Expected value and projection of a normal random variable onto a linear span...? I just wanted to clarify a part of a proof which used the fact a random variable has zero mean.
Suppose $X, Z_{s_1},\dots,Z_{s_n}$ are all jointly normal random variables for all $s_i \leq t$ and $n \geq 1$.
Define $\mathcal{L}(Z,t)$ as set of random variables of the form
$$c_1 Z_{s_1}+ \dots + c_n Z_{s_n}$$
for $c_i \in \mathbb{R}$ and $s_i \leq t$, $n \geq 1$, along with all their limit points (in $L^2(P)$).
Define $\tilde{X} = X - \mathcal{P}_L (X)$ where $\mathcal{P}_L$ is the projection onto the space $\mathcal{L}(Z,t)$.
Then the proof use $\mathbb{E}[\tilde{X}] = 0$.
My understanding of this, is that the projection $\mathcal{P}_L (X)$ coincides with the conditional expectation
$$\mathcal{P}_L (X) = E[X|\mathcal{F}_L]$$
where $\mathcal{F}_L$ is the $\sigma$-field generated by the random variables in $\mathcal{L}(Z,t)$. And hence by law of total expectation we have
$$\mathbb{E}[\tilde{X}] = \mathbb{E}[X] - \mathbb{E}[E[X|\mathcal{F}_L]] = \mathbb{E}[X]-\mathbb{E}[X] =0$$
Is this correct? The proof doesn't really explain, and the relationship between projection and probability spaces is still a bit confusing to me.
 A: Your argument is not valid. Conditional expectation is the projection on the space of all random variables measurable w.r.t. $\sigma \{Z_s:s\leq t\}$, not the projection on to the vector space spanned by $ \{Z_s:s\leq t\}$. The former space contains many nonlinear functions of $ \{Z_s:s\leq t\}$ like $Z_t^{2}$.
A: As @Kavi Rama Murthy pointed out, it holds that
$$
\mathcal{L}(Z,t) \neq L^2(\sigma(Z_s;s\leq t)).
$$ But if we assume additionally that $$(X,Z_s;s\leq t)$$
is jointly normally distributed, it holds that
$$
P_{\mathcal{L}(Z,t) }X = P_{L^2(\sigma(Z_s;s\leq t))}X.
$$ To see this, notice that if $(X,Y_1,\ldots ,Y_n)$ is jointly normal, then
$$
\operatorname{Cov}(X,Y_i) = 0,\,\forall i\leq n \Leftrightarrow X\text{ and }(Y_i)_{i\leq n} \text{ are independent.}
$$ It holds that
$$
\operatorname{Cov}(X-P_{\mathcal{L}(Z,t) }X, Z_s) = 0,\quad\forall s\leq t,
$$ by the definition of the projection. And since $(X,Z_s;s\leq t)$ is jointly normal, so is $(X-P_{\mathcal{L}(Z,t) }X,Z_s;s\leq t).$ This implies that
$$
X-P_{\mathcal{L}(Z,t) }X  \perp \!\!\! \perp   (Z_s)_{s\leq t}.
$$ Therefore, we have
$$
P_{L^2(\sigma(Z_s;s\leq t))}[X-P_{\mathcal{L}(Z,t) }X]=P_{L^2(\sigma(Z_s;s\leq t))}X-P_{\mathcal{L}(Z,t) }X=0,
$$as desired. In your notation, $\mathcal{F}_L=\sigma(Z_s;s\leq t)$ and $P_{L^2(\sigma(Z_s;s\leq t))}X = E[X|\mathcal{F}_L].$
