Expected value and projection of a normal random variable onto a linear span...?

Question

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.

Answer 1

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 $P_{L^2(\sigma(Z_s;s\leq t))}X = E[X|\mathcal{F}_L].$

Answer 2

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 $Z_t^{2}$.