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If random variable $X$ has a probability distribution of $f(x)$ and random variable $Y$ has a probability distribution $g(x)$ then $(f*g)(x)$, the convolution of $f$ and $g$, is the probability distribution of $X+Y$. This is the only intuition I have for what convolution means.

Are there any other intuitive models for the process of convolution?

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As Liviu Nicolaescu and isomorphismes's answer showed it's related with Fourier Transform. Also Hisenberg Uncertainty Principle, Wave packets, and Feynman's Path Integral.

http://www.med.harvard.edu/jpnm/physics/didactics/improc/intro/fourier3.html

(wikipedia search Wave packet)

http://en.wikipedia.org/wiki/Path_integral_formulation

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Whenever there is a binary operation $A\times B\to C$, it can be extended to a binary operation on measures $M(A)\times M(B)\to M(C)$. In the particular case when $A=B=C$ is a group, and $G\times G\to G$ is the group law, this is the convolution operation. In the same way one defines the convolution of a measure on the group and a measure on the action space in the case of group actions.

One should keep in mind that the natural entities for defining the convolution operation are measures, not functions. Measure and function spaces have different functorial properties (covariant and contravariant, respectively, which means that a map $A\to B$ induces a map in the same direction $M(A)\to M(B)$ between the measure spaces, and a map in the opposite direction $F(B)\to F(A)$ between the function spaces. Whenever one talks about convolutions of functions, these functions in reality are always treated as densities with respect to certain reference measures (most often the countable measure in the discrete case and the Lebesgue or Haar measure in the continuous case).

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The intuition of the convolution integral is not the most easy to grasp. Searching the net is rather unhelpful. The discussion on Wikipedia is extensive, but for many of us too complicated. Gilbert Strang has an illuminating and nice example in his book «Calculus». However, his explanation is not «in depth».

The example is as follows: You put mony in the bank as a stream with constant flow rate $r$ (dollars/year). The yearly interest rate is $p$. The interest rate calculated continously is e^pt. Thus, in a infinite short time interval $dt$ your deposit is $dr$. Now, after $t$ years, how much money do you have in the bank? If a deposit $r$ was made at time $a$, the deposit will, after the time interval $(t-a)$, have increased to $dr\cdot e^{(t-a)}$. This will apply to any depost $dr$ at any time $a$. To obtain the total sum, we just add all deposits and all interest: Thus, we get a convolution integral: $\int (re^{(t-a)} )dt$. Now, here the function $r$ is a constant. However, the reasoning will be the same if $r$ and $p$ varies, i.e. are functions. Thus, the money in the bank after $t$ years is the convolution of the mony that streams in with the interest rate.

In partial differential equations, the integral formulas used when analysing inhomogenous wave and transport equation are convolution integrals. At time t, you just add up all previous effects of the source funcion, just like the interest rate and deposits in the example above, and you use this sum not as a new initial condition at the time t. As these problems are linear, you just add up the convolution at time t with the initial condition at time 0.

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Algebraic meaning of convolution in analysis can be seen from the following formula:

$$\int_{-\infty}^{+\infty} f(x)*g(x)\ dx=\left(\int_{-\infty}^{+\infty} f(x)\right)\left(\int_{-\infty}^{+\infty} g(x)\right)$$

So, in short, convolution is the product of integrals.

This is also true in discrete calculus where discrete convolution is used when multiplying sums, and in particular n-times repeated convolution is used in binomial theorem to express the power of a binomial.

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    $\begingroup$ Ignoring the extraneous first $x$, this formula is only correct if the functions are integrable. Convolution makes sense if both functions are square-integrable but not integrable, in which case this equation is incorrect (since Fubini does not apply). But in any case, what intuition do you glean from knowing, for a fixed $f$, that $T_f(g)=f*g$ has the property you stated? The operator $S_f(g)=g(x)\int_{-\infty}^{\infty}f(t)dt$ has the same property but is just a constant times the identity... $\endgroup$ Jun 16, 2011 at 20:42
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