Because I do not want to exaggerate this thread Show that $E(Y\mid X=x)$ is a linear function in $x$ I continue my special problem here.
In order to make the setting clear I'll give some information. $Y$ and $X$ are bivariate normal distributed with expectation vector $\mu=(\mu_Y,\mu_X)^T$ and covariance matrix $\Sigma=\begin{pmatrix}\sigma_Y^2 & p_{XY}\\p_{XY} & \sigma_X^2\end{pmatrix}$ and the task is to show that the conditional expectation $E(Y\mid X=x)$ is a linear function in $x$. My calculation came to the fact, that the conditional density is given by $$ f_{Y\mid X}(y\mid x)=\frac{\sigma_X}{\sqrt{2\pi}\sqrt{\sigma_Y^2\sigma_X^2-p_{XY}^2}}\exp\left(-\frac{1}{2(\sigma_Y^2\sigma_X^2-p_{XY}^2)}\cdot(\sigma_X^2(y-\mu_Y)^2-2p_{XY}(x-\mu_X)(y-\mu_Y)+\sigma_Y^2(x-\mu_X)^2)+\frac{1}{2}\frac{(x-\mu_X)^2}{\sigma_X^2}\right). $$
Update
Now I made a lot of calculations and brought my $f_{Y|X}$ into the form $$ f_{Y|X}(y|x)=\frac{1}{\sqrt{2\pi}\frac{\sqrt{\lvert\Sigma}}{\sigma_X}}\exp\left(-\frac{1}{2\frac{\lvert\Sigma}{\sigma_X^2}}\cdot\left(y-\left(\frac{p_{XY}\mu_X}{\sigma_X^2}-\frac{p_{XY}x}{\sigma_X^2}-\mu_Y\right)\right)^2+r(x)\right), $$ whereat $r$ is a function only depending on $x$.
This is - apart from the summand $r(x)$ - the density of the normal distribution with expectation $\lambda(x):=\frac{p_{XY}\mu_X}{\sigma_X^2}-\frac{p_{XY}x}{\sigma_X^2}-\mu_Y$ and variance $\theta^2:=\frac{\lvert\Sigma\rvert}{\sigma_X^2}$.
Now do I see it right, that it has to be $r(x)=0$, because of the condition
$$ \int f_{Y|X}(y|x)\, dy=1, $$ i.e. $$ \int f_{Y|X}(y|x)\, dy\\ =\exp(r(x))\cdot\overbrace{\frac{1}{\sqrt{2\pi}\frac{\sqrt{\lvert\Sigma}}{\sigma_X}}\int\exp\left(-\frac{1}{2\frac{\lvert\Sigma}{\sigma_X^2}}\cdot\left(y-\left(\frac{p_{XY}\mu_X}{\sigma_X^2}-\frac{p_{XY}x}{\sigma_X^2}-\mu_Y\right)\right)^2\right)\, dy}^{=1}\\ =1\\ \Leftrightarrow \exp(r(x))=1\\ \Leftrightarrow r(x)=0? $$
If yes, it is shown that $Y|X=x\sim\mathcal{N}(\lambda(x),\theta^2)$ and so $$ E(Y|X=x)=\lambda(x) $$ and $\lambda$ is a linear function in $x$.
Can anybody tell me if my proof is correct, please?
With regards
math12
Hint: simplify your problem by assuming $\mu_X = \mu_Y = 0$ and $\sigma_X = \sigma_Y = 1$. It should not be hard to argue that if the original relationship is linear, then this new relationship will also be linear.
EDIT: because $Z = (X - \mu_X)/\sigma_X$ is a linear relation that makes a variable standard Normal.
Once you've done that: isolate terms $A,B$ in $-Ay^2 + By$ in the exponent, then you will get $a= B/(2A)$.
Note that the "free term" $C$ in the expression
$$exp(-Ay^2 + By + C) $$ does not matter. I wouldn't even bother checking for it. Why? Because by definition, this expression is a conditional density. Therefore, it must integrate to 1. Therefore, $(-Ay^2 + By + C) $ must already be a complete square.
EDITED: the new calculation seems ok.
However, I still don't understand why you keep dragging $\mu, \sigma$ around. Mathematics is about simplifying things!