- #1
chingkui
- 181
- 2
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.
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.