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If I have 2 bivariately normally distributed variables $x,y$, with correlation $\rho$, I find after a long and tedious calculation that I'd rather not copy here, that

$$E(y|x)=\mu_y+\sigma_y \rho \frac{x-\mu_x}{\sigma_x}$$

I cannot find anywhere whether it is correct or not, is it?

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Your expression for the conditional mean is correct.

Suppose $\mathbf{X}\sim N_{p}(\mathbf{\mu},\mathbf{\Sigma})$. Partition $\mathbf{X}$ into two subvcetors $\mathbf{X}^{(1)}$ and $\mathbf{X}^{(2)}$ of dimensions $q\times 1$ and $(p-q)\times 1$, and similarly partition the mean vector and covariance matrix such that

$\mathbf{\mu}=\left(\begin{array}{c} \mu_{1}\\ \mu_{2} \end{array}\right) $ and $\Sigma = \left(\begin{array}{c|c} \Sigma_{11}&\Sigma_{12}\\ \hline \Sigma_{21}&\Sigma_{22} \end{array} \right) $.

Then, it is well known that, the conditional distribution of $$X^{(2)}|X^{(1)}=\mathbf{x}\;\;\sim N_{p-q}\left(\mu_{2}+\Sigma_{21}\Sigma_{11}^{-1}(x-\mu_{1}), \Sigma_{22}-\Sigma_{21}\Sigma_{11}^{-1}\Sigma_{12}\right)$$

In the present problen, the random vector $\left(\begin{array}{c} X\\ Y \end{array}\right)\sim N_{2}(\mu, \Sigma)$, where $\mu = \left(\begin{array}{c} \mu_{x}\\ \mu_{y} \end{array}\right)$ and $\Sigma = \left(\begin{array}{cc} \sigma_{x}^{2}&\rho\sigma_{x}\sigma_{y}\\ \rho\sigma_{y}\sigma_{x}&\sigma_{y}^{2}\\ \end{array}\right)$, with $\Sigma_{xx} = \sigma_{x}^{2}$, $\Sigma_{xy}=\Sigma_{yx}=\rho \sigma_{x}\sigma_{y}$, $\Sigma_{yy} = \sigma_{y}^{2}$. Hope, you could see the conditional mean, you have obtained.

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This is correct. First suppose that $X,Y \sim N(0,1)$ are standard normal distributed with correlation $ \operatorname{Cor}(x,y) = \rho$. Take a look at the densities $ f^{X,Y}(x,y)$ and $f^X(x)$ to compute that $$f^{X,Y}(x,y) = c \exp (- \frac{(x - \rho y) ^2}{2 (1-\rho ^ 2)}) f ^ X(x),$$ where $c$ does not depend on $x$ or $y$. This leads us to $\mathbb P ^{Y | X} = N(\rho X, 1- \rho ^2)$ yielding $\mathbb E (Y | X) = \rho X$.

Now if $X \sim N(\mu _X, \sigma _X)$ and $Y \sim N(\mu_Y, \sigma_Y)$ use the following transformation: \begin{align} \mathbb E (Y | X) &= \mathbb E Y + E( \frac{Y - \mathbb EY}{\sigma _ Y} | X) \\ &= \mathbb E Y + E( \frac{Y - \mathbb EY}{\sigma _ Y} | \frac{X - \mathbb EX}{\sigma _ X}) \\ &= \mu _ Y + \sigma _X \rho \frac{ X - \mathbb E X}{\sigma _ X} \end{align}

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