Span of Jointly Gaussian Random Variables

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

-
$\mathcal{L}^2(\Omega, \sigma(Y_1, \dots, Y_n), P)$ is infinite dimensional. –  Nate Eldredge Mar 29 '12 at 17:00
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.
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.