Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$ 
Question: Given that $X$ and $Y$ are two random variables  satisfying $X\sim N(0,a)$ and $Y\sim N(0,b)$ for some $a,b>0$. Assume that $X$ and $Y$ have correlation $\rho.$ 
  Calculate 
  $$\mathbb{E}(X-Y \mid 2X+Y).$$

I tried to use the fact that if $A$ and $B$ are independent, then $\mathbb{E}(A\mid B) = \mathbb{E}(A)$ and uncorrelated implies independence in jointly normal distribution. 
So, I attempted to express $X-Y$ as a linear combination of $2X+Y$ and $Z$ where $\operatorname{Cov}(2X+Y,Z) = 0.$
But I am not able to do so.
Any hint is appreciated.
 A: Choose $A$ such that $(X-Y)-A(2X+Y)$ is independent of $2X+Y$. For this  need $E[((X-Y)-A(2X+Y)) (2X+Y)]=0$  and this is certainly possible. Now $E(X-Y|2X+Y)=E(((X-Y)-A(2X+Y)+A(2X+Y)|2X+Y)=0+A(2X+Y)$.
A: The joint distribution of $(Z_1,Z_2)\equiv(X-Y,2X+Y)$ is $\mathcal{N}(0,\Sigma)$, where
$$
\Sigma=\begin{bmatrix}
a+b-2\rho\sqrt{ab} & 2a-b-\rho\sqrt{ab} \\
2a-b-\rho\sqrt{ab} & 4a+b+4\rho\sqrt{ab}
\end{bmatrix}.
$$
Then the conditional distribution of $Z_1$ given $Z_2$ is 
$$
Z_1\mid Z_2=z\sim \mathcal{N}(\Sigma_{12}\Sigma_{22}^{-1}z,\,\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}).
$$
A: We use two property:
First: $E(2X+Y|2X+Y)=2X+Y$ 
Second:  $(X-dY,2X+Y)$ is bi-variate normal(for $d\neq - \frac{1}{2}$), if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent(by set $\rho=0$ in bivarite distribution of joint $(X-dY,2X+Y)$
Correlations_and_independence). so $E(X-dY|2X+Y)=E(X-dY)=0$.
$$E(2X+Y|2X+Y)=2X+Y$$
so 
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y)  \hspace{1cm} (1)$$
For first step let $\rho=0$
$$cou(X-2\frac{a}{b} Y,2X+Y)=2Var(X)-2\frac{a}{b} Var(Y)=2a-2\frac{a}{b}b=0$$
so since $X-2\frac{a}{b} Y$ and $2X+Y$$ are normal so they are independent.
in hence 
$$E(X-2\frac{a}{b} Y|2X+Y)=E(X-2\frac{a}{b} Y)=0$$
$$E(X|2X+Y)=2\frac{a}{b} E(Y|2X+Y)\hspace{1cm} (2)$$
combine (1) and (2)
$$E(X|2X+Y)=\frac{2\frac{a}{b}}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$ 
$$E(Y|2X+Y)=\frac{1}{1+4\frac{a}{b}}\bigg(2X+Y\bigg)$$ 
so 
$$E(X -Y|2X+Y)=(\frac{2\frac{a}{b}}{1+4\frac{a}{b}}-\frac{1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)=(\frac{2\frac{a}{b}-1}{1+4\frac{a}{b}})\bigg(2X+Y\bigg)$$
**Now for general case ** $\rho \in[-1,1]$
if $$d=\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}} \hspace{1cm} (3)$$
$$cou(X-dY,2X+Y)=2a-db+(1-2d)\rho \sqrt{a} \sqrt{b}$$
$$=2a+\rho \sqrt{a} \sqrt{b}-d(b+2\rho \sqrt{a} \sqrt{b})=0$$
so $$E(X-dY|2X+Y)=E(X-dY)=0$$ and in hence 
$$E(X|2X+Y)=dE(Y|2X+Y) \hspace{1cm} (4)$$
Combine (4) and (1) 
$$E(Y|2X+Y)=2X+Y-2E(X|2X+Y)=2X+Y-2dE(Y|2X+Y)$$ so
$$E(Y|2X+Y)=\frac{1}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (5)$$ and 
$$E(X|2X+Y)=dE(Y|2X+Y)=\frac{d}{1+2d}\bigg(2X+Y\bigg) \hspace{1cm} (6)$$
(5) and (6)
$$E(X-Y|2X+Y)=\frac{d-1}{1+2d}\bigg(2X+Y\bigg)$$
$$=\frac{\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}-1}{1+2\frac{2a+\rho\sqrt{a} \sqrt{b}}{b+2\rho \sqrt{a} \sqrt{b}}}\bigg(2X+Y\bigg)$$
$$=\frac{2a-b-\rho \sqrt{a} \sqrt{b}}{b+4a+4\rho \sqrt{a} \sqrt{b}}\bigg(2X+Y\bigg)$$
detail for  "@Student" 
I explain now  why I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent. 
