Let $Z, X$ be random vectors. If $E(Z\vert X) = 0$ then is it impossible for $Z$ to include a non-zero constant? That is, it is impossible for $Z$ to look like $$ \begin{pmatrix} 1 & z_{12} & z_{13} &... & z_{1t} \end{pmatrix} $$ instead of $$ \begin{pmatrix} z_{11} & z_{12} & z_{13} &... & z_{1t} \end{pmatrix} $$ letting $Z, X$ be $(1 \times t)$, and where $z_it$ are random variables.
I believe the answer is trivially yes, because if $Z$ contains a non-zero constant -- say $1$ -- then that constant must show up in the expectation, hence ruling out the zero vector?
I also feel like I'm overlooking something or that I may be understanding expectations/random vectors improperly..
Thanks.
Just let $\mathbf o=(0,0,...)$ be the constant zero vector.
Then let $\mathbf Z$ contain at least one non-zero constant member, for example: $\mathbf Z=(1,Z_2,...)$. It is then clearly impossible for $\mathsf E(\mathbf Z\mid \mathbf X) = \mathbf o$.