Meaning of $ E(u|v) = p\,v$

Question

I am having problem understanding the intuition behind this: if $u$ and $v$ are jointly normal (with zero mean), then $E(u|v) = p\,v$, for some parameter $p$ ?

Answer 1

It means that $u$ and $v$ aren't independent, and they have a covariance of $p$. For instance, if you know from the context that $u = -v$ always holds, then $u$ and $v$ have a covariance of $-1$, and the fact that $E(u \mid v) = -v$ should be quite apparent.

Answer 2

Let $u$ and $v$ are jointly normal each with a zero mean and correlation $\rho$, where $|\rho|<1$, that is, \begin{align*} \rho =\frac{E(uv)}{\sqrt{E(u^2)}\sqrt{E(v^2)}}. \end{align*} Then $v$ and \begin{align*} X := \frac{u}{\sqrt{E(u^2)}}-\rho\frac{v}{\sqrt{E(v^2)}} \end{align*} are independent normal random variables (for two jointly normal random variable, if the covariance is zero, then they are independent). Therefore, \begin{align*} E(u\mid v) &= E\left(\sqrt{E(u^2)}X + \rho\frac{\sqrt{E(u^2)}}{\sqrt{E(v^2)}}v \mid v \right)\\ &= E\left(\sqrt{E(u^2)}X \mid v \right) + \rho\frac{\sqrt{E(u^2)}}{\sqrt{E(v^2)}} v \\ &= \sqrt{E(u^2)}E(X) + \rho\frac{\sqrt{E(u^2)}}{\sqrt{E(v^2)}}v \\ &= pv, \end{align*} where \begin{align*} p = \rho\frac{\sqrt{E(u^2)}}{\sqrt{E(v^2)}}. \end{align*}