Convolutions

Math Probability

Discrete Case

Let’s create a function expressing the probability two functions results have a sum of $s$.

$$\sum _{x=-\infty }^{\infty }f(x)g(s-x)$$

Let’s unpack this formula. The inside of the sum finds the probability of a single case where $f$ and $g$ adds to $s$. By using a summation, we can run through every possible case that this happens.

This operation is called a discrete convolution. Convolutions are notated as

$$[f\ast g](s)$$

Continuous Case

Extending the previous equation over to a continuous function, we can attain a definition like this:

$$[f\ast g](s)={\int }_{-\infty }^{\infty }f(x)g(s-x)dx$$

Naturally, we’d expect this to be a probably density function of $f+g$. This is from the same effect as the discrete convolution, except we talk about this for an infinitely small point and probability densities.

Summary

Convolutions return the probability or probability density of adding two functions together (this depends on the type of function you use).

They are defined by:

Discrete:

$$[f\ast g](s)=\sum _{x=-\infty }^{\infty }f(x)g(s-x)$$

Continuous:

$$[f\ast g](s)={\int }_{-\infty }^{\infty }f(x)g(s-x)dx$$