How to prove the output of Linear Filtering a Gaussian Process is still Gaussian?

  • Thread starter chingkui
  • Start date
193
1

Main Question or Discussion Point

How to prove the output of Linear Filtering a Gaussian Process is still Gaussian? It has been stated in many books I read, but none of them actually prove it. One even stated that "The technical mechinery to prove this property is beyond the scope of this book..."
By definition, a Gaussian process is a function x such that for any finite integer k, and for any arbitary time t1, t2, ..., tk, that x(t1), x(t2), ...., x(tk) are jointly Gaussian RV.
To linear filter the process x(t) means just to convolute it with a function h(t), i.e., the output y(t)=h(t)*x(t)=integrate(h(t-s)x(s)ds)
To prove the statement is to prove that for any m>0, and any time t1,..., tm, that y(t1),...,y(tm) are jointly Gaussian.
Is it that difficult to prove? What does one need to prove it?
It is quite obvious that a Gaussian RV remains Gaussian after linear filtering it, but for a Gaussian process, I am not sure what to use to prove that. Does anyone know how? Thanks.
 

Answers and Replies

I suspect that the a formal proof will be quite technical. However, to understand it intuitively, one may try to discretise the integral in the definition of y(t). The decretised integral is a linear combination of jointly Gaussian variables, and is therefore Gaussian. Each of the y(ti) can be discretised this way, and each is a linear combination of a set of jointly Gaussian variables, and {y(ti)} is therefore jointly Gaussian.

Of course this is far from being a proof. But I think this theorem belongs to the category where it can be easily understood but not easily proved.
 

Related Threads for: How to prove the output of Linear Filtering a Gaussian Process is still Gaussian?

Replies
13
Views
2K
Replies
12
Views
865
  • Last Post
Replies
16
Views
6K
  • Last Post
Replies
6
Views
653
  • Last Post
Replies
2
Views
1K
Replies
4
Views
1K
  • Last Post
Replies
2
Views
2K
Top