let $(X,Y)$ is a bivariate normal random variable with $\mathbb E(X)=\mu_1$,$\text{ Var}(X)=\sigma_1^2$, $\mathbb E(Y)=\mu_2$,$\text{ Var}(Y)=\sigma_2^2$ and $\rho(X,Y)=\rho$($\rho$ is the correlation function ). I want to find Find the conditional expectation $\mathbb E(X\mid aX+bY)$ and know the following calculations is correct or no? The question is: Is the calculation correct?
The importance of this question is all of following question are special case of this so we can answer it easily use the results of this question. $\color{red}{1)}\mathbb E( cX + dY | aX + bY = u), \text{i.i.d case}$here , $\color{red}{2)}\mathbb E(X \, | \, 3X + 4Y],\text{i.i.d case}$here. $\color{red}{3)}\mathbb{E}(X-Y \mid 2X+Y),\text{correlated case}$here $\color{red}{4)}\mathbb{E}(X\mid X+Y),\text{i.i.d case}$here. $\color{red}{5)}\mathbb{E}(3X+Y|X-Y=1),\text{correlated case}$here $\color{red}{6)}\mathbb{E}(X\mid 2X + Y),\text{i.i.d case}$here. $\color{red}{7)}\mathbb E(2X+Y|X-Y=1),\text{i.i.d case}$here.
I want to find $d$ such that $\text{cou}(X-dY, aX+bY)=0$ and by Correlations_and_independence conclude $X-dY$ and $aX+bY$ are independent.
$$0=\text{cou}(X-dY, aX+bY)=a\text{Var}(X)-db \text{Var}(Y)+(b-ad) \text{cou}(X,Y)=a\sigma_1^2-db\sigma_2^2 +(b-ad) \rho \sigma_1 \sigma_2=(a\sigma_1^2 +b \rho \sigma_1 \sigma_2) -d(b\sigma_2^2+a \rho \sigma_1 \sigma_2)$$ so $d=\frac{a\sigma_1^2 +b \rho \sigma_1 \sigma_2}{b\sigma_2^2+a \rho \sigma_1 \sigma_2}$
So I want to use following properties \begin{eqnarray} \left\{ \begin{array}{c} \mathbb E(aX+bY\mid aX+bY) =aX+bY \\ \mathbb E(X-dY \mid aX+bY) = \mathbb E(X-dY)=\mu_1 -d \mu_2 \end{array} \right. \end{eqnarray} and find $E(X|aX+bY)$.
\begin{eqnarray} \left\{ \begin{array}{c} a\mathbb E(X \mid aX+bY)+b\mathbb E(Y\mid aX+bY) =aX+bY \\ \mathbb E(X \mid aX+bY)-d \mathbb E(Y \mid aX+bY)=\mu_1 -d \mu_2 \end{array} \right. \end{eqnarray}
\begin{eqnarray} \left\{ \begin{array}{c} \mathbb E(Y \mid aX+bY) &=&\frac{(aX+bY)-a(\mu_1 -d \mu_2)}{b+ad} \\ \mathbb E(X \mid aX+bY) &=&\frac{(aX+bY)+\frac{b}{d}(\mu_1 -d \mu_2)}{a+\frac{b}{d}} = \frac{d(aX+bY)+b(\mu_1 -d \mu_2)}{b+ad} \end{array} \right. \end{eqnarray}
$$\mathbb E(X\mid aX+bY)=\frac{a\sigma_1^2+b \rho \sigma_1 \sigma_2}{ a^2 \sigma_1^2+b^2 \sigma_2^2 +2ab\sigma_1 \sigma_2}(aX+bY)+\frac{b^2 \sigma_2^2 +2ab\sigma_1 \sigma_2}{ a^2 \sigma_1^2+b^2 \sigma_2^2 +2ab\sigma_1 \sigma_2}(\mu_1 -\frac{a\sigma_1^2 +b \rho \sigma_1 \sigma_2}{b\sigma_2^2+a \rho \sigma_1 \sigma_2} \mu_2)$$
Thanks in advance for any help you are able to provide