the probability they add to $4$, you look at all ways of getting a $4$: roll a $1$ and $3$, $2$ and $2$, or $3$ and $1$. So the chance that you roll a $4$ is $$P(D_1 = 1) \cdot P(D_2 =3) + P(D_1 = 2)\cdot P(D_2 =2) + P(D_1 = 3)\cdot P(D_2 =1).$$