simple conditional expectation question I came across the following in an article that I was reading.  I cannot prove to myself that it is true or not?  Any help would be greatly appreciated.  
Is is true that $E[x|y]=\rho y$, $E[x|z]=0$ imply $E[x|y, z]=ρy $?
Or do these assumptions imply the following:  $E[E[x|y, z]|z]=ρy $, which is maybe what the authors meant to write. 
 A: The claim: $E[x|y]=\rho y$, $E[x|z]=0$ imply $E[x|y, z]=ρy $ is not true. Consider the following example. Let $y$ and $z$ be independent variables such that $E[z] = 1$ and $E[y] = 0$. Define $x= \rho y z$. While  $E[x|y]=\rho y$, $E[x|z]=0$, we get $E[x|y, z]=ρy z$
A: This result does not hold in general. It appears to be an often-seen fallacy, so let's provide a detailed counter-example.  
Assume that the three variables $X,Y,Z$ follow jointly a tri-variate normal distribution. All three have zero means and unitary variances. $X$ is correlated with $Y$, with correlation coefficient $\rho_{xy}$, but $X$ is uncorrelated with $Z$, $\rho_{xz}=0$. But also, $Y$ is correlated with $Z$, with correlation coefficient $r_{yz}$. 
For two variables from a multivariate normal distribution, we have that
$$E(X\mid Y) = \mu_x + \rho_{xy}\frac {\sigma_X}{\sigma_Y}(Y-\mu_y)$$
$$E(X\mid Z) = \mu_x + \rho_{xz}\frac {\sigma_X}{\sigma_Z}(Z-\mu_z)$$
Given our assumptions we obtain
$$E(X\mid Y) = \rho_{xy}Y$$
and also
$$E(X\mid Z) = 0$$
so the premises of the question are satisfied.  
The variance-covariance matrix of this tri-variate normal is 
$$\Sigma = \left[\begin{matrix} 1 & \rho_{xy} & 0\\
\rho_{xy} & 1 & r_{yz}\\
0 & r_{yz} & 1 \\
\end{matrix}\right]$$
The formula for the conditional expected value $E(X \mid \{Y,Z\})$ is 
$$E(X \mid \{Y,Z\}) = [\rho_{xy} \;\;\; 0]\left[\begin{matrix} 1 & r_{yz}\\ r_{yz} & 1 \\ \end{matrix} \right]^{-1}\left[\begin{matrix} Y \\ Z \\ \end{matrix} \right]$$
where we have already taken into account that the means are zero.
Performing the inversion and matrix multiplication  we obtain
$$E(X \mid \{Y,Z\}) = \frac {\rho_{xy}Y -\rho_{xy}r_{yz} Z}{1-r^2_{yz}}$$
Only if $r_{yz} = 0$ will we obtain $E(X \mid \{Y,Z\}) = E(X\mid Y)$.  
Intuitively, even though in our example $Z$ is uncorrelated with $X$ (and more, fully independent in this Normality case), still, the joint realization of $\{Y,Z\}$ provides different information than $Y$ alone as regards $X$. In a sense, the existence of correlation between $Y$ and $Z$ makes the realization of $Z$ to "color" the realization of $Y$ differently, and so it changes the indirect information we receive on $X$ through $Y$.
Since I do not know the article mentioned, the OP should look for additional assumptions that may make the result hold in that particular case, before concluding that the authors have fallen victim to the fallacy.

PS: As for the suggestion by the OP, since the sigma-algebra from $\{Y,Z\}$ is no smaller than the sigma-algebra from $Z$ alone, by the law of itereated expectations we have that
$$E[E[X\mid Y,Z]\mid Z]= E[X\mid Z] = 0$$