1)$(X-dY,2X+Y)$ is bi-variate normal for $d\neq \frac{-1}{2}$
I can write
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y 
\end{bmatrix}
=\begin{bmatrix}
1 & -d \\
2 & 1 
\end{bmatrix}
\begin{bmatrix}
X \\
Y 
\end{bmatrix}
\end{eqnarray}
By linear-transformation-of-gaussian-random-variable I think 
\begin{eqnarray}
\begin{bmatrix}
X-dY \\
2X+Y 
\end{bmatrix}
\end{eqnarray}
is bi-variate normal.
2) Now by Correlations_and_independence I think if $Cou(X-dY,2X+Y)=0$ so $X-dY$ and $2X+Y$ are independent. wikipedia: "In general, random variables may be uncorrelated but statistically dependent. But if a random vector has a multivariate normal distribution then any two or more of its components that are uncorrelated are independent".
A: By @Kavi Rama Murthy answer (and me  in other answer)
$$E(X-Y|2X+Y)=A(2X+Y)$$ 
Now By the Projection property ,$E(X-Y|2X+Y)$ minimized 
$$E(X-Y-g(2X+Y))^2$$ conditional-expectation-as-best-predictor
I want to find $A$ by minimizing $E(X-Y-A(2X+Y))^2$
$$E(X-Y-A(2X+Y))^2=E((1-2A)X-(1+A)Y)^2$$
$$=E((1-2A)X)^2+E((1+A)Y)^2
-2E((1-2A)X (1+A)Y)2$$
$$=(1-2A)^2E(X)^2+(1+A)^2E(Y)^2
-2(1-2A)(1+A)E(X Y)$$
$$=(1-2A)^2a+(1+A)^2b
-2(1-2A)(1+A)cou(X Y)$$
$$=(1-2A)^2a+(1+A)^2b
-2(1-2A)(1+A)\rho \sqrt{a}\sqrt{b}$$
by derivation $\frac{d}{dA}$ and equal to $0$
$$\frac{d}{dA} E((1-2A)X-(1+A)Y)^2=0$$
$$\Leftrightarrow$$
$$0=
-4(1-2A)a+2(1+A)b-2(-2)(1+A)\rho \sqrt{a}\sqrt{b}-2(1-2A)\rho \sqrt{a}\sqrt{b}$$
$$\Leftrightarrow$$
$$0=\bigg(
-4a+2b+4\rho \sqrt{a}\sqrt{b}-2\rho\sqrt{a}\sqrt{b}
\bigg)+\bigg(
8a+2b+4\rho \sqrt{a}\sqrt{b}+4\rho \sqrt{a}\sqrt{b}
\bigg)A$$
$$\Leftrightarrow$$
$$0=\bigg(
-4a+2b+2\rho \sqrt{a}\sqrt{b}
\bigg)+\bigg(
8a+2b+8\rho \sqrt{a}\sqrt{b}
\bigg)A$$
$$\Leftrightarrow$$
$$A=\frac{2a-b-\rho \sqrt{a}\sqrt{b}}{4a+b+4\rho \sqrt{a}\sqrt{b}}$$
