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Inquisitive_Mind
#2
Aug7-05, 11:00 AM
P: 11
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.