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It is possible to define an algebraic structure to the set of all continuous probability densities under certain operation ?

Example: Let $D = \{f(x_1,...,x_n) \mbox{ | } \int f(x_1,...,x_n)dx_1,...,dx_n = 1 \}$

This set posses any algebraic structure under certain operations such as multiplication, division, composition or any other special operation ?

I'm just curious about this.

Dr Richard Clare
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    convolution – A. Webb Mar 28 '17 at 19:07
  • Convolution is an interesting operation and we can find the density of the sum of R.V's using convolution, and that density is also in the the set D. So under convolution, (D,*) is a group ? what is the identity element ?, for unbounded domain, the uniform does not belongs to D, but for bounded domain we can take the uniform distribution as the identity. – Dr Richard Clare Mar 28 '17 at 19:21
  • In wikipedia I was reading now and found this: The convolution defines a product on the linear space of integrable functions. This product satisfies the following algebraic properties, which formally mean that the space of integrable functions with the product given by convolution is a commutative algebra without identity (Strichartz 1994, §3.3). Other linear spaces of functions, such as the space of continuous functions of compact support, are closed under the convolution, and so also form commutative algebras. – Dr Richard Clare Mar 28 '17 at 19:51

1 Answers1

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One way you can do it is by treating $f_{X}(x)=f(x)$ as the pdf of a random variable. Then as others pointed out in the comments, $Z=X+Y$ is essentially defined by convolution of $f_{X}, f_{Y}$. The role of the identity should be given by the pointed measure $\delta_{0}$. But mind this is not a a continuous pdf anymore.

Similarly you can define the random variable $XY$ and try to compute $f_{XY}$ using so called "Jacobian method". It is easy to see that under this operation you get a commutative ring with is far from an integral domain. You may want to introduce division into the picture by using distributions; but it complicates the theory considerably. In general the calculus of distributions is not easy to compute in practice even without the constraint of being probability distributions.

It should be noted that this ring is useful for a lot of practice purposes in statistics. I am others in the site can provide more interesting examples (like the ones coming from order statistics).

Bombyx mori
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