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.

  • $\begingroup$ You should probably add the [self-study] tag & read its wiki for this. $\endgroup$ Oct 20, 2015 at 17:36

2 Answers 2


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 $E[x|y, z]=ρy z$

  • 1
    $\begingroup$ (+1) Great compact counter-example. $\endgroup$ Oct 20, 2015 at 19:03

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

$$E[E[X\mid Y,Z]\mid Z]= E[X\mid Z] = 0$$

  • $\begingroup$ Thanks so much for this wonderful answer. What I thought was a mistake, and what you have confirmed, actually came from a textbook. I am going to revisit the original paper cited for the result. Also, the normality assumption was made in the textbook, which makes your answer all the more wonderful and helpful!!! $\endgroup$ Oct 20, 2015 at 19:50

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