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Why do we use $x - y$ rather than $x + y$ in the definition of the convolution? Is it just convention? (If we are thinking of convolutions as weighted averages, for instance against "good kernels," it should make no difference.)

Why $(f * g) (x) = \int f(y) g(x - y) dy$ rather than $(f * g) (x) = \int f(y) g(x + y) dy$?

Edit: I'm finding it really hard to choose a best answer. There are at least three very good ones here.

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A shift-invariant linear operator $T$ is completely determined by its impulse response $T(\delta) = f$ (where $\delta$ is the Dirac delta function). You can show that for any function $g$, $T(g) = f*g$. This motivates the definition of convolution. –  littleO Mar 16 at 19:19
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The Cross-correlation does use x+y. –  Navin Mar 17 at 2:33

5 Answers 5

Intuitively, and abusing the notation a bit, you can consider the convolution as

$$ (f*g)(x) = \int_{p+q=x} f(p)g(q) $$

This makes it clear that $f*g = g*f$. On the other hand with your alternative definition we would get $$ (f*'g)(x) = \int_{q-p=x} f(p)g(q) $$ and therefore $(f*'g)(x) = (g*'f)(-x)$, which is untidy for no good reason.

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Consider the discrete analogue: Given two functions $a:\>k\mapsto a(k)$ and $b:\>l\mapsto b(l)$ we are collecting (i.e., summing up) for given $r$ all products $a(k)\,b(l)$ where $k+l=r$. This is the right thing to do, e.g., when multiplying two power series $$a(z):=\sum_{k=0}^\infty a_k z^k, \quad b(z):=\sum_{l=0}^\infty b_lz^l\ .$$ Then $c(z):=a(z)b(z)$ can be written as $c(z)=\sum_{r=0}^\infty c_r z^r$ with $$c_r:=\sum\nolimits_{k+l=r} a_k b_l=\sum_{l=0}^r a_{r-l}\, b_l\qquad(r\geq0)\ .$$ This is expressed by saying that the sequence $c:=(c_r)_{r\geq0}$ is the convolution of the two sequences $a:=(a_k)_{k\geq0}$ and $b:=(b_l)_{l\geq0}$, in short: $c=a*b$.

A similar argument can be put forward when dealing with the sum of two independent random variables $X$ and $Y$ having probabilities $p_k$ and $q_l$ of assuming the values $k$ and $l$, respectively.

Translating this into a continuous setting we have $$(f*g)(x)=\int_{-\infty}^\infty f(x-t)\,g(t)\ dt\ ,$$ assuming that the integral on the right hand side makes sense.

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I think this is an intriguing answer. I agree that the algebraic rule for computing the coefficients of the product of two power series and convolution are very similar. Based on your connection, it seems to me that convolution therefore defines a different "natural multiplication" between functions if we consider functions $\mathbb{R} \to \mathbb{R}$ as generalized power series in which the exponents can take on values in $\mathbb{R}$. –  AreaMan Mar 17 at 3:14
    
I am aware that such "series" would never converge (in the traditional sense) unless they were countably supported, but oddly enough this helps me understand the definition for (continuous) convolution. Thank you. (It also makes me wonder if the algebraic theory for such objects, defined for instance for complex functions on groups with some nice measure, is nearly as rich as the theory of polynomial rings...) –  AreaMan Mar 17 at 3:21

For one thing, we want convolution (a linear operator) to be connmutative. That holds for the traditional definition

$$(f * g) (x) = \int f(y) g(x - y) dy =_{(z=x-y)}= \int g(z) f(x-z) dz = (g * f) (x)$$

I doesn't hold with the other

$$(f * g) (x) = \int f(y) g(x + y) dy = \int g(z) f(z-x)dz$$

Other nice property we'd miss is the convolution theorem.

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You could think of simple examples as this:

Impulse response $g(x)$ is zero except for $x=10$, $g(10) = 1$. This could mean "dog is barking 10 seconds after he has seen a cat".

Then the convolution could be explained as: The volume at which the dog is barking at time t is the amount of cats he has seen 10 seconds before time $t$. Which is $t$ minus $10$ seconds.

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In addition to other useful remarks, it might be worthwhile to note that thinking in terms of representations of a topological group $G$ (on topological vector spaces $V$) shows what "convolution" must be, in the following way. For simplicity, suppose $G$ is unimodular, in the sense that left and right Haar measures are the same.

Let $G\times V\to V$ be a continuous group respresentation, so including associativity $g(hv)=(gh)v$ and that the identity of $G$ acts trivially. For a broad class of topological vector spaces $V$ (quasi-complete locally convex, including Hilbert, Banach, Frechet, LF, their weak duals...) compactly-supported continuous functions $f$ on $G$ act by integrating (e.g., Gelfand-Pettis "weak" integrals suffice) $$ f\cdot v \;=\; \int_G f(g)\;gv\;dg $$ If we characterize an operation $f*F$ by requiring $(f*f)v=f(Fv)$, we are led to to an expression (or two) for that convolution: first, $$ f(Fv) \;=\; \int_G f(g)\;g (Fv)\;dg \;=\; \int_G f(g)\,g\Big(\int_G F(h)v\;dh\Big)dg \;=\; \int\int f(g)\,F(h)\,ghv\;dh\,dg $$ There are at least two reasonable choices now: replace $g$ by $gh^{-1}$, or replace $h$ by $g^{-1}h$. In the former, we have $$ f(Fv) \;=\; \int\int f(gh^{-1})\,F(h)\;gv\;dh\,dg \;=\; \int\Big(\int f(gh^{-1})\,F(h)\,dh\Big)\,gv\;dg \;=\; \Big(g\to \int f(gh^{-1})\,F(h)\,dg\Big)\cdot v $$ which shows that $$ (f*F)(g)\;=\; \int_G f(gh^{-1})\,F(h)\;dh $$ For the real numbers, this gives the $x-y$ rather than $x+y$. But, to my mind, a larger point is that we can deduce what convolution is, rather than "guessing" a "definition" and "checking" whether or not it works as we hope.

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