Calculate $\mathbb{E}(X-Y\mid 2X+Y).$ if $X\sim N(0,a)$ and $Y\sim N(0,b)$

Question

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.

Answer 1

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 $E(X-Y|2X+Y)=E(((X-Y)-A(2X+Y)+A(2X+Y)|2X+Y)=0+A(2X+Y)$.

Answer 2

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".

Answer 3

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 $0$

$$\frac{d}{dA} E((1-2A)X-(1+A)Y)^2=0$$ $$\Leftrightarrow$$

$$0= -4(1-2A)a+2(1+A)b-2(-2)(1+A)\rho \sqrt{a}\sqrt{b}-2(1-2A)\rho \sqrt{a}\sqrt{b}$$

$$\Leftrightarrow$$

$$0=\bigg( -4a+2b+4\rho \sqrt{a}\sqrt{b}-2\rho\sqrt{a}\sqrt{b} \bigg)+\bigg( 8a+2b+4\rho \sqrt{a}\sqrt{b}+4\rho \sqrt{a}\sqrt{b} \bigg)A$$

$$\Leftrightarrow$$

$$0=\bigg( -4a+2b+2\rho \sqrt{a}\sqrt{b} \bigg)+\bigg( 8a+2b+8\rho \sqrt{a}\sqrt{b} \bigg)A$$ $$\Leftrightarrow$$

$$A=\frac{2a-b-\rho \sqrt{a}\sqrt{b}}{4a+b+4\rho \sqrt{a}\sqrt{b}}$$

Answer 4

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 $$ Z_1\mid Z_2=z\sim \mathcal{N}(\Sigma_{12}\Sigma_{22}^{-1}z,\,\Sigma_{11}-\Sigma_{12}\Sigma_{22}^{-1}\Sigma_{21}). $$