Is it true that if $Y\_1,Y\_2,\cdots,Y\_n$ are jointly Gaussian with $E[Y\_i]=0~\forall i$ then 


$$\mathrm{span}(Y\_1,Y\_2,\cdots,Y\_n) = \mathcal{L}^2(\Omega,\sigma(Y\_1,Y\_2,\cdots,Y\_n),P) ?$$


I wanted to know if this is true based on the following evidence:


If $Y\_1,Y\_2,\cdots,Y\_n$ are jointly Gaussian with $E[Y\_i]=0~\forall i$, then the Conditional 
Expectation defined by


$$E[Y\_1 | Y\_2, Y\_3, \cdots, Y\_n] := E[Y\_1 | \sigma(Y\_2, Y\_3, \cdots, Y\_n)]$$ 


is just the projection of $Y\_1$ on $\mathcal{L}^2(\Omega,\sigma(Y\_2,\cdots,Y\_n),P)$ as 
$Y\_1 \in \mathcal{L}^2(\Omega,\sigma(Y\_1,Y\_2,\cdots,Y\_n),P)$.


On the other hand it can also be shown that the conditional expectation is just the projection of $Y\_1$ on $\mathrm{span}(Y\_2,Y\_3,\cdots,Y\_n)$ (for example, this is used in 
the derivation of the Kalman FIlter) so that the conditional expectation is just a linear
combination of $Y\_i$, i.e.


$$E[Y\_1 | \sigma(Y\_2, Y\_3, \cdots, Y\_n)] = \sum\_{i=2}^n \alpha\_i Y\_i ~a.s.$$


for some $\alpha\_i \in \mathbb{R}$.


I have asked a similar question at stackexchange but did not get much help.


Thanks,
Phanindra



Span of Jointly Gaussian Random Variables
It is certainly not true, there are $L^2$ variables measurable in the $(Y\_i)$ which are not Gaussian (e.g. the sign of $Y\_1$) whereas the span of the $(Y\_i)$ is composed of Gaussians, which is the definition of being jointly Gaussian.


That some vector has the same projection on two sub-spaces is rather weak evidence that those two sub-spaces are equal ...


OTOH, you can look into Skorokhod embedding, which allows you to realize any centered, square-integrable random variable as the value of Brownian motion at a well-chosen stopping time. This is not the same as what you were asking, but it does tell you that $L^2$ is somehow not *that* far from being spanned by Gaussian variables.


